Question: Given the parabola $y=ax^2+3ax+c$ ($a>0$) intersects the y-axis at point C and the x-axis at points A and B, with point A being to the left of point B. The coordinates of point B are $(1, 0)$, and $OC=3OB$. If point E lies on the x-axis and point P lies on the parabola, is there a parallelogram with vertices A, C, E, and P that has AC as one of its sides? If such a parallelogram exists, provide the coordinates of point P; if it does not exist, explain why.

Solution: \textbf{Solution:} There exist $\boxed{P_1(-3, -3)}$, $\boxed{P_2\left(\frac{-3-\sqrt{41}}{2}, 3\right)}$, $\boxed{P_3\left(\frac{-3+\sqrt{41}}{2}, 3\right)}$.


Question: As shown in Figure 1, the parabola \(y = -x^2 + bx + c\) intersects the x-axis at points A and B, and intersects the y-axis at point C \((0, 3)\). The axis of symmetry is the line \(x=1\), intersecting the x-axis at point D, and the vertex of the parabola is point E. What is the analytical expression of this parabola?

Solution:
\textbf{Solution:} Since the parabola $y=-x^{2}+bx+c$ intersects the y-axis at point C(0, 3),\\
let $x=0$, then we have $c=3$,\\
Since the axis of symmetry is the line $x=1$,\\
it follows that $-\frac{b}{2\times (-1)}=1$,\\
thus, $b=2$,\\
therefore, the equation of the parabola is $y=-x^{2}+2x+3=\boxed{-x^{2}+2x+3}$.


Question: As shown in Figure 1, the parabola $y=-x^2+bx+c$ intersects the x-axis at points A and B, and intersects the y-axis at point C $(0,3)$. The axis of symmetry is the line $x=1$, which intersects the x-axis at point D. The vertex is point E. Connect AC, CE, AE. What is the area of $\triangle ACE$?

Solution:
\textbf{Solution:} As illustrated in figure 1, let the point of intersection between line AE and the y-axis be denoted as H,\\
given the equation of the parabola as $y=-x^2+2x+3$,\\
setting $y=0$, yields $-x^2+2x+3=0$,\\
$\therefore x=-1$ or $x=3$,\\
$\therefore A(-1,0)$,\\
when $x=1$, $y=-1+2+3=4$,\\
$\therefore E(1,4)$,\\
$\therefore$ the equation of line AE is $y=2x+2$,\\
$\therefore H(0,2)$,\\
$\therefore CH=3-2=1$,\\
$\therefore S_{\triangle ACE}=\frac{1}{2}CH\cdot|x_E-x_A|=\frac{1}{2}\times 1\times 2=\boxed{1};$


Question: As shown in Figure 1, the parabola $y=-x^2+bx+c$ intersects the x-axis at points A and B, and intersects the y-axis at point C (0, 3). The axis of symmetry is the line $x=1$, which intersects the x-axis at point D, and the vertex is point E. As shown in Figure 2, point F is on the y-axis, and $OF=\sqrt{2}$, point N is a moving point on the parabola within the first quadrant, and is to the right of the parabola's axis of symmetry. The line ON intersects the axis of symmetry at point G, and line GF is drawn. If GF bisects $\angle OGE$, find the coordinates of point N.

Solution:
\textbf{Solution:} Let $FP \perp DE$ at point P through point F, thus $FP=1$. Draw $FQ \perp ON$ at Q through point F, \\
since $GF$ bisects $\angle OGE$, \\
therefore $FQ=FP=1$, \\
in $\triangle FQO$, $OF= \sqrt{2}$, \\
according to the Pythagorean theorem, we get $OQ=\sqrt{OF^{2}-FQ^{2}}=1$, \\
thus $OQ=FQ$, \\
thus $\angle FOQ=45^\circ$, \\
thus $\angle BON=90^\circ-45^\circ=45^\circ$, \\
draw $QM \perp OB$ at M through point Q, $OM=QM$ \\
therefore, the equation of $ON$ is $y=x$, \\
since point N lies on the parabola $y=-x^{2}+2x+3$, \\
by solving the system $\begin{cases}y=x\\ y=-x^{2}+2x+3\end{cases}$, \\
we obtain $\left\{\begin{array}{l}x=\frac{1+\sqrt{13}}{2}\\ y=\frac{1+\sqrt{13}}{2}\end{array}\right\}$ or $\left\{\begin{array}{l}x=\frac{1-\sqrt{13}}{2}\\ y=\frac{1-\sqrt{13}}{2}\end{array}\right\}$ (since point N is to the right of the axis of symmetry $x=1$, the latter is discarded), \\
thus, the coordinates of point N are: $\left(\boxed{\frac{1+\sqrt{13}}{2}}, \boxed{\frac{1+\sqrt{13}}{2}}\right)$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $l\colon y=kx-1$ ($k\neq 0$) intersects with the graph of the function $y=\frac{m}{x}$ ($x>0$) at point $A(3,2)$. What are the values of $k$ and $m$?

Solution:
\textbf{Solution:} Since the line $l\colon y=kx-1$ ($k\neq 0$) intersects with the graph of the function $y=\frac{m}{x}$ ($x>0$) at point $A(3,2)$,\\
we have $2=3k-1$ and $2=\frac{m}{3}$,\\
solving these equations, we find $k=\boxed{1}, m=\boxed{6}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $l\colon y=kx-1$ ($k\neq 0$) intersects the graph of the function $y=\frac{m}{x}$ ($x>0$) at point $A(3,2)$. After the line $l$ is translated upwards along the y-axis by $t$ ($t>0$) units, the resulting line intersects the $x$ axis and the $y$ axis at points $P$ and $Q$, respectively, and intersects the graph of the function $y=\frac{m}{x}$ ($x>0$) at point $C$.

Solution:
\textbf{Solution:}\\
(1) Given the equation of line $l$ as $y=x-1$ and the equation of the hyperbola as $y=\frac{6}{x}$,\\
when $t=2$, the shifted equation of line $l$ is $y=x+1$,\\
when $x=0$, $y=1$, and when $y=0$, $x=-1$,\\
$\therefore$ points $P(-1, 0), Q(0, 1)$,\\
combining $y=x+1$ and $y=\frac{6}{x}$ yields:\\
$\begin{cases}
y=x+1\\
y=\frac{6}{x}
\end{cases}$, solving gives:\\
$\begin{cases}
x_1=2\\
y_1=3
\end{cases}, \begin{cases}
x_2=-3\\
y_2=-2
\end{cases}$ (discarded),\\
$\therefore$ point $C(2, 3)$,\\
$\therefore QC=\sqrt{(2-0)^2+(3-1)^2}=\boxed{2\sqrt{2}}$;\\
(2) $1+\frac{\sqrt{2}}{2}<t<2$


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=\frac{1}{2}x^{2}+bx+c$ (where $b$ and $c$ are constants) has its vertex at the coordinates $\left(\frac{3}{2}, -\frac{25}{8}\right)$, intersects the $x$-axis at points $A$ and $B$ (with point $A$ to the left of point $B$), and intersects the $y$-axis at point $C$. Point $C$ and point $D$ are symmetric about the $x$-axis. Connect $AD$ and draw line $BD$. What are the values of $b$ and $c$?

Solution:
\textbf{Solution:} Given the parabola $y=\frac{1}{2}x^{2}+bx+c$ has its vertex at $\left(\frac{3}{2}, -\frac{25}{8}\right)$,\\
thus, the equation of the parabola is $y=\frac{1}{2}\left(x-\frac{3}{2}\right)^{2}-\frac{25}{8}$\\
Simplifying, we obtain $y=\frac{1}{2}x^{2}-\frac{3}{2}x-2$,\\
thus, $b=\boxed{-\frac{3}{2}}$ and $c=\boxed{-2}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the parabola $y=\frac{1}{2}x^{2}+bx+c$ (where $b$ and $c$ are constants) has its vertex at $\left(\frac{3}{2}, -\frac{25}{8}\right)$, intersects the x-axis at points $A$ and $B$ (with point $A$ to the left of point $B$), and intersects the y-axis at point $C$. Point $C$ and point $D$ are symmetric about the x-axis. Line $AD$ is drawn, and line $BD$ is constructed. What are the coordinates of points $A$ and $B$?

Solution:
\textbf{Solution:} Let $\frac{1}{2}x^{2}-\frac{3}{2}x-2=0$,\\
we rearrange it to get: $x^{2}-3x-4=0$,\\
from which we find $x_{1}=-1$ and $x_{2}=4$,\\
$\therefore \boxed{A\left(-1, 0\right)}$ and $\boxed{B\left(4, 0\right)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the parabola $y=\frac{1}{2}x^{2}+bx+c$ (where $b$ and $c$ are constants) has its vertex at $\left(\frac{3}{2}, -\frac{25}{8}\right)$ and intersects the $x$-axis at points $A$ and $B$ (with point $A$ to the left of point $B$) and intersects the $y$-axis at point $C$. Point $C$ and point $D$ are symmetric about the $x$-axis. Connect points $A$ and $D$, and draw line $BD$. What is the equation of line $BD$?

Solution:
\textbf{Solution:} Let $x=0$, then $y=-2$, hence $C(0, -2)$,\\
since point C and point D are symmetric with respect to the x-axis,\\
therefore $D(0, 2)$,\\
let the equation of the line $BD$ be $y=mx+n$, by the problem statement, we have\\
\[
\begin{cases}
n=2\\
4m+n=0
\end{cases}
\], solving this we get \[
\begin{cases}
m=-\frac{1}{2}\\
n=2
\end{cases}
\],\\
thus, the equation of the line $BD$ is $\boxed{y=-\frac{1}{2}x+2}$.


Question: In the Cartesian coordinate system, let us consider a parabola $y=\frac{1}{2}x^{2}+bx+c$ (where $b$ and $c$ are constants). The vertex of the parabola is given by the coordinates $\left(\frac{3}{2}, -\frac{25}{8}\right)$. This parabola intersects the $x$-axis at points $A$ and $B$ (with $A$ located to the left of $B$) and intersects the $y$-axis at point $C$. Point $C$ and point $D$ are symmetric with respect to the $x$-axis. The line segment $AD$ is drawn, and so is the straight line $BD$. When point $P$ is on the parabola and point $Q$ is on the line $BD$, and the quadrilateral formed by points $C$, $D$, $P$, and $Q$ is a parallelogram, write down the coordinates of point $Q$ directly.

Solution: \textbf{Solution:} The coordinates of point $Q$ are $\boxed{\left(2, 1\right)}$, $\left(1-\sqrt{17}, \frac{3+\sqrt{17}}{2}\right)$, $\left(1+\sqrt{17}, \frac{3-\sqrt{17}}{2}\right)$, $\left(-2, 3\right)$.


Question: As shown in the figure, the graph of the linear function \(y=-2x+8\) intersects the graph of the function \(y=\frac{k}{x}\) (\(x>0\)) at points \(A(m, 6)\) and \(B(n, 2)\). The line \(AC\) is perpendicular to the \(y\)-axis at point \(C\), and the line \(BD\) is perpendicular to the \(x\)-axis at point \(D\). What is the value of \(k\)?

Solution:
\textbf{Solution:} Since the graph of the linear function $y=-2x+8$ passes through the points $A(m,6)$ and $B(n,2)$,\\
it follows that $-2m+8=6$ and $-2n+8=2$,\\
from which we find $m=1$ and $n=3$.\\
Since the graph of the function $y=\frac{k}{x}$ (for $x>0$) passes through the points $A(1,6)$ and $B(3,2)$,\\
it follows that $k=\boxed{6}$.


Question: As shown in the figure, the graph of the linear function $y=-2x+8$ intersects with the graph of the function $y=\frac{k}{x}$ (where $x>0$) at points $A(m, 6)$ and $B(n, 2)$. Line segment $AC$ is perpendicular to the $y$-axis at point $C$, and line segment $BD$ is perpendicular to the $x$-axis at point $D$. Based on the graph, directly write down the range of values of $x$ for which $-2x+8-\frac{k}{x}<0$.

Solution: \textbf{Solution:} The range of values for $x$ is \boxed{0<x<1} or \boxed{x>3}.


Question: As illustrated in the figure, the graph of the linear function $y=-2x+8$ intersects with the graph of the function $y=\frac{k}{x}$ ($x>0$) at points $A(m,6)$ and $B(n,2)$. $AC$ is perpendicular to the $y$-axis at point $C$, and $BD$ is perpendicular to the $x$-axis at point $D$. Let $P$ be a point on the line segment $AB$. Lines $PC$ and $PD$ are drawn. If the areas of triangles $\triangle PCA$ and $\triangle PDB$ are equal, find the coordinates of point $P$.

Solution:
\textbf{Solution:} Let the coordinates of point $P$ on the line $y=-2x+8$ be $(x, -2x+8)$, \\
by the equality of areas of $\triangle PCA$ and $\triangle PDB$, we get: $\frac{1}{2}AC\times |y_A-y_P|=\frac{1}{2}BD\times |y_B-y_P|$, \\
$\therefore \frac{1}{2}\times 1\times[6-(-2x+8)]=\frac{1}{2}\times 2\times(3-x)$, \\
Solving this yields: $x=2$, \\
thus $y=-2x+8=4$, \\
$\therefore$ the coordinates of point $P$ are $\boxed{(2,4)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the vertices $B$, $C$ of rectangle $ABCD$ are on the positive x-axis, with $AB=8$ and $BC=6$. The diagonals $AC$ and $BD$ intersect at point $E$. The graph of the inverse proportion function $y=\frac{k}{x}(x>0)$ passes through point $E$ and intersects $AB$ and $CD$ at points $F$ and $G$, respectively. If $OC=10$, what is the value of $k$?

Solution:
\[
\textbf{Solution:} \quad \text{Given that in rectangle ABCD, AB = 8, BC = 6, and OC = 10,} \\
\therefore OB = OC - BC = 4, \\
\therefore B(4,0), A(4,8), C(10,0), \\
\text{since diagonals AC and BD intersect at point E,} \\
\therefore \text{point E is the midpoint of AC,} \\
\therefore E(7,4), \\
\text{substituting } E(7,4) \text{ into } y = \frac{k}{x} \text{ gives } k = 7 \times 4 = \boxed{28}.
\]


Question: As shown in the figure, in the Cartesian coordinate system, the vertices $B$ and $C$ of rectangle $ABCD$ are located on the positive half of the $x$-axis, with $AB=8$ and $BC=6$. The diagonals $AC$ and $BD$ intersect at point $E$. The graph of the inverse proportion function $y=\frac{k}{x}$ (for $x>0$) passes through point $E$ and intersects $AB$ and $CD$ at points $F$ and $G$, respectively. Connect $EG$. If $BF+BE=11$, what is the area of $\triangle CEG$?

Solution:
\textbf{Solution:}
Given that $AC=\sqrt{AB^2+BC^2}=\sqrt{6^2+8^2}=10$,\\
hence $BE=EC=5$,\\
Given that $BF+BE=11$,\\
hence $BF=6$,\\
Let $OB=t$, then $F(t,6)$, and $E(t+3,4)$,\\
Since the graph of the inverse proportion function $y=\frac{36}{x}$ passes through points E and F,\\
thus $6t=4(t+3)$,\\
Solving this, we find $t=6$,\\
thus $k=6t=36$,\\
Hence, the equation of the inverse proportion function is $y=\frac{36}{x}$,\\
when $x=12$, $y=\frac{36}{12}=3$,\\
thus $G(12,3)$,\\
therefore, the area of $\triangle CEG$ equals $\frac{1}{2}\times 3\times 3=\boxed{\frac{9}{2}}$.


Question: As shown in the figure, it's known that the point $A(-1,-3)$ lies on the line $l: y=kx-2$, and point $M(m,y_{1})$ is a moving point on the parabola $y=ax^{2}-4ax+2\ (a\neq 0)$. If the parabola intersects the line $l$ at point $A$.

Solution:
\textbf{Solution:}

(1) Substituting the point $A(-1,-3)$ into the line $l: y=kx-2$ yields\\
$-k-2=-3$, solving this equation gives $k=1$,\\
substituting the point $A(-1,-3)$ into the parabola $y=ax^2-4ax+2$ yields\\
$a+4a+2=-3$, solving this equation gives $a=-1$;\\
(2) Since $a=-1$ and $k=1$,\\
the parabola is $y=-x^2+4x+2$ and the line $l: y=x-2$,\\
consider $M(m,y_1)$ as a moving point on the parabola $y=-x^2+4x+2$,\\
hence $y_1=-m^2+4m+2$,\\
since line segment $MN$ is parallel to the y-axis and intersects line $l$ at point $N$,\\
we have $N(m,m-2)$,\\
$MN=-m^2+3m+4=-(m-\frac{3}{2})^2+\frac{25}{4}$,\\
when $m=\frac{3}{2}$, the maximal length of the line segment $MN$ is $\boxed{\frac{25}{4}}$, at this point the coordinates of $M$ are $\left(\frac{3}{2},\frac{23}{4}\right)$.


Question: Given that point $A(-1, -3)$ lies on the line $l: y=kx-2$, and point $M(m, y_1)$ is a moving point on the parabola $y=ax^2-4ax+2\ (a\neq 0)$. Point $B(x_2, y_2)$ is the intersection of the parabola and line $l$ in the first quadrant. If $y_1 \leq y_2$, directly write out the range of values for $m$.

Solution: \textbf{Solution}: We have $\boxed{m\le 0}$ or $m\ge 4}$.


Question: In the Cartesian coordinate system, the graph of the quadratic function \(y=ax^2+bx+c (a \ne 0)\) intersects the x-axis at points \(A(-3,0)\) and \(B(1,0)\), and intersects the y-axis at point \(C(0, -3)\). The vertex is \(D\), and its axis of symmetry intersects the x-axis at point \(E\). What is the analytic expression of the quadratic function?

Solution:
\textbf{Solution:} Let $y=ax^{2}+bx+c\ (a\ne 0)$, \\
since $A(-3,0)$, $B(1,0)$, $C(0, -3)$, \\
therefore $\begin{cases}
(-3)^{2}a+(-3)b+c=0\\
a+b+c=0\\
c=-3
\end{cases}$, \\
hence $\begin{cases}
a=1\\
b=2\\
c=-3
\end{cases}$, \\
therefore the quadratic function is: \boxed{y=x^{2}+2x-3}.


Question: In the Cartesian coordinate system, the graph of the quadratic function $y=ax^2+bx+c\ (a\ne 0)$ intersects the x-axis at points $A(-3,0)$ and $B(1,0)$, and intersects the y-axis at point $C(0, -3)$. The vertex is denoted as $D$, and its axis of symmetry intersects the x-axis at point $E$. Let point $P$ be a point on the parabola in the third quadrant. The area of $\triangle APC$ is denoted as $S$. Find the maximum value of $S$ and the coordinates of point $P$ at this time?

Solution:
\textbf{Solution:} Suppose the equation of line $AC$ is: $y=kx+m$,\\
since $A(-3,0)$, $B(0, -3)$,\\
then $\left\{\begin{array}{l}
-3k+m=0\\
m=-3
\end{array}\right.$,\\
thus $k=-1$, $m=-3$,\\
therefore $y=-x-3$.\\
Draw a perpendicular line from point $P$ to the x-axis intersecting $AC$ at $G$, let $P\left(x, -x-3\right)$,\\
then $G\left(x, -x-3\right)$.\\
Since point $P$ is in the third quadrant,\\
therefore $PG=-x-3-\left(-x^2+2x-3\right)$\\
$=-x-3-x^2-2x+3$\\
$=-x^2-3x$,\\
thus the area of $\triangle APC=\frac{1}{2}PG\cdot OA$\\
$=\frac{1}{2}\left(-x^2-3x\right)\times 3$\\
$=-\frac{3}{2}x^2-\frac{9}{2}x$\\
$=-\frac{3}{2}\left(x+\frac{3}{2}\right)^2+\frac{27}{8}$.\\
Therefore, when $x=-\frac{3}{2}$, the maximum area $S_{\text{max}}=\boxed{\frac{27}{8}}$, at this point $P\left(-\frac{3}{2}, -\frac{15}{4}\right)$.


Question: In the Cartesian coordinate system, the graph of the quadratic function $y=ax^2+bx+c\ (a \neq 0)$ intersects the x-axis at points $A(-3,0)$ and $B(1,0)$, and intersects the y-axis at point $C(0, -3)$. The vertex of the parabola is $D$, and its axis of symmetry intersects the x-axis at point $E$. On the segment $AC$, is there a point $F$ such that triangle $\triangle AEF$ is an isosceles triangle? If such a point exists, provide the coordinates of point $F$ directly; if it does not exist, please explain the reason.

Solution: \textbf{Solution}: Let $F_1(-1, -2)$, $F_2(\sqrt{2}-3, -\sqrt{2})$, and $F_3(-2, -1)$. $\boxed{F_1(-1, -2)}$, $\boxed{F_2(\sqrt{2}-3, -\sqrt{2})}$, and $\boxed{F_3(-2, -1)}$.


Question: As shown in the figure, the coordinates of points A and B are respectively $(-2,0)$ and $(0,3)$. Rotating line segment AB $90^\circ$ counterclockwise around point B gives line segment BC. Draw line CD $\perp$ OB with the foot of the perpendicular being D. The graph of the inverse proportion function $y=\frac{k}{x}$ passes through point C. Write down the coordinates of point C directly, and find the equation of the inverse proportion function.

Solution:
\textbf{Solution:} We have $C(3,1)$; $k=3\times1=3$,\\
$\therefore$ the equation of the inverse proportion function is $\boxed{y=\frac{3}{x}}$.


Question: As shown in the figure, the coordinates of points A and B are $(-2,0)$ and $(0,3)$, respectively. Segment AB is rotated counterclockwise by $90^\circ$ about point B to obtain segment BC. Through point C, line CD is drawn perpendicular to OB, with D being the foot of the perpendicular. The graph of the inverse proportion function $y=\frac{k}{x}$ passes through point C. Point P is on the graph of the inverse proportion function $y=\frac{k}{x}$. When the area of $\triangle PCD$ is 3, find the coordinates of point P.

Solution:
\textbf{Solution:} Let $P\left( \frac{3}{m}, m \right)$,\\
since $CD \perp y$-axis, and $CD = 3$,\\
from the area of $\triangle PCD$ being 3, we have: $\frac{1}{2}CD \cdot |m-1|=3$,\\
thus, $\frac{1}{2} \times 3 |m-1| = 3$,\\
thus, $m-1 = \pm 2$,\\
thus, $m=3$ or $m=-1$,\\
when $m=3$, $\frac{3}{m}=1$; when $m=-1$, $\frac{3}{m}=-3$,\\
thus, the coordinates of point P are $\boxed{(1, 3)}$ or $\boxed{(-3, -1)}$.


Question: In the Cartesian coordinate system, the parabola $y = ax^2 - 6ax - 16a \, (a \neq 0)$ intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C$. The line $BC$ is drawn, and it is known that point C is at $(0,4)$. Find the coordinates of points $A$ and $B$, and the equation of the parabola?

Solution:
\textbf{Solution:}

Given $y=-\frac{1}{4}x^2+\frac{3}{2}x+4$ passes through point $C(0, 4)$,\\
$\therefore -16a=4$,\\
solving for $a$ gives $a=-\frac{1}{4}$,\\
$\therefore y=-\frac{1}{4}x^2+\frac{3}{2}x+4$;\\
Let $y=0$, i.e., $-\frac{1}{4}x^2+\frac{3}{2}x+4=0$,\\
solving yields: $x_1=-2, x_2=8$\\
$\therefore \boxed{A(-2, 0)}, \boxed{B(8, 0)}$


Question: As shown in the figure, the graph of the linear function $y=kx+2$ intersects the x-axis and the y-axis at points $A$ and $B$, respectively, and intersects the graph of the inverse proportion function $y=\frac{n}{x}$ $(n \neq 0)$ at point $C$ in the first quadrant. If $OA=OB$, and $B$ is the midpoint of segment $AC$, find the analytical expressions of the linear function and the inverse proportion function.

Solution:
\textbf{Solution:} Construct $CD\perp x$-axis at $D$,\\
since the graph of the linear function $y=kx+2$ intersects the $y$-axis at point $B$,\\
thus $B(0, 2)$,\\
therefore $OB=2$,\\
therefore $OA=OB=2$,\\
thus $A\left(-2, 0\right)$,\\
substituting $A\left(-2, 0\right)$ into $y=kx+2$ gives $k=1$,\\
therefore, the equation of the linear function is $y=x+2$;\\
Furthermore, since $B$ is the midpoint of the segment $AC$, and $OB\parallel CD$,\\
thus $OB$ is the median of $\triangle ACD$,\\
thus $OA=OD=2$, $CD=2OB=4$,\\
thus $C(2, 4)$,\\
since the graph of the inverse proportionality function $y=\frac{n}{x}(n\ne 0)$ intersects in the first quadrant at point $C$,\\
thus $n=2\times 4=8$,\\
thus, the equation of the inverse proportionality function is $y=\frac{8}{x}$\\
Therefore, the equations of the linear function and the inverse proportionality function are $\boxed{y=x+2}$ and $\boxed{y=\frac{8}{x}}$, respectively;


Question: As shown in the figure, the graph of the linear function $y=kx+2$ intersects the x-axis and the y-axis at points $A$ and $B$, respectively. It also intersects the graph of the inverse proportion function $y=\frac{n}{x} (n\ne 0)$ at point $C$ in the first quadrant. If $OA=OB$ and $B$ is the midpoint of the line segment $AC$. Directly write down the solution set of the inequality $kx+2<\frac{n}{x}$.

Solution: \textbf{Solution:} The solution set of $x<-4$ or $0<x<2$ is $x\in(-\infty, -4)\cup(0, 2)$. Therefore, the solution set for the inequality $kx+2<\frac{n}{x}$ is $x\in\boxed{(-\infty, -4)\cup(0, 2)}$.


Question: As shown in the figure, the graph of the quadratic function $y=-x^2+bx+c$ intersects the $x$ axis at points $A$ and $B$, and intersects the $y$ axis at point $C(0, 6)$, and also passes through the point $(1, 10)$. Determine: What is the expression of this parabola?

Solution:
Solution: Since the graph of the quadratic function $y=-x^2+bx+c$ passes through the points $C(0, 6)$ and $(1, 10)$,
\[
\begin{cases}
c=6\\
-1+b+c=10
\end{cases}
\]
Solving this system of equations gives
\[
\begin{cases}
b=5\\
c=6
\end{cases}
\]
Therefore, the expression for the quadratic function is $y=-x^2+5x+6=\boxed{-x^2+5x+6}$.


Question: As shown in the figure, the graph of the quadratic function $y=-x^2+bx+c$ intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C(0, 6)$, and passes through the point $(1, 10)$. Find: What is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} We have the quadratic function $y=-x^2+5x+6$ intersects the $x$-axis at points A and B,\\
therefore $-x^2+5x+6=0$, solving this gives $x_1=-1$, $x_2=6$,\\
thus, the coordinates of the two intersection points of the quadratic function $y=-x^2+5x+6$ with the $x$-axis are $A(-1, 0)$, $B(6, 0)$.\\
Therefore, the area of $\triangle ABC$ is $\frac{1}{2}AB\times OC=\frac{1}{2}\times |6-(-1)|\times 6=\boxed{21}$.


Question: As shown in the diagram, the graph of the linear function $y=mx+n$ and the graph of the inverse proportion function $y=\frac{k}{x}$ intersect at point $A(-2,2)$ and point $B(1,a)$. What are the expressions of these two functions?

Solution:
\textbf{Solution:} Let's substitute $A(-2,2)$ into $y=\frac{k}{x}$ to obtain $k=-2\times 2=-4$,\\
$\therefore$ the equation for the inverse proportion function is $y=-\frac{4}{x}$,\\
Substituting $B(1,a)$ into $y=-\frac{4}{x}$ gives $a=-4$,\\
$\therefore \boxed{B(1,-4)}$,\\
Substituting $A(-2,2)$ and $B(1,-4)$ into $y=mx+n$ we get $\left\{\begin{array}{l}-2m+n=2 \\ m+n=-4\end{array}\right.$,\\
Solving this system gives $\left\{\begin{array}{l}m=-2 \\ n=-2\end{array}\right.$,\\
$\therefore$ the equation for the linear function is $\boxed{y=-2x-2}$.


Question: As shown in the figure, the graph of the linear function $y=mx+n$ intersects with the graph of the inverse proportion function $y=\frac{k}{x}$ at points $A\left(-2,2\right)$ and $B\left(1, a\right)$. What is the solution set for the inequality $mx+n>\frac{k}{x}$?

Solution:
\textbf{Solution:} The solution set of the inequality $mx+n>\frac{k}{x}$ is $x<-2$ or $0<x<10$.

Therefore, the answer is $x<-2$ or $\boxed{0<x<10}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\frac{m}{x}$ at points $A(1,n)$ and $B(-3,-2)$. Try to determine the analytical expressions of the above-mentioned inverse proportion function and linear function.

Solution:
\textbf{Solution}: Let's substitute point $B(-3, -2)$ into the inverse proportion function $y=\frac{m}{x}$, we get: $m=-3\times (-2)=6$,\\
thus, the analytic expression of the inverse proportion function is $y=\frac{6}{x}$,\\
substituting point $A(1, n)$ into the inverse proportion function $y=\frac{6}{x}$, we find: $n=\frac{6}{1}=6$, that is $A(1, 6)$,\\
substituting points $A(1, 6)$ and $B(-3, -2)$ into the linear function $y=kx+b$, we get: $\left\{\begin{array}{l}k+b=6 \\ -3k+b=-2\end{array}\right.$,\\
from which we obtain $\left\{\begin{array}{l}k=2 \\ b=4\end{array}\right.$,\\
thus, the analytic expression of the linear function is $\boxed{y=2x+4}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\frac{m}{x}$ at points $A(1,n)$ and $B(-3,-2)$. What is the area of $\triangle AOB$?

Solution:
\textbf{Solution:} As shown in the diagram, let the intersection of $AB$ and the $x$-axis be $C$, \\
for the linear function $y=2x+4$, \\
when $y=0$, $2x+4=0$, solving this we find $x=-2$, i.e., $C(-2,0)$, \\
thus $S_{\triangle AOB}=S_{\triangle AOC}+S_{\triangle BOC}=\frac{1}{2}\times 2\times 6+\frac{1}{2}\times 2\times 2=\boxed{8}$, \\
meaning the area of $\triangle AOB$ is $\boxed{8}$.


Question: As shown in the figure, in the Cartesian coordinate system, $O$ is the origin, and point $A(3, 3)$ is on the graph of the inverse proportional function $y = \frac{k}{x}$. What is the value of $k$?

Solution:
\textbf{Solution}: Since the point $A(3, 3)$ lies on the graph of the inverse proportion function $y=\frac{k}{x}$, \\
it follows that $k=3\times 3=\boxed{9}$


Question: As shown in the figure, in the Cartesian coordinate system, $O$ is the origin. The point $A(3, 3)$ lies on the graph of the inverse proportion function $y = \frac{k}{x}$. Through point $A$, draw $AB \perp x$-axis at point $B$ and $AC \perp y$-axis at point $C$. Point $D$ is on the graph of the function in the fourth quadrant. Connect $OD$ and $CD$. If the area of $\triangle OCD$ is equal to the area of quadrilateral $ABOC$, find the coordinates of point $D$.

Solution:
\textbf{Solution:} Since point $A(3, 3)$ lies on the graph of the inverse proportion function $y=\frac{9}{x}$,\\
it follows that ${S}_{\text{quadrilateral} ABOC}=3\times 3=9$,\\
Since $D$ is on the graph of the inverse proportion function $y=\frac{9}{x}$,\\
let $D\left(x, \frac{9}{x}\right)$,\\
thus ${S}_{\triangle OCD}=\frac{1}{2}OC\cdot\left|x_{D}\right|=\frac{1}{2}\times 3\times \left(-x_{D}\right)$,\\
leading to $-\frac{3}{2}x_{D}=9$,\\
solving this gives $x_{D}=-6$,\\
therefore \boxed{D\left(-6, -\frac{3}{2}\right)}.


Question: As shown in the figure, the line $y=kx$ intersects the parabola $y=x^2+c$ at points A and B, where the coordinates of point B are $(2, 2)$. Find the equations of the line $AB$ and the parabola.

Solution:
Solution: Since the line $y=kx$ intersects the parabola $y=x^{2}+c$ at point $B(2, 2)$,\\
therefore $2k=2$ and $4+c=2$,\\
thus $k=1$ and $c=-2$,\\
therefore the equation of line $AB$ is \boxed{y=x}, and the equation of the parabola is \boxed{y=x^{2}-2}.


Question: As shown in the figure, the line $y=kx$ intersects the parabola $y=x^2+c$ at points A and B, where the coordinates of point B are $(2, 2)$. Point C is a point on the parabola, and the area of $\triangle ABC$ is 3. Find the coordinates of point C.

Solution:
\textbf{Solution:} Solve the system of equations $\begin{cases}y=kx\\ y=x^{2}+c\end{cases}$, yielding $\begin{cases}x=2\\ y=2\end{cases}$ or $\begin{cases}x=-1\\ y=-1\end{cases}$.\\
When C is above line AB, construct $CD\perp x$ axis intersecting AB at D, with the coordinates of point C as $\left(t, t^{2}+c\right)$, hence the coordinates of point D are $\left(t, kt\right)$,\\
$\therefore CD=t^{2}+c-kt$,\\
$\therefore S_{\triangle ABC}=S_{\triangle DBC}-S_{\triangle ADC}=\frac{1}{2}CD\times (2-t)-\frac{1}{2}CD\times (-1-t)=\frac{3}{2}CD=3$,\\
$\therefore CD=2$,\\
$\therefore t^{2}+c-kt=2$,\\
yielding $t_{1}=\frac{1-\sqrt{17}}{2}$, $t_{2}=\frac{1+\sqrt{17}}{2}$,\\
$\therefore t^{2}+c=\frac{5-\sqrt{17}}{2}$ or $\frac{5+\sqrt{17}}{2}$,\\
$\therefore C_{1}\left(\frac{1-\sqrt{17}}{2}, \frac{5-\sqrt{17}}{2}\right)$, $C_{2}\left(\frac{1+\sqrt{17}}{2}, \frac{5+\sqrt{17}}{2}\right)$,\\
When C is below line AB, construct $CD\perp x$ axis intersecting AB at D, with the coordinates of point C as $\left(t, t^{2}+c\right)$, hence the coordinates of point D are $\left(t, kt\right)$,\\
$\therefore CD=-t^{2}-c+kt$,\\
$\therefore S_{\triangle ABC}=S_{\triangle BBC}+S_{\triangle ADC}=\frac{1}{2}CD(2-t)+\frac{1}{2}CD(t+1)=\frac{3}{2}CD=3$,\\
$\therefore CD=2$,\\
$\therefore t-t^{2}+c=2$,\\
yielding $t_{1}=0$, $t_{2}=1$,\\
$\therefore t^{2}+c=-2$ or $-1$,\\
$\therefore C_{3}\left(0, -2\right)$, $C_{4}\left(1, -1\right)$,\\
$\therefore$ the coordinates of point C are $\boxed{C_{1}\left(\frac{1-\sqrt{17}}{2}, \frac{5-\sqrt{17}}{2}\right)}$, $\boxed{C_{2}\left(\frac{1+\sqrt{17}}{2}, \frac{5+\sqrt{17}}{2}\right)}$, $\boxed{C_{3}\left(0, -2\right)}$, $\boxed{C_{4}\left(1, -1\right)}$.


Question: In the figure, the line $y=kx$ intersects with the parabola $y=x^2+c$ at points $A$ and $B$, where the coordinates of point $B$ are $(2, 2)$. Point $P$ is on the parabola, point $Q$ is on line $AB$, and point $M$ is within the coordinate plane. When the quadrilateral with vertices $A$, $P$, $Q$, $M$ forms a square, what are the coordinates of point $M$?

Solution: \textbf{Solution:} Let $M_1(1, -1)$, $M_2(0, -2)$, $M_3(-1, 1)$, and $M_4(-1, -3)$. $\boxed{M_1(1, -1)}$, $\boxed{M_2(0, -2)}$, $\boxed{M_3(-1, 1)}$, and $\boxed{M_4(-1, -3)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y = 2x + 6$ intersects the x-axis at point A and intersects the y-axis at point C. The parabola $y = -2x^2 + bx + c$ passes through points A and C, and intersects the x-axis at another point B. Find the analytical equation of the parabola.

Solution:
\textbf{Solution:} Let's solve for $y=2x+6$. When $x=0$, $y=6$, and when $y=0$, $x=-3$,\\
$\therefore C(0, 6)$ and $A(-3, 0)$,\\
$\because$ the parabola $y=-2x^{2}+bx+c$ passes through points A and C,\\
$\therefore \left\{\begin{array}{l}
-18-3b+c=0 \\
c=6
\end{array}\right.$,\\
Solving this, we get $\left\{\begin{array}{l}
b=-4 \\
c=6
\end{array}\right.$,\\
$\therefore$ the equation of the parabola is $y=-2x^{2}-4x+6=\boxed{-2x^{2}-4x+6}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the line $y=2x+6$ intersects the x-axis at point A and intersects the y-axis at point C. The parabola $y=-2x^2+bx+c$ passes through points A and C and intersects the x-axis at another point B. There is a moving point E on the parabola above line $AC$. Connect $BE$ and let it intersect line $AC$ at point F. When $EF=\dfrac{1}{2}BF$, what are the coordinates of point E?

Solution:
\textbf{Solution:} Let $-2x^2-4x+6=0$, we find $x_1=-3$, $x_2=1$, therefore $B(1, 0)$,

Suppose the x-coordinate of point E is t, then $E\left(t, -2t^2-4t+6\right)$,

As shown in the diagram, draw $EH \perp x$ axis at point H, and draw $FG \perp x$ axis at point G, then $EH \parallel FG$, therefore $\triangle BFG \sim \triangle BEH$,

Since $EF=\frac{1}{2}BF$,

Therefore $\frac{BF}{BE}=\frac{BG}{BH}=\frac{FG}{EH}=\frac{2}{3}$,

Since $BH=1-t$,

Therefore $BG=\frac{2}{3}BH=\frac{2}{3}-\frac{2}{3}t$,

Therefore the x-coordinate of point F is $\frac{1}{3}+\frac{2}{3}t$,

Therefore $F\left(\frac{1}{3}+\frac{2}{3}t, \frac{20}{3}+\frac{4}{3}t\right)$,

Therefore $-2t^2-4t+6=\frac{3}{2}\left(\frac{20}{3}+\frac{4}{3}t\right)$,

Therefore $t^2+3t+2=0$,

We find $t_1=-2$, $t_2=-1$,

When $t_1=-2$, $-2t^2-4t+6=6$,

When $t_2=-1$, $-2t^2-4t+6=8$,

Therefore $E_1(-2, 6)$, $E_2(-1, 8)$,

Hence, the coordinates of point E are $\boxed{E_1(-2, 6)}$ or $\boxed{E_2(-1, 8)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=2x+6$ intersects the x-axis at point A and the y-axis at point C. The parabola $y=-2x^{2}+bx+c$ passes through points A and C, and intersects the x-axis at another point B. If point E is located to the left of the axis of symmetry, point M is a point on the axis of symmetry of the parabola, and point N is a point on the parabola, when the quadrilateral formed by points M, N, E, B is a rhombus, directly write down the coordinates of point M.

Solution: \textbf{Solution:} The coordinates of point M are \boxed{(-1, -4)}, \boxed{(-1, -16)}, or \boxed{(-1, 0)}.


Question: As shown in the figure, it is known that the parabola $y=-(x+1)^{2}+4$ intersects with the x-axis at points A and B, and intersects with the y-axis at point C. Find the coordinates of points A, B, and C.

Solution:
\textbf{Solution:} The parabola $y=-(x+1)^{2}+4$ intersects the x-axis at points A and B, and intersects the y-axis at point C.\\
$\because y=-(x+1)^{2}+4=-x^{2}-2x+3$,\\
Let $x=0$, then $y=3$,\\
$\therefore \boxed{C(0, 3)}$,\\
Let $y=0$,\\
then $-(x+1)^{2}+4=0$,\\
solving for $x$ gives $x_{1}=1, x_{2}=-3$,\\
$\therefore \boxed{A(-3, 0)}, \boxed{B(1, 0)}$.


Question: As shown in the figure, it is known that the parabola $y=-(x+1)^{2}+4$ intersects the x-axis at points A and B, and intersects the y-axis at point C. Is there a point P on the parabola such that the area of $\triangle PAB$ is 8? If it exists, write down the coordinates of point P directly; if not, explain the reason.

Solution: \textbf{Solution:} Given that, the coordinates of point P are \boxed{(-1, 4)}, \boxed{(-1+2\sqrt{2}, -4)}, or \boxed{(-1-2\sqrt{2}, -4)}.


Question: As shown in the figure, the parabola $y=ax^2+3x+c\ (a \ne 0)$ intersects the $x$-axis at points $A(-2, 0)$ and $B$, and intersects the $y$-axis at point $C(0, 8)$. Point $P$ is a moving point on the parabola above line $BC$. Connect $CP$ and $PB$. The line $BC$ intersects the axis of symmetry $l$ of the parabola at point $E$. What is the analytic expression of the parabola?

Solution:
\textbf{Solution:} Let's substitute $A(-2, 0)$, and $C(0, 8)$ into $y=ax^2+3x+c$,\\
hence $\begin{cases} 4a-6+c=0 \\ c=8 \end{cases}$,\\
solving this, we get $\begin{cases} a=-\frac{1}{2} \\ c=8 \end{cases}$,\\
therefore $y=-\frac{1}{2}x^2+3x+8=\boxed{-\frac{1}{2}x^2+3x+8}$.


Question: As shown in the figure, the parabola $y=ax^2+3x+c\ (a\neq 0)$ intersects the x-axis at points $A(-2, 0)$ and $B$, and intersects the y-axis at point $C(0, 8)$. Point $P$ is a moving point on the parabola above the line $BC$. Connect $CP$ and $PB$. The line $BC$ intersects the axis of symmetry $l$ of the parabola at point $E$. What is the maximum area of $\triangle BCP$?

Solution:
\textbf{Solution:} Let $y=0$, then $-\frac{1}{2}x^2+3x+8=0$,\\
we get $x=-2$ or $x=8$,\\
$\therefore B(8, 0)$,\\
Let the equation of line $BC$ be $y=kx+b$,\\
$\therefore \left\{\begin{array}{l} b=8 \\ 8k+b=0 \end{array}\right.$,\\
Solving, we get $\left\{\begin{array}{l} k=-1 \\ b=8 \end{array}\right.$,\\
$\therefore y=-x+8$,\\
Draw a line through $P$ parallel to the $y$ axis intersecting $BC$ at $G$,\\
Let $P(t, -\frac{1}{2}t^2+3t+8)$, then $G(t, -t+8)$,\\
$\therefore PG=-\frac{1}{2}t^2+3t+8+t-8=-\frac{1}{2}t^2+4t$,\\
$\therefore S_{\triangle CBP}=\frac{1}{2}\times 8\times \left(-\frac{1}{2}t^2+4t\right)=-2t^2+16t=-2(t-4)^2+32$,\\
$\therefore$ when $t=4$, the area of $\triangle BCP$ reaches its maximum value, which is $\boxed{32}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+3$ ($a\neq 0$) intersects the x-axis at points $A(1,0)$ and $B(-3,0)$ and intersects the y-axis at point $C$. Connect $BC$ and intersect the axis of symmetry of the parabola at point $E$, with the vertex being point $D$. What is the equation of the parabola?

Solution:
\textbf{Solution:} Since the parabola $y=ax^{2}+bx+3$ ($a\neq 0$) intersects the x-axis at points A $(1,0)$ and B $(-3,0)$,\\
it follows that $\begin{cases} a+b+3=0 \\ 9a-3b+3=0 \end{cases}$,\\
solving these equations yields $\begin{cases} a=-1 \\ b=-2 \end{cases}$,\\
therefore, the equation of the parabola is $\boxed{y=-x^{2}-2x+3}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+3$ ($a\neq 0$) intersects the x-axis at points $A(1,0)$ and $B(-3,0)$, and intersects the y-axis at point $C$. Line $BC$ is drawn, intersecting the axis of symmetry of the parabola at point $E$, with the vertex being point $D$. What is the area of $\triangle BOC$?

Solution:
\textbf{Solution:} Given equation (1), $y=-x^{2}-2x+3$,\\
$\therefore$ the coordinates of point C are $(0,3)$,\\
$\therefore$ OC$=3$,\\
$\because$ the coordinates of point B are $(-3,0)$,\\
$\therefore$ OB$=3$,\\
$\because$ $\angle$BOC$=90^\circ$,\\
$\therefore$ the area of $\triangle BOC$ is $\frac{OB \cdot OC}{2}=\frac{3\times 3}{2}=\boxed{\frac{9}{2}}$.


Question: As shown in the figure, the parabola $y=x^2+bx+c$ (where \(b\) and \(c\) are constants) has its vertex at \(C\), and intersects the \(x\)-axis at points \(A\) and \(B\), with \(A(1, 0)\) and \(AB = 4\). Point \(P\) is a moving point on segment \(AB\), and a line through \(P\) parallel to \(BC\) intersects \(AC\) at point \(Q\). What is the equation of this parabola?

Solution:
\textbf{Solution:} Since point A is (1,0) and AB = 4,
it follows that the coordinates of point B are (-3,0).
Substituting points A(1,0) and B(-3,0) into the function's equation yields:
\[
\begin{cases}
0 = 1 + b + c\\
0 = 9 - 3b + c
\end{cases},
\]
from which we find that $b=2$ and $c=-3$,
thus, the equation of the parabola is \boxed{y=x^{2}+2x-3}.


Question: As shown in the figure, the parabola $y=x^2+bx+c$ (where $b$ and $c$ are constants) has the vertex $C$, intersects the x-axis at points $A$ and $B$, with $A(1, 0)$, and $AB=4$. Let point $P$ be a moving point on segment $AB$, and draw $PQ\parallel BC$ intersecting $AC$ at point $Q$. Determine the maximum area of $\triangle CPQ$ and the coordinates of point $P$ at that maximum.

Solution:
\textbf{Solution:} From equation (1), we get the parabola equation as $y=x^{2}+2x-3$,\\
the vertex form is: $y=(x+1)^{2}-4$,\\
thus the coordinates of point C are: $(-1,-4)$,\\
from B $(-3,0)$ and C $(-1,-4)$, we can find the equation of line BC as: $y=-2x-6$,\\
from A $(1,0)$ and C $(-1,-4)$, we can find the equation of line AC as: $y=2x-2$,\\
since $PQ\parallel BC$,\\
let the equation of line PQ be: $y=-2x+n$, intersecting the x-axis at P$\left(\frac{n}{2}, 0\right)$,\\
from $\begin{cases}y=-2x+n\\ y=2x-2\end{cases}$, we find: $Q\left(\frac{n+2}{4}, \frac{n-2}{2}\right)$,\\
since $P$ is on the segment AB,\\
therefore $-3<\frac{n}{2}<1$,\\
therefore the range of n is $-6<n<2$,\\
then the area of $\triangle CPQ$ equals the area of $\triangle CPA$ minus the area of $\triangle APQ$\\
$=\frac{1}{2}\times \left(1-\frac{n}{2}\right)\times 4-\frac{1}{2}\times \left(1-\frac{n}{2}\right)\times \left(-\frac{n-2}{2}\right)$\\
$=-\frac{1}{8}(n+2)^{2}+2$\\
therefore, when $n=-2$, i.e., P $(-1,0)$, the area of $\triangle CPQ$ is maximized, with the maximum value being $\boxed{2}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ ($k\neq 0$) intersects with the graph of the inverse proportion function $y=\frac{6}{x}$ ($x>0$) at points $A(m, 6)$ and $B(3, n)$. Find the explicit expression of the linear function.

Solution:
\textbf{Solution:} Given points A(1, 6) and B(3, 2) are on the graph of the inverse proportional function $y=\frac{6}{x}$ ($x>0$),\\
$\therefore$ 6m=3n=6,\\
$\therefore$ m=1, n=2,\\
Also, since points A(1, 6) and B(3, 2) are on the graph of the linear function $y=kx+b$ ($k\neq 0$),\\
$\therefore$ $\left\{\begin{array}{l}6=k+b \\ 2=3k+b \end{array}\right.$,\\
Solving these, we get $\left\{\begin{array}{l}k=-2 \\ b=8 \end{array}\right.$,\\
Thus, the equation of the linear function is: $\boxed{y=-2x+8}$;


Question: As shown in the figure, the graph of the linear function $y=kx+b$ ($k\ne 0$) intersects with the graph of the inverse proportion function $y=\frac{6}{x}$ ($x>0$) at points $A(m, 6)$ and $B(3, n)$. Based on the graph, directly write the range of $x$ values for which $kx+b<\frac{6}{x}$ holds.

Solution:
\textbf{Solution:} For the values of $x$ that satisfy $kx+b<\frac{6}{x}$, the range of $x$ is $0<x<1$ or $x>3$, \\
thus the answer is $x \in \boxed{(0,1) \cup (3, +\infty)}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ (where $k\ne 0$) intersects with the graph of the inverse proportion function $y=\frac{6}{x}$ (where $x>0$) at points $A(m, 6)$ and $B(3, n)$. Find the area of triangle $\triangle AOB$.

Solution:
\textbf{Solution:} As shown in the diagram, lines are drawn from point A and point B perpendicular to the x-axis, meeting the x-axis at points E and C respectively. The line AB intersects the x-axis at point D.\\
Let $-2x+8=0$, we get $x=4$, that is, D(4,0).\\
Since A(1,6) and B(3,2),\\
it follows that AE=6 and BC=2,\\
therefore, the area of triangle AOB is calculated as $S_{\triangle AOB}=S_{\triangle AOD}-S_{\triangle BOD}=\frac{1}{2}\times4\times6-\frac{1}{2}\times4\times2=\boxed{8}$.


Question: As shown in the graph, within the Cartesian coordinate system, the parabola $y = \frac{1}{2}x^{2} + bx + c$ passes through point $A(-4,0)$. Point $M$ is the vertex of the parabola, and point $B$ is on the $y$-axis, with $OA=OB$. The line $AB$ intersects with the parabola at point $C(2,6)$ in the first quadrant. Find the equation of the parabola and the coordinates of the vertex $M$?

Solution:
\textbf{Solution:} Let's substitute $A(-4,0)$, $C(2,6)$ into $y=\frac{1}{2}x^{2}+bx+c$, we get: $\left\{\begin{array}{l}0=\frac{1}{2}\times 16-4b+c \\ 6=\frac{1}{2}\times 4+2b+c\end{array}\right.$. Solving these, we find $\left\{\begin{array}{l}b=2 \\ c=0\end{array}\right.$, \\
$\therefore$ the equation of the parabola is $y=\frac{1}{2}x^{2}+2x=\frac{1}{2}(x+2)^{2}-2$, \\
$\therefore$ the coordinates of the vertex $M$ are $\boxed{(-2，-2)}$;


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=\frac{1}{2}x^{2}+bx+c$ passes through point $A(-4, 0)$. Point $M$ is the vertex of the parabola, and point $B$ is on the $y$-axis, with $OA=OB$. The line $AB$ intersects the parabola at point $C(2, 6)$ in the first quadrant. Find the equation of line $AB$ and the value of $\sin\angle ABO$; connect $OC$. If a line through point $O$ intersects segment $AC$ at point $P$, dividing the area of triangle $AOC$ into two parts in the ratio $1:2$, find the coordinates of point $P$.

Solution:
\textbf{Solution:} Given that $A(-4,0)$, \\
therefore $OA=4$, \\
since $OA=OB$, \\
thus $OB=4$, \\
therefore, $B(0,4)$, \\
Let the equation of line AB be $y=kx+b$, \\
Substituting $A(-4,0)$ and $B(0,4)$ into the equation yields: $\left\{\begin{array}{l}0=-4k+b \\ 4=b\end{array}\right.$, \\
Solving this, we get $\left\{\begin{array}{l}k=1 \\ b=4\end{array}\right.$, \\
Therefore, the equation of line AB is $y=x+4$, \\
In $\triangle AOB$, $AB=\sqrt{OA^2+OB^2}=4\sqrt{2}$, \\
thus $\sin\angle ABO=\frac{OA}{AB}=\frac{4}{4\sqrt{2}}=\frac{\sqrt{2}}{2}$. \\
A line through the point O intersects segment AC at point P, dividing the area of triangle AOC into two parts with a ratio of 1:2. Draw $PQ\perp x$-axis at Q, and draw $CH\perp x$-axis at H, \\
We consider two cases: \\
(1) If $S_{\triangle AOP}:S_{\triangle COP}=1:2$, \\
since $S_{\triangle AOP}:S_{\triangle AOC}=1:3$, \\
therefore $PQ:CH=1:3$, \\
given that $C(2,6)$, \\
thus $CH=6$, \\
therefore $PQ=2$, i.e., $y_P=2$, \\
In $y=x+4$, setting $y=2$ gives, $2=x+4$, \\
thus $x=-2$, \\
therefore $P(-2,2)$; \\
(2) If $S_{\triangle COP}:S_{\triangle AOP}=1:2$, \\
since $S_{\triangle COP}:S_{\triangle AOP}=1:2$, \\
therefore $S_{\triangle AOP}:S_{\triangle AOC}=2:3$, \\
thus $PQ:CH=2:3$, \\
given that $CH=6$, \\
therefore $PQ=4$, i.e., $y_P=4$, \\
In $y=x+4$, setting $y=4$ gives, $4=x+4$, \\
thus $x=0$, \\
therefore $P(0,4)$; \\
In summary, a line through the point O intersects segment AC at point P, dividing the area of triangle AOC into two parts with a ratio of 1:2, then the coordinates of P are \boxed{(-2,2)} or \boxed{(0,4)}.


Question: As shown in the graph, in the Cartesian coordinate system, the parabola $y=\frac{1}{2}x^{2}+bx+c$ passes through point $A(-4,0)$. Point $M$ is the vertex of the parabola, and point $B$ is on the $y$-axis, with $OA=OB$. The line $AB$ intersects the parabola in the first quadrant at point $C(2,6)$. Is there a point $N$ in the coordinate plane such that the quadrilateral with vertices $A$, $O$, $C$, and $N$ is a parallelogram? If it exists, please write the coordinates of point $N$ directly; if not, please explain why.

Solution: \textbf{Solution:} There exists, the coordinates of point $N$ are: $\boxed{(6,6)}$, $\boxed{(-2,6)}$, or $\boxed{(-6,-6)}$.


Question: As shown in the figure, the parabola $y=-x^2+bx+c$ passes through point $A(3, 0)$ and point $B(0, 3)$. Find the expression of the parabola and the coordinates of vertex $C$.

Solution:
\textbf{Solution:} Substitute $A\left(3, 0\right)$ and $B\left(0, 3\right)$ into the parabola equation $y=-{x}^{2}+bx+c$ and get:
\[
\begin{cases}
3b+c-9=0\\
c=3
\end{cases}
\]
Solving these, we find $b=2$ and $c=3$;\\
Therefore, the equation of the parabola is: $y=-{x}^{2}+2x+3$\\
Since $y=-{x}^{2}+2x+3=-{\left(x-1\right)}^{2}+4$,\\
Therefore, the coordinates of the vertex $C$ of the parabola are: $\boxed{\left(1, 4\right)}$


Question: As shown in the figure, the graph of the linear function $y=kx+2$ intersects with the graph of the inverse proportion function $y=\frac{m}{x}$ (where $x>0$) at point $P$. Point $P$ is located in the first quadrant, with $PA\perp x$-axis at point A. The graph of the linear function intersects the x-axis and y-axis at points $C$ and $D$, respectively, and ${S}_{\triangle COD}=1$, $\frac{CO}{OA}=\frac{1}{2}$. What are the coordinates of point $D$?

Solution:
\textbf{Solution:} In the equation $y=kx+2$, let $x=0$, then $y=2$. \\
$\therefore$ The coordinates of point D are $\boxed{(0, 2)}$.


Question: As shown in the figure, the graph of the linear function $y=kx+2$ intersects with the graph of the hyperbolic function $y=\frac{m}{x}$ ($x>0$) at point P, which lies in the first quadrant. $PA$ is perpendicular to the $x$-axis at point A. The graph of the linear function intersects the $x$-axis and $y$-axis at points C and D, respectively, and the area of $\triangle COD$ is 1, and $\frac{CO}{OA}=\frac{1}{2}$. Find the analytical expressions of the linear function and the hyperbolic function?

Solution:
\textbf{Solution:} Given $PA\perp CA$, $DO\perp CO$,\\
$\therefore \angle PAC=\angle COD=90^\circ$\\
Since $\angle DCO=\angle DCO$,\\
$\therefore Rt\triangle PAC\sim Rt\triangle DOC$\\
Given $\frac{CO}{OA}=\frac{1}{2}$, and $OD=2$,\\
$\therefore \frac{OD}{PA}=\frac{CO}{CA}=\frac{1}{3}$\\
We find $PA=6$\\
Given ${S}_{\triangle COD}=1$, we have: $\frac{1}{2}OC\cdot OD=1$,\\
We find $OC=1$\\
$\therefore OA=2$, $P(2, 6)$\\
Substitute $P(2, 6)$ into $y=kx+2$ and $y=\frac{m}{x}$\\
Thus, the linear function is $y=2x+2$, and the inverse proportion function is $\boxed{y=\frac{12}{x}}\left(x>0\right)$


Question: As shown in the figure, the graph of the linear function $y=kx+2$ intersects with the graph of the inverse proportion function $y=\frac{m}{x}$ at point P, which lies in the first quadrant. $PA\perp x$ axis at point A, and the graph of the linear function intersects the x-axis and y-axis at points C and D, respectively, with $S_{\triangle COD}=1$ and $\frac{CO}{OA}=\frac{1}{2}$. Based on the figure, directly write the range of $x$ for which $x>0$ and the value of the linear function is greater than or equal to the value of the inverse proportion function.

Solution: \textbf{Solution:} According to the graph, we directly write that for $x>0$, the range of x for which the value of the linear function is greater than or equal to the value of the inverse proportion function is $x\ge \boxed{2}$.


Question: As shown in Figure (1), it is known that the line $y = \frac{4}{3}x + 4$ intersects the x-axis at point A and the y-axis at point C. A parabola $y = ax^2 + bx + c$ passes through points A and C, and intersects the x-axis at another point, B, with its axis of symmetry being the line $x = -1$. Let D be a moving point on the parabola in the second quadrant with its x-coordinate being $m$. What is the equation of the parabola?

Solution:
\textbf{Solution:} Substituting $x=0$ into $y=\frac{4}{3}x+4$, we get $y=4$.\\
$\therefore$ The coordinates of point C are $(0,4)$.\\
Substituting $y=0$ into $y=\frac{4}{3}x+4$, we get $x=-3$.\\
$\therefore$ The coordinates of point A are $(-3,0)$.\\
$\because$ The axis of symmetry for the parabola is the line $x=-1$,\\
$\therefore$ $-\frac{b}{2a}=-1$, that is, $b=2a$.\\
Forming a system of equations from the given information, we get\\
$\begin{cases}
9a-3b+c=0,\\
b=2a,\\
c=4.
\end{cases}$ Solving this system, we find $\begin{cases}
a=-\frac{4}{3},\\
b=-\frac{8}{3},\\
c=4.
\end{cases}$\\
$\therefore$ The equation of the parabola is $y=-\frac{4}{3}x^2-\frac{8}{3}x+4$.\\
Hence, the expression for the parabola is $\boxed{y=-\frac{4}{3}x^2-\frac{8}{3}x+4}$.


Question: As shown in Figure (1), it's known that the line $y=\frac{4}{3}x+4$ intersects the x-axis at point A and the y-axis at point C. The parabola $y=ax^2+bx+c$ passes through points A and C, also intersects the x-axis at another point B, with its axis of symmetry being the line $x=-1$. Point D is a moving point on the parabola in the second quadrant, and the x-coordinate of D is denoted as $m$. What is the maximum value of the area $S$ of quadrilateral $ABCD$ and the coordinates of D at this maximum?

Solution:
\textbf{Solution:} Connect $OD$,\\
since point B is symmetric to point $A(-3,0)$ with respect to the line $x=-1$,\\
therefore, the coordinates of point B are $(1,0)$.\\
According to the problem, the coordinates of point D are $(m, -\frac{4}{3}m^{2}-\frac{8}{3}m+4)$.\\
Therefore, the area $S_{\text{quadrilateral} ABCD}$ of quadrilateral $ABCD$ equals $S_{\triangle OAD}+S_{\triangle OCD}+S_{\triangle OBC}$\\
$=\frac{1}{2}\times3\times(-\frac{4}{3}m^{2}-\frac{8}{3}m+4)+\frac{1}{2}\times4\times(-m)+\frac{1}{2}\times1\times4$\\
$=-2m^{2}-6m+8=-2(m+\frac{3}{2})^{2}+\frac{25}{2}$\\
Since $-2<0$,\\
therefore, when $m=-\frac{3}{2}$, the maximum area of quadrilateral $ABCD$ is $\boxed{\frac{25}{2}}$.\\
At this point, the coordinates of point D are $\boxed{(-\frac{3}{2}, 5)}$.


Question: As shown in Figure (1), it is known that the line $y=\frac{4}{3}x+4$ intersects the x-axis at point A and the y-axis at point C. The parabola $y=ax^2+bx+c$ passes through points A and C, and has another intersection with the x-axis at point B. The axis of symmetry is the line $x=-1$. D is a moving point on the parabola in the second quadrant, and the x-coordinate of point D is denoted by m. A perpendicular line is drawn from point D to the y-axis, creating a perpendicular foot at point E, as shown in Figure (2). Is there a point D that makes $\triangle CDE$ similar to $\triangle AOC$? If so, find the value of the x-coordinate m of point D; if not, please explain why.

Solution:
\textbf{Solution:} Existence proven. From the given information, we have: $OA=3$, $OC=4$, $DE=-m$,\\
since $\triangle CDE$ is similar to $\triangle AOC$, $\angle AOC=\angle CED=90^\circ$\\
therefore $\triangle CDE\sim \triangle ACO$ or $\triangle CDE\sim \triangle CAO$\\
thus $\frac{CE}{DE}=\frac{OA}{OC}=\frac{3}{4}$ or $\frac{CE}{DE}=\frac{OC}{OA}=\frac{4}{3}$\\
therefore $CE=-\frac{3}{4}m$ or $-\frac{4}{3}m$\\
hence, point $D$'s coordinates are $(m, -\frac{3}{4}m+4)$ or $(m, \frac{3}{4}m+4)$ or $(m, -\frac{4}{3}m+4)$ or $(m, \frac{4}{3}m+4)$\\
Substituting $(m, -\frac{3}{4}m+4)$ into $y=-\frac{4}{3}x^2-\frac{8}{3}x+4$, yields: $-\frac{3}{4}m+4=-\frac{4}{3}m^2-\frac{8}{3}m+4$\\
Solving gives $m=\boxed{-\frac{23}{16}}$;\\
Substituting $(m, \frac{3}{4}m+4)$ into $y=-\frac{4}{3}x^2-\frac{8}{3}x+4$, yields: $\frac{3}{4}m+4=-\frac{4}{3}m^2-\frac{8}{3}m+4$\\
Solving gives $m=\boxed{-\frac{41}{16}}$;\\
Substituting $(m, -\frac{4}{3}m+4)$ into $y=-\frac{4}{3}x^2-\frac{8}{3}x+4$, yields: $-\frac{4}{3}m+4=-\frac{4}{3}m^2-\frac{8}{3}m+4$\\
Solving gives $m=\boxed{-1}$;\\
Substituting $(m, \frac{4}{3}m+4)$ into $y=-\frac{4}{3}x^2-\frac{8}{3}x+4$, yields: $\frac{4}{3}m+4=-\frac{4}{3}m^2-\frac{8}{3}m+4$\\
Solving gives $m=-3$, which is discarded;\\
therefore, the possible values for $m$ are $\boxed{-\frac{23}{16}}$, $\boxed{-1}$, or $\boxed{-\frac{41}{16}}$.


Question: As shown in the figure, the graph of the inverse proportion function $y=\frac{k_1}{x}$ intersects with the graph of the direct proportion function $y=k_2x$ at points $A(a, 1)$ and B. Point $M(a-3, a)$ lies on the graph of the inverse proportion function. Connect $OM$ and $BM$, where $BM$ intersects the y-axis at point N. Determine the analytic expression of the inverse proportion function.

Solution:
\textbf{Solution:} Since points A$(4,1)$ and M$(1,4)$ lie on the graph of an inverse variation function, \\
it follows that the equation of the inverse variation function is $\boxed{y=\frac{4}{x}}$.


Question: As shown in the figure, the graph of the inverse proportion function $y = \frac{k_1}{x}$ intersects with the graph of the direct proportion function $y = k_2x$ at points $A(a, 1)$ and $B$. The point $M(a-3, a)$ lies on the graph of the inverse proportion function. Connect $OM$ and extend $BM$ to intersect the y-axis at point $N$. What is the area of $\triangle BOM$?

Solution:
\textbf{Solution:} Since the graph of the inverse proportion function $y=\frac{k_1}{x}$ intersects with the graph of the direct proportion function $y=k_2x$ at points A and B, and given A(4, 1),\\
it follows that the coordinates of point B are $(-4, -1)$,\\
Assuming the equation of line $BM$ is $y=mx+b$,\\
Substituting points $B(-4, -1)$ and $M(1, 4)$ we have\\
$\left\{
\begin{array}{ll}
-4m+b=-1\\
m+b=4
\end{array}
\right.$,\\
Solving this gives $\left\{
\begin{array}{ll}
m=1\\
b=3
\end{array}
\right.$,\\
therefore, the equation of line $BM$ is $y=x+3$,\\
when $x=0$, $y=3$,\\
thus, the coordinates of point N are (0, 3),\\
As shown in the figure, vertical lines to the y-axis are drawn through points M and B, with the feet of the perpendiculars being points P and Q, respectively,\\
thus, $PM=1$, $BQ=4$,\\
Therefore, $S_{\triangle BOM}=S_{\triangle BON}+S_{\triangle MON}=\frac{1}{2}\times 3\times 4+\frac{1}{2}\times 3\times 1=\boxed{\frac{15}{2}}$.


Question: As shown in the figure, it is known that the vertex of the parabola $y=ax^2+bx+c$ is $A(4, 3)$, and it intersects the y-axis at point $B(0, -5)$. The axis of symmetry is line $l$, and point $M$ is the midpoint of segment $AB$. What is the equation of the parabola?

Solution:
\textbf{Solution:} Let the function be expressed as: $y=a(x-4)^2+3$, \\
Substituting the coordinates of point B into the equation and solving, we obtain: $a=-\frac{1}{2}$, \\
Therefore, the equation of the parabola is: $y=-\frac{1}{2}x^2+4x-5=\boxed{-\frac{1}{2}x^2+4x-5}$.


Question: As shown in the diagram, given the parabola $y = ax^2 + bx + c$ with its vertex at $A(4, 3)$ and intersecting the y-axis at point $B(0, -5)$, with the line of symmetry being the line $l$, and the point $M$ being the midpoint of segment $AB$. Find the coordinates of point $M$ and the equation of line $AB$.

Solution:
\textbf{Solution:} Let the coordinates of midpoint $M$ be $(x, y)$,\\
since $A(4, 3), B(0, -5)$,\\
thus $x-0=4-x$, i.e., $x=\frac{4+0}{2}=2$,\\
Similarly, we get $y=\frac{-5+3}{2}=-1$,\\
therefore the midpoint $\boxed{M(2, -1)}$,\\
Let the equation of the line $AB$ be: $y=kx-5$,\\
Substituting the coordinates of point $A$ into the equation yields: $3=4k-5$, solving gives: $k=2$,\\
Hence, the equation of the line $AB$ is: $\boxed{y=2x-5}$.


Question: As shown in the figure, it is known that the parabola $y=ax^2+bx+c$ has its vertex at $A(4,3)$ and intersects the $y$-axis at point $B(0,-5)$. The axis of symmetry is the line $l$. Let point $M$ be the midpoint of segment $AB$. Assume that $P$ and $Q$ are moving points on the parabola and the axis of symmetry $l$ respectively. When the quadrilateral with vertices $A$, $P$, $Q$, $M$ forms a parallelogram, determine the coordinates of points $P$ and $Q$.

Solution:
\textbf{Solution:} Let point $Q(4, s)$ and point $P(m, -\frac{1}{2}m^2+4m-5)$,\\
(1) When $AM$ is one side of the parallelogram,\\
When point Q is below A,\\
Point A is moved 2 units to the left and 4 units down to get M,\\
Similarly, point $P(m, -\frac{1}{2}m^2+4m-5)$ moved 2 units to the left and 4 units down to get $Q(4, s)$\\
i.e., $m-2=4, -\frac{1}{2}m^2+4m-5-4=s$\\
Solving yields: $m=6, s=-3,$\\
Therefore, when point Q is above point A, $AQ=MP=2,$\\
Similarly, the coordinates of point Q are found to be $(4, 5),$\\
(2) When $AM$ is the diagonal of the parallelogram,\\
By the midpoint theorem, $4+2=m+4, 3-1=-\frac{1}{2}m^2+4m-5+s$\\
Solving yields: $m=2, s=1$\\
Therefore, the coordinates of points $P, Q$ are respectively $\left(2, 1\right), \left(4, 1\right)$\\
In summary, the coordinates of $P, Q$ are respectively $\left(6, -3\right), \left(4, 5\right)$ or $\boxed{\left(2, 1\right)}, \boxed{\left(4, 1\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\frac{k_2}{x}$ at points $A(-2, 3)$ and $B(m, -2)$. What are the expressions for the functions corresponding to $y_1$ and $y_2$?

Solution:
\textbf{Solution:} Let point $A\left(-2, 3\right)$ be substituted into the equation of the inverse proportion function to find: $k=-6$,\\
$\therefore y_{2}=-\frac{6}{x}$,\\
$\because$ point B is on the graph of the inverse proportion function,\\
$\therefore -2m=-6$, which gives: $m=3$,\\
$\therefore B\left(3, -2\right)$,\\
Substituting points A and B into the equation of the line gives: $\left\{\begin{array}{l}
-2k_{1}+b=3 \\
3k_{1}+b=-2
\end{array}\right.$, solving this yields: $\left\{\begin{array}{l}
k_{1}=-1 \\
b=1
\end{array}\right.$,\\
$\therefore y_{1}=\boxed{-x+1}$;


Question: As shown in the figure, in the Cartesian coordinate system, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\frac{k_2}{x}$ at points $A(-2, 3)$ and $B(m, -2)$. A line through point B, $BP\parallel x$-axis, intersects the y-axis at point P. What is the area of $\triangle ABP$?

Solution:
\textbf{Solution:} From (1) we have: $A(-2, 3)$, $B(3, -2)$,\\
since $BP\parallel x$-axis,\\
therefore, $BP=3$,\\
thus, the distance from point A to PB is $3-(-2)=5$,\\
hence, the area of $\triangle ABP$ is $\frac{1}{2}\times 3\times 5=\boxed{\frac{15}{2}}$;


Question: As shown in the Cartesian coordinate system, the line $y_1=k_1x+b$ intersects the hyperbola $y_2=\frac{k_2}{x}$ at points $A(-2, 3)$ and $B(m, -2)$. Based on the graph of the function, directly write the solution set of the inequality $k_1x+b<\frac{k_2}{x}$ with respect to $x$.

Solution:
Solution:
According to the graph of the function, we can directly write the solution set of the inequality $k_1x + b < \frac{k_2}{x}$ with respect to $x$ as $-2 < x < 0$ or $x > 3$.

Hence, the answer is $x \in \boxed{(-2, 0) \cup (3, +\infty)}$.


Question: As shown in the figure, the parabola $y = ax^2 + bx + \frac{5}{2}$ intersects the line $AB$ at points $A(-1, 0)$ and $B(4, \frac{5}{2})$. Point $D$ is a moving point on the parabola between points $A$ and $B$ (not coinciding with points $A$ and $B$). The line $CD$ is parallel to the y-axis and intersects the line $AB$ at point $C$. Connect $AD$ and $BD$. What is the analytic expression of the parabola?

Solution:
\textbf{Solution:} Solve: The parabola $y=a{x}^{2}+bx+\frac{5}{2}$ intersects the line $AB$ at point $A\left(-1, 0\right)$ and $B\left(4, \frac{5}{2}\right)$,\\
$\therefore \left\{\begin{array}{l}
0=a(-1)^{2}-b+\frac{5}{2}\\
\frac{5}{2}=16a+4b+\frac{5}{2}
\end{array}\right.$,\\
Solving, we get: $\left\{\begin{array}{l}
a=-\frac{1}{2}\\
b=2
\end{array}\right.$,\\
$\therefore$ the equation of the parabola is $\boxed{y=-\frac{1}{2}{x}^{2}+2x+\frac{5}{2}}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+\frac{5}{2}$ intersects with line segment $AB$ at points $A(-1,0)$ and $B(4,\frac{5}{2})$. Point $D$ is a moving point on the parabola between points $A$ and $B$ (not coinciding with $A$ or $B$), the line $CD$ is parallel to the y-axis and intersects line $AB$ at point $C$, and lines $AD$ and $BD$ are connected. Let the x-coordinate of point $D$ be $m$, and the area of $\triangle ADB$ be $S$. Determine the function of $S$ in terms of $m$, and find the coordinates of point $C$ when $S$ reaches its maximum value.

Solution:
\textbf{Solution:} As shown in the figure, construct $BF\perp DE$ at point F passing through point B.\\
Let the equation of line $AB$ be: $y=kx+m$,\\
then: $\left\{\begin{array}{l}
0=-k+m \\
\frac{5}{2}=4k+m
\end{array}\right.$,\\
solving this gives: $\left\{\begin{array}{l}
k=\frac{1}{2} \\
m=\frac{1}{2}
\end{array}\right.$,\\
$\therefore y=\frac{1}{2}x+\frac{1}{2}$,\\
$\because$ the abscissa of point D is m and $CD\parallel y$-axis,\\
$\therefore$ the coordinate of point C is $\left(m, \frac{1}{2}m+\frac{1}{2}\right)$, and the ordinate of point D is $\left(-\frac{1}{2}m^{2}+2m+\frac{5}{2}\right)$,\\
$\therefore AE=m+1, BF=4-m$, $CD=-\frac{1}{2}m^{2}+\frac{3}{2}m+2$,\\
$\therefore S=\frac{1}{2}CD\cdot (AE+BF)=\frac{1}{2}\times \left(-\frac{1}{2}m^{2}+\frac{3}{2}m+2\right)\times (m+1+4-m)$\\
$=-\frac{5}{4}\left(m-\frac{3}{2}\right)^{2}+\frac{125}{16}$,\\
$\therefore$ when $m=\frac{3}{2}$, $S$ reaches its maximum value $\boxed{\frac{125}{16}}$, at this point $C\left(\frac{3}{2}, \frac{5}{4}\right)$.


Question: As shown in the diagram, the parabola $y=ax^2+bx+\frac{5}{2}$ intersects line $AB$ at points $A(-1, 0)$ and $B(4, \frac{5}{2})$. Point $D$ is a moving point on the parabola between points $A$ and $B$ (not coinciding with points $A$ and $B$), and line $CD$ is parallel to the y-axis, intersecting line $AB$ at point $C$. Lines $AD$ and $BD$ are drawn. Point $D$ is the vertex of the parabola, point $P$ is a moving point on the parabola, and point $Q$ is a moving point on line $AB$. When the quadrilateral with vertices at points $P$, $Q$, $C$, and $D$ forms a parallelogram, determine the coordinates of point $Q$.

Solution:
Solution: Suppose there exist points P and Q such that the quadrilateral formed by the vertices P, Q, C, and D is a parallelogram.\\
$\because y=-\frac{1}{2}x^2+2x+\frac{5}{2}=-\frac{1}{2}(x-2)^2+\frac{9}{2}$, point D is the vertex of the parabola, $CD\parallel y$-axis,\\
$\therefore D(2, \frac{9}{2})$, $C(2, \frac{3}{2})$,\\
(1) When $CD$ is a side of the parallelogram, $PQ\parallel CD, PQ=CD$ \\\\
Let $P(x, -\frac{1}{2}x^2+2x+\frac{5}{2})$, then: $Q(x, \frac{1}{2}x+\frac{1}{2})$\\
As shown in the figure, when P is above Q: $PQ=-\frac{1}{2}x^2+2x+\frac{5}{2}-\frac{1}{2}x-\frac{1}{2}=3$\\
Solving, $x=1$ or $x=2$ (discard),\\
$\therefore \boxed{Q\left(1, 1\right)}$;\\
When P is below Q: $PQ=\frac{1}{2}x+\frac{1}{2}+\frac{1}{2}x^2-2x-\frac{5}{2}=3$,\\
Solving: $x=5$ or $x=-2$,\\
$\therefore \boxed{Q\left(5, 3\right)}$ or \boxed{Q\left(-2, -\frac{1}{2}\right)};\\
(2) When $CD$ is a diagonal of the parallelogram,\\
Draw $PE\perp CD$ at point E from point P, and draw $QF\perp CD$ at point F from point Q, then $PE=QF, DE=FC$.\\
Let $P(x, -\frac{1}{2}x^2+2x+\frac{5}{2})$, then $E(2, -\frac{1}{2}x^2+2x+\frac{5}{2})$,\\
$\therefore Q(4-x, \frac{5}{2}-\frac{1}{2}x)$, $F(2, \frac{5}{2}-\frac{1}{2}x)$,\\
From $DE=CF$, we get $\frac{9}{2}-(\frac{5}{2}-\frac{1}{2}x^2+2x+\frac{5}{2})=\frac{5}{2}-\frac{1}{2}x-\frac{3}{2}$,\\
Solving, $x=1$ or $x=2$ (discard),\\
$\therefore \boxed{Q\left(3, 2\right)}$\\
In summary: when \boxed{Q\left(1, 1\right)} or \boxed{Q\left(5, 3\right)} or \boxed{Q\left(-2, -\frac{1}{2}\right)} or \boxed{Q\left(3, 2\right)}, the quadrilateral formed by the vertices P, Q, C, and D is a parallelogram.


Question: As shown in the figure, the parabola $y=ax^2+bx+3$ intersects the x-axis at points $A$ and $B$ (with $A$ to the left of $B$) and intersects the y-axis at point $C$, with $\tan\angle CAB=3$. The hyperbola $y=\frac{k}{x}$ (with $k\neq 0$) passes through the vertex $D$ of the parabola $y=ax^2+bx+3$, where the x-coordinate of point $D$ is 1. Find the equations of the parabola and the hyperbola.

Solution:
\textbf{Solution:} Let us set $x=0$, which yields: $y=3$,\\
Therefore, the coordinates of point C are $\left(0, 3\right)$,\\
Therefore, $OC=3$.\\
Since $\tan\angle CAB=3$,\\
Therefore, $\frac{OC}{OA}=3$, that is $\frac{3}{OA}=3$,\\
Therefore, $OA=1$.\\
Therefore, the coordinates of point A are $\left(-1, 0\right)$.\\
Since the axis of symmetry for the parabola is $x=1$,\\
Therefore, the coordinates of point B are $\left(3, 0\right)$.\\
Substituting $A\left(-1, 0\right)$ and $B\left(3, 0\right)$ yields: $\left\{\begin{array}{l}a-b+3=0 \\ 9a+3b+3=0\end{array}\right.$,\\
Solving this, we find: $a=-1, b=2$,\\
Therefore, the equation of the parabola is $y=-x^{2}+2x+3$.\\
Substituting $x=1$ yields: $y=-1+2+3=4$.\\
Therefore, the coordinates of point D are $\left(1, 4\right)$.\\
Substituting $\left(1, 4\right)$ into the equation of the inverse proportion function yields: $4=\frac{k}{1}$,\\
Solving this, we find: $k=4$.\\
Therefore, the equation of the inverse proportion function is $y=\frac{4}{x}$.\\
$\therefore \boxed{y=-x^{2}+2x+3}$ and $\boxed{y=\frac{4}{x}}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+3$ intersects the x-axis at points $A$ and $B$ (with $A$ to the left of $B$), and intersects the y-axis at point $C$, where $\tan\angle CAB=3$. The hyperbola $y=\frac{k}{x}$ (with $k\ne 0$) passes through the vertex $D$ of the parabola $y=ax^2+bx+3$, with the x-coordinate of point $D$ being 1. Let point $P$ be a moving point on the parabola, located in the first quadrant. Connect points $BP$ and $CP$. Find the coordinates of point $P$ and the maximum value of the quadrilateral $ABPC$ when it reaches its maximum value.

Solution:
\textbf{Solution:} As shown in Figure 1: Connect $BC$, and draw $PE\perp AB$ through point P, intersecting $BC$ at point E.\\
$\because AB=4, OC=3$,\\
$\therefore {S}_{\triangle ABC}=\frac{1}{2}AB\times OC=\frac{1}{2}\times 4\times 3=6$.\\
Let the equation of line BC be $y=kx+b$, substituting $(3, 0)$ and $(0, 3)$ into it gives: $3k+b=0$, $b=3$,\\
solving this yields $b=3, k=-1$,\\
$\therefore$ the equation of line BC is $y=-x+3$.\\
Let the coordinates of point P be $\left(x, -{x}^{2}+2x+3\right)$, then the coordinates of point E are $\left(x, -x+3\right)$.\\
$\therefore PE=-{x}^{2}+3x$,\\
$\therefore {S}_{\triangle PBC}=\frac{1}{2}PE\cdot OB=\frac{1}{2}\times 3\times \left(-{x}^{2}+3x\right)=-\frac{3}{2}{x}^{2}+\frac{9}{2}x$.\\
$\therefore {S}_{\text{quadrilateral} ABPC}=-\frac{3}{2}{x}^{2}+\frac{9}{2}x+6=-\frac{3}{2}{\left(x-\frac{3}{2}\right)}^{2}+\frac{75}{8}$,\\
Substituting $x=\frac{3}{2}$ into the equation of the parabola yields: $y=\frac{15}{4}$,\\
$\therefore$ The coordinates of point P are $\left(\frac{3}{2}, \frac{15}{4}\right)$, ${S}_{\text{quadrilateral ABPC}\text{ maximum}}=\boxed{\frac{75}{8}}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+3$ intersects the x-axis at points $A$ and $B$ (with $A$ on the left side of $B$) and intersects the y-axis at point $C$, with $\tan\angle CAB=3$; the hyperbola $y=\frac{k}{x}$ ($k\ne 0$) passes through the vertex $D$ of the parabola $y=ax^2+bx+3$, where the x-coordinate of point $D$ is 1. If there exists a point $Q$ on this parabola and hyperbola such that $QB=QC$, what are the coordinates of point $Q$?

Solution:
\textbf{Solution:} As shown in Figure 2: Connect $BC$, and draw $OE\perp BC$ through point O with E as the foot of the perpendicular.\\
$\because QB=QC$,\\
$\therefore$ point Q lies on the perpendicular bisector of $BC$.\\
$\because OE\perp BC$, $OB=OC$,\\
$\therefore EC=BE$,\\
$\therefore OE$ is the perpendicular bisector of $BC$,\\
$\therefore$ point Q lies on $OE$,\\
$\because OE$ bisects $BC$ perpendicularly,\\
$\therefore$ the equation of line OE is $y=x$.\\
Solving $y=x$ together with $y=\frac{4}{x}$ yields $\left\{\begin{array}{l}y=x \\ y=\frac{4}{x}\end{array}\right.$,\\
which gives $\left\{\begin{array}{l}x=2 \\ y=2\end{array}\right.$ or $\left\{\begin{array}{l}x=-2 \\ y=-2\end{array}\right.$\\
$\therefore$ the coordinates of point Q are $\left(2, 2\right)$ or $\left(-2, -2\right)$,\\
Solving $y=x$ together with $y=-x^{2}+2x+3$ yields $\left\{\begin{array}{l}y=x \\ y=-x^{2}+2x+3\end{array}\right.$,\\
which gives $\left\{\begin{array}{l}x=\frac{1+\sqrt{13}}{2} \\ y=\frac{1+\sqrt{13}}{2}\end{array}\right.$ or $\left\{\begin{array}{l}x=\frac{1-\sqrt{13}}{2} \\ y=\frac{1-\sqrt{13}}{2}\end{array}\right.$,\\
$\therefore$ the coordinates of point Q are $\left(\frac{1+\sqrt{13}}{2}, \frac{1+\sqrt{13}}{2}\right)$ or $\left(\frac{1-\sqrt{13}}{2}, \frac{1-\sqrt{13}}{2}\right)$.\\
In summary, the coordinates of point Q are $\boxed{\left(2, 2\right)}$ or $\boxed{\left(-2, -2\right)}$ or $\boxed{\left(\frac{1+\sqrt{13}}{2}, \frac{1+\sqrt{13}}{2}\right)}$ or $\boxed{\left(\frac{1-\sqrt{13}}{2}, \frac{1-\sqrt{13}}{2}\right)}$.


Question: Given the diagram, it is known that the line $AB$ intersects the parabola $C: y=ax^2+2x+c$ at points $A(-1, 0)$ and $B(2, 3)$. What is the function expression of the parabola $C$?

Solution:
\textbf{Solution:} By substituting the points $(-1,0)$ and $(2,3)$ into $y=ax^{2}+2x+c$, we obtain
\[
\begin{cases}
a-2+c=0\\
4a+4+c=3
\end{cases},
\]
from which we get
\[
\begin{cases}
a=-1\\
c=3
\end{cases}.
\]
Thus, the equation of parabola $C$ is: \boxed{y=-x^{2}+2x+3}.


Question: As shown in the figure, it is known that line $AB$ intersects the parabola $C\colon y=ax^2+2x+c$ at points $A(-1, 0)$ and $B(2, 3)$. If point $M$ is a moving point on the parabola above line $AB$, and a parallelogram $MANB$ is formed with $MA$ and $MB$ as two adjacent sides, when the area of the parallelogram $MANB$ is maximized, find the area $S$ of the parallelogram $MANB$ at this time and the coordinates of point $M$?

Solution:
\textbf{Solution:} As shown in Figure 1, draw line $MH\perp x$-axis at point $H$ through point $M$, and intersect line $AB$ at point $K$.\\
Substituting the points $\left(-1,0\right)$ and $\left(2,3\right)$ into $y=kx+b$, we get
\[
\begin{cases}
-k+b=0\\
2k+b=3
\end{cases},
\]
from which we find
\[
\begin{cases}
k=1\\
b=1
\end{cases},
\]
thus $\therefore y_{AB}=x+1$.\\
Let $M$ be the point $\left(a,-a^{2}+2a+3\right)$, then $K$ is $\left(a,a+1\right)$,\\
then $MK=-a^{2}+2a+3-(a+1)=-\left(a-\frac{1}{2}\right)^{2}+\frac{9}{4}$.\\
According to the properties of quadratic functions, when $a=\frac{1}{2}$, $MK$ has the maximum length $\frac{9}{4}$.\\
$\therefore \text{The maximum area of } \triangle AMB=S_{\triangle AMK}+S_{\triangle BMK}$\\
$=\frac{1}{2}MK\cdot AH+\frac{1}{2}MK\cdot \left(x_{B}-x_{H}\right)$\\
$=\frac{1}{2}MK\cdot \left(x_{B}-x_{A}\right)$\\
$=\frac{1}{2}\times \frac{9}{4}\times 3$\\
$=\frac{27}{8}$,\\
$\therefore$ For parallelogram $MANB$ constructed with sides $MA$ and $MB$ being adjacent sides, when the area of parallelogram $MANB$ is maximized,\\
$\text{The maximum area } S_{\text{max}}=2S_{\text{max } \triangle AMB}=2\times \frac{27}{8}=\boxed{\frac{27}{4}}$, with $\boxed{M\left(\frac{1}{2}, \frac{15}{4}\right)}$;


Question: As shown in the figure, it is known that the line $AB$ intersects the parabola $C\colon y=ax^2+2x+c$ at points $A(-1, 0)$ and $B(2, 3)$. Is there a fixed point $F$ on the axis of symmetry of the parabola $C$ such that the distance from any point $P$ on the parabola $C$ to point $F$ is equal to the distance to the line $y=\frac{17}{4}$? If such a point exists, determine the coordinates of the fixed point $F$; if not, please explain why.

Solution:
\textbf{Solution:}

There exists a point $F$, \\
$\because y = -x^2 + 2x + 3 = - (x - 1)^2 + 4$, \\
$\therefore$ the axis of symmetry is the line $x = 1$, \\
when $y = 0$, $x_1 = -1$, $x_2 = 3$, \\
$\therefore$ the parabola intersects the positive $x$-axis at point $C(3, 0)$, \\
Perpendiculars to the line $y = \frac{17}{4}$ through points $B$, $C$ are drawn, with foot of perpendiculars being $N$, $H$, respectively, \\
There exists a point $F$ on the axis of symmetry of the parabola such that for any point $P$ on the parabola $C$, the distance from $P$ to point $F$ is equal to its distance to the line $y = \frac{17}{4}$. Let $F(1,t)$, connect $BF$, $CF$, \\
then $BF = BN = \frac{17}{4} - 3 = \frac{5}{4}$, $CF = CH = \frac{17}{4}$, \\
Based on the problem statement, we can form the equations: $\begin{cases}
(2 - 1)^2 + (t - 3)^2 = \left(\frac{5}{4}\right)^2 \\
(3 - 1)^2 + t^2 = \left(\frac{17}{4}\right)^2
\end{cases}$, \\
Solving, we get $t = \frac{15}{4}$, \\
$\therefore F\left(1, \boxed{\frac{15}{4}}\right)$.


Question: As shown in the figure, it is known that the line $y_1=x+m$ intersects the x-axis and y-axis at points A and B, respectively, and intersects the hyperbola $y_2=\frac{k}{x}$ (where $x<0$) at points C and D, with the coordinates of point C being $(-1, 2)$. Find the expressions for the line and the hyperbola respectively.

Solution:
\textbf{Solution:} Given that point $C(-1, 2)$ lies on the graph of $y_1=x+m$,\\
it follows that $2=-1+m$,\\
solving this gives $m=3$, hence $y_1=x+3$.\\
Given that $C(-1, 2)$ also lies on the graph of $y_2=\frac{k}{x}$ (for $x<0$),\\
it follows that $2=\frac{k}{-1}$, solving for $k$ gives $k=-2$,\\
therefore, \boxed{y_1=x+3} and \boxed{y_2=\frac{-2}{x} \left(x<0\right)}.


Question: As shown in the figure, it is known that the line $y_1=x+m$ intersects the x-axis and y-axis at points A and B, respectively, and intersects the hyperbola $y_2=\frac{k}{x}$ ($x<0$) at points C and D, where the coordinates of point C are $(-1, 2)$. Find the coordinates of point D.

Solution:
\textbf{Solution:} Solving the systems of equations simultaneously, we get
\[
\begin{cases}
y=x+3\\
y=\frac{-2}{x}
\end{cases}
\]
which yields
\[
\begin{cases}
x=-1\\
y=2
\end{cases}
\]
or
\[
\begin{cases}
x=-2\\
y=1
\end{cases}
\]
Since the coordinates of point C are $(-1, 2)$,\\
therefore, the coordinates of point D are $\boxed{(-2, 1)}$.


Question: As shown in the figure, it is known that the line $y_1 = x + m$ intersects the x-axis and y-axis at points A and B, respectively, and intersects the hyperbola $y_2 = \frac{k}{x} \, (x < 0)$ at points C and D, with the coordinates of point C being $(-1, 2)$. Using the graph, directly write out: when $x$ takes values in what range does $y_1 > y_2$?

Solution: \textbf{Solution}: It can be known from the graph that when $-2<x<-1$, $y_1>y_2$. Therefore, when $x$ takes values in the range $(-2,-1)$, $y_1>y_2$ holds. Hence, the answer is $x\in\boxed{(-2,-1)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=\frac{1}{2}x^{2}+bx+c$ passes through points $A\left(0, 2\right)$ and $B\left(1, \frac{3}{2}\right)$. What is the equation of the parabola?

Solution:
\textbf{Solution:} Let us substitute points $A\left(0, 2\right)$ and $B\left(1, \frac{3}{2}\right)$ into $y=\frac{1}{2}x^{2}+bx+c$,\\
we get: $\begin{cases}c=2\\ \frac{1}{2}+b+c=\frac{3}{2}\end{cases}$,\\
Solving these, we find: $\begin{cases}b=-1\\ c=2\end{cases}$,\\
Therefore, the equation of the parabola is $\boxed{y=\frac{1}{2}x^{2}-x+2}$;


Question: As shown in the diagram, in the Cartesian coordinate system, the parabola $y=\frac{1}{2}x^{2}+bx+c$ passes through points $A\left(0, 2\right)$ and $B\left(1, \frac{3}{2}\right)$. Given that point C is symmetric to point A with respect to the axis of symmetry of this parabola, what are the coordinates of point C?

Solution:
\textbf{Solution:} Since $y=\frac{1}{2}x^{2}-x+2=\frac{1}{2}(x-1)^{2}+\frac{3}{2}$,\\
the axis of symmetry of the parabola is the line $x=1$,\\
and since point C is symmetric to point A with respect to this axis of the parabola,\\
the coordinates of point C are $\boxed{(2, 2)}$;


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=\frac{1}{2}x^{2}+bx+c$ passes through points $A(0, 2)$ and $B(1, \frac{3}{2})$. Point $D$ is on the parabola with an x-coordinate of $4$. Denote the segment of the parabola between points $A$ and $D$ (including points $A$ and $D$) as graph $G$. If graph $G$ is translated downwards by $t$ units ($t>0$) and intersects line $BC$ at only one point, find the range of values for $t$.

Solution:
\textbf{Solution:} Let the equation of line BC be $y=mx+n$,\\
substituting $B\left(1, \frac{3}{2}\right)$ and $C\left(2, 2\right)$ into $y=mx+n$,\\
we get: $\left\{\begin{array}{l}
m+n=\frac{3}{2}\\
2m+n=2
\end{array}\right.$,\\
solving this, we find: $\left\{\begin{array}{l}
m=\frac{1}{2}\\
n=1
\end{array}\right.$,\\
$\therefore$ the equation of line BC is $y=\frac{1}{2}x+1$,\\
when $x=0$, $y=\frac{1}{2}x+1=1$,\\
$\therefore$ when graph G is translated downwards by 1 unit, point A lies on line BC,\\
when $x=4$, $y=\frac{1}{2}x+1=3$,\\
since for $x=4$, $y=\frac{1}{2}x^2-x+2=6$,\\
$\therefore$ when graph G is translated downwards by 3 units, point D lies on line BC,\\
$\therefore$ when $1<t\le 3$, after graph G is translated downwards by t ($t>0$) units, it intersects line BC at only one point,\\
the desired range of values for $t$ is $t\in\boxed{(1,3]}$.


Question: In a Cartesian coordinate system, point \(O\) is the origin. The parabola \(y=ax^2+bx+3\) passes through point \(B(-3, 0)\), point \(C(1, 0)\), and intersects the y-axis at point \(A\). What are the values of \(a\) and \(b\)?

Solution:
Solution: Since the parabola $y=ax^2+bx+3$ passes through points $B(-3, 0)$ and $C(1, 0)$,
\[
\therefore
\begin{cases}
9a-3b+3=0 \\
a+b+3=0
\end{cases}
\]
\[
\therefore
\begin{cases}
a=-1 \\
b=-2
\end{cases}
\]
Hence, $a=\boxed{-1}$, $b=\boxed{-2}$.


Question: In the Cartesian coordinate system, point O is the origin. The parabola $y = ax^2 + bx + 3$ passes through points $B(-3, 0)$ and $C(1, 0)$, and intersects the y-axis at point A. As shown in Figure 1, point P lies on the parabola. A line through point P parallel to the y-axis intersects the line segment $AB$ at point D. Denoting the x-coordinate of point P as t, and the length of segment $PD$ as d, find the functional expression of d in terms of t (it is not required to state the domain of the independent variable t).

Solution:
\textbf{Solution:}

\[
\begin{aligned}
&\because a=-1, b=-2 \\
&\therefore y=x^2-2x+3 \\
&\text{when} x=0, y=3 \\
&\therefore A(0, 3) \\
&\text{Let the line} AB: y=kx+b' \\
&\because B(-3, 0) \\
&\therefore
\left\{
\begin{array}{l}
b'=3 \\
-3k+b'=0
\end{array}
\right. \\
&\therefore
\left\{
\begin{array}{l}
k=1 \\
b'=3
\end{array}
\right. \\
&\therefore \text{The equation of line} AB \text{is} y=x+3 \\
&\because DP \parallel \text{the y-axis}, \\
&\therefore P, O \text{have the same x-coordinate} t \\
&\therefore p(t, -t^2-2t+3), O(t, t+3) \\
&\therefore d=-t^2-2t+3-(t+3) \\
&=-t^2-3t \\
&\therefore d=\boxed{-t^2-3t}
\end{aligned}
\]


Question: In the Cartesian coordinate system, point O is the origin. The parabola $y=a{x}^{2}+bx+3$ passes through point $B(-3, 0)$ and point $C(1, 0)$, and intersects the y-axis at point A. As shown in the figure, under the given conditions, connect $CD$, draw $AE\perp CD$ at point F where $AE=CD$, draw $EG\perp AB$ at point G, draw $GH\parallel DP$ intersecting the x-axis at point H, and connect $FH$ and $FO$. If $\angle HFO=90^\circ$, find the coordinates of point P.

Solution:
\textbf{Solution:}

(1) Construct $EL \perp OA$ at L, extend $PD$ to intersect the x-axis at T.\\
$\therefore \angle AFC = \angle AOC = 90^\circ$\\
$\because \angle 1 = \angle 2$\\
$\therefore \angle EAL = \angle DCT$\\
$\because AE = CD$\\
$\therefore \triangle AEL \cong \triangle CDT$\\
$\therefore EL = DT$, $AL = CT$\\
$\because D(t, t+3)$\\
$\therefore EL = t+3$, $AL = CT = 1-t$\\
$\because OL = OA - AL = 3 - (1 - t) = 2 + t$\\
$\therefore E(-t-3, 2+t)$\\
(2) Construct $EW \parallel y$-axis, intersecting $AB$ at W, and $OB$ at V, extend $GE$ to intersect the x-axis at R.\\
Thus, $W(-t-3, -t)$, $V(-t-3, 0)$, $WE = -2t-2$, $WV = -t$\\
From the coordinates of points A and B, we get: $OA = OB$\\
$\therefore \angle ABO = 45^\circ$,\\
$\therefore \angle BWV = \angle WEG = \angle GRB = \angle ABO = 45^\circ$\\
$\therefore WG = GE$, $WV = BV$, $GR = BG$\\
$\therefore$ By Pythagorean theorem, $GW = \frac{\sqrt{2}}{2}WE = -\sqrt{2}(t+1)$, $BW = \sqrt{2}WV = -\sqrt{2}t$,\\
$\therefore BG = BW - GW = \sqrt{2}$\\
By Pythagorean theorem, $BR = \sqrt{2}BG = 2$\\
Also, point H is the midpoint of $BR$, $OB = 3$\\
$\therefore G(-2, 1)$, $R(-1, 0)$, $H(-2, 0)$\\
(3) Extend $GE$ to intersect the x-axis at R, connect $DH$ and $OE$.\\
$\because D(t, t+3), G(-2, 1), E(-t-3, 2+t), R(-1, 0)$\\
$\therefore DG = ER$, $OR = GH = 1$, $\angle DGH = \angle FRO = 135^\circ$\\
$\therefore \triangle DGH \cong \triangle ERO$\\
$\therefore DH = EO$, $\angle GDH = \angle REO$\\
Connect $AC$\\
$\because \angle HFO = \angle DFE = 90^\circ$,\\
$\therefore \angle DFH = \angle EFO$\\
$\because EG \perp AB$, $AF \perp CD$, $\angle DAF = \angle EAG$\\
$\therefore \angle ADF = \angle AEG$,\\
$\therefore \angle GDF = \angle FER$\\
$\because \angle GDH = \angle REO$\\
$\therefore \angle FEO = \angle FDH$,\\
$\because DH = EO$\\
$\therefore \triangle DHF \cong \triangle EOF$\\
$\therefore DF = EF$\\
$\because AE = CD$\\
$\therefore AF = CF$\\
$\therefore \angle FAC = 45^\circ$\\
In right triangle $AOC$, $\tan\angle OAC = \frac{1}{3}$\\
$\because \angle BAO = \angle FAC = 45^\circ$\\
$\therefore \angle GAE = \angle OAC$\\
$\therefore \tan\angle GAE = \frac{EG}{AG} = \frac{1}{3}$\\
That is, $\frac{\sqrt{2}}{2}(-2t-2) = \frac{2}{3}\sqrt{2}$\\
Solving for $t$, we get: $t = -\frac{5}{3}$\\
$\therefore \boxed{P\left(-\frac{5}{3}, \frac{32}{9}\right)}$


Question: In Figure 1, within rectangle $ABCD$, $AC$ and $BD$ intersect at point $O$, and $E$ is a point on $AD$, with $CE$ and $BD$ intersecting at point $F$. Given that $AE = CD$, and $BD \perp CE$, (1) find the degree measure of $\angle DEC$. (2) As shown in Figure 2, connect $AF$. When $BC = 3$, find the length of $AF$.

Solution:
\textbf{Solution:}

(1)\\
Since AE = CE,\\
it follows that $\angle$EAC = $\angle$ECA,\\
Since AD $\parallel$ BC,\\
it follows that $\angle$EAC = $\angle$ACB,\\
Since OB = OC,\\
it follows that $\angle$ACB = $\angle$DBC,\\
Hence, $\angle$ACB = $\angle$DBC = $\angle$ECA,\\
Since BD $\perp$ CE,\\
it follows that $\angle$BFC = 90$^\circ$,\\
Therefore, $\angle$ECA + $\angle$ACB + $\angle$DBC = 90$^\circ$,\\
Therefore, $\angle$ECA = $\angle$ACB = $\angle$DBC = 30$^\circ$,\\
Therefore, $\angle$DEC = $\angle$ECB = \boxed{60^\circ},\\

(2) Draw FH $\perp$ AD at point H,\\
In $\triangle$BCD, $\angle$DBC = 30$^\circ$, BC = 3,\\
it follows that BD = 2$\sqrt{3}$,\\
In $\triangle$BFC, $\angle$FBC = 30$^\circ$, BC = 3,\\
it follows that BF = $\frac{3}{2}\sqrt{3}$,\\
Therefore, DF = BD - BF = $\frac{\sqrt{3}}{2}$,\\
In $\triangle$DHF, $\angle$FDH = $\angle$DBC = 30$^\circ$,\\
it follows that FH = $\frac{1}{2}$DF = $\frac{\sqrt{3}}{4}$,\\
Therefore, HD = $\sqrt{3}$HF = $\frac{3}{4}$,\\
Therefore, AH = AD - DH = 3 - $\frac{3}{4}$ = $\frac{9}{4}$,\\
Therefore, AF = $\sqrt{AH^2 + HF^2} = \sqrt{\left(\frac{9}{4}\right)^2 + \left(\frac{\sqrt{3}}{4}\right)^2} = \boxed{\frac{\sqrt{21}}{2}}$


Question: As shown in Figure 1, within rectangle $ABCD$, $AC$ and $BD$ intersect at point $O$, and point $E$ is on $AD$, with $CE$ and $BD$ intersecting at point $F$. Let $\frac{DE}{AD}=k (0<k<1)$, denote the area of $\triangle CBF$ as $S_{1}$, and the area of quadrilateral $ABFE$ as $S_{2}$. Find the maximum value of $\frac{S_{2}}{S_{1}}$.

Solution:
:\textbf{Solution:}

Let $S_{\triangle DEF}=x$,

since AD$\parallel$BC,

hence $\triangle DEF\sim \triangle CFB$,

thus $\frac{S_{\triangle DEF}}{S_{\triangle CFB}}=\left(\frac{DE}{BC}\right)^2=k^2$,

therefore $S_1=\frac{x}{k^2}$,

since $\frac{DF}{BF}=\frac{DE}{BC}=\frac{DE}{AD}=k$,

hence $\frac{DF}{BD}=\frac{k}{k+1}$,

since FH$\perp$AD,

therefore $\angle FHD=90^\circ$,

thus $\angle FHD=\angle BAD$,

moreover,

since $\angle FDH=\angle BDA$,

hence $\triangle FDH\sim \triangle BDA$,

thus $\frac{FH}{AB}=\frac{DF}{BD}=\frac{k}{k+1}$,

therefore $\frac{S_{\triangle DEF}}{S_{\triangle DAB}}=\frac{\frac{1}{2}DE\cdot FH}{\frac{1}{2}AD\cdot AB}=\frac{DE}{AD}\cdot \frac{FH}{AB}=\frac{k^2}{k+1}$,

therefore $S_{\triangle DAB}=\frac{(k+1)x}{k^2}$,

thus $S_2=S_{\triangle DAB}-S_{\triangle DEF}=\frac{(k+1-k^2)x}{k^2}$,

hence $\frac{S_2}{S_1}=k+1-k^2=-\left(k-\frac{1}{2}\right)^2+\frac{5}{4}$,

therefore, when $k=\frac{1}{2}$, the maximum value of $\frac{S_2}{S_1}$ is $\boxed{\frac{5}{4}}$.


Question: As shown in Figure 1, it is known that point $M$ is the vertex of the parabola $y=-\left(x-b\right)^2+4b+1$, and the line $y=mx+5$ intersects with the positive x-axis and y-axis at points $A$ and $B$ respectively. If the parabolic graph also passes through points $A$ and $B$, and if $mx+5>-\left(x-b\right)^2+4b+1$, according to the graph, write down the range of values for $x$.

Solution:
\textbf{Solution:} As shown in Figure 1,\\
the line $y=mx+5$ intersects the y-axis at point $B$,\\
$\therefore$ the coordinates of point $B$ are $(0,5)$,\\
Since $B$ lies on the parabola,\\
$\therefore 5 = -\left(0-b\right)^2 + 4b + 1 = 5$,\\
solving for $b$ yields $b=2$,\\
The analytical expression for the quadratic function is $y=-\left(x-2\right)^2+9$,\\
When $y=0$, $-\left(x-2\right)^2+9=0$,\\
solving for $x$ yields $x_1=5$, $x_2=-1$,\\
$\therefore A(5,0)$,\\
From the graph, we find that when $mx+5 > -\left(x-b\right)^2 + 4b + 1$, the range of $x$ is $x<0$ or $x>5$;\\
Therefore, the range of $x$ is $\boxed{x<0}$ or $\boxed{x>5}$.


Question: Given that point $M$ is the vertex of the quadratic function $y=-(x-b)^2+4b+1$, and the line $y=mx+5$ intersects the positive half of the $x$-axis and the $y$-axis at points $A$ and $B$ respectively. As shown in Figure 2, the coordinates of point $A$ are $(5,0)$, and point $M$ is inside $\triangle AOB$. If points $C\left(\frac{1}{4},y_1\right)$ and $D\left(\frac{3}{4},y_2\right)$ are both on the graph of the quadratic function, compare the sizes of $y_1$ and $y_2$.

Solution:
\textbf{Solution:}

Substituting $A(5,0)$ into $y=mx+5$ yields, \\
$0=5m+5$, \\
solving for $m$ gives $m=-1$, \\
$\therefore y=-x+5$, \\
$\because M(b,4b+1)$ is inside $\triangle AOB$, \\
$\therefore \left\{\begin{array}{l}0<b<5 \\ 0<4b+1<-b+5\end{array}\right.$, \\
solving yields $0<b<\frac{4}{5}$, \\
When points C and D are symmetrical about the axis of symmetry, $b=\frac{\frac{1}{4}+\frac{3}{4}}{2}=\frac{1}{2}$, \\
$\therefore$ for $0<b<\frac{1}{2}$, \boxed{y_1>y_2}, \\
for $b=\frac{1}{2}$, \boxed{y_1=y_2}, \\
and for $\frac{1}{2}<b<\frac{4}{5}$, \boxed{y_1<y_2}.


Question: Given the quadratic function $y=x^{2}-4x+3$, let its intersection points with the $x$-axis be A and B (with point A to the left of point B), and its intersection point with the $y$-axis be C. Find the coordinates of points A, B, and C.

Solution:
\textbf{Solution:} Since $y=x^{2}-4x+3=(x-1)(x-3)=0$, \\
we find $x_{1}=1, x_{2}=3$, \\
thus, the intersections of the quadratic function $y=x^{2}-4x+3$ and the x-axis are $\boxed{A(1,0)}$ and $\boxed{B(3,0)}$.\\
Setting $x=0$, we have $y=3$, which means the coordinates of point $C$ are $\boxed{(0,3)}$.


Question: As shown in the figure, given the quadratic function $y = x^{2} - 4x + 3$, let the points where its graph intersects the x-axis be A and B (with point A to the left of point B), and let the point where it intersects the y-axis be C. Find: What is the area of $\triangle ABC$?

Solution:
\textbf{Solution}: From (1), we know $A(1,0)$, $B(3,0)$, $C(0,3)$,\\
thus, the area of $\triangle ABC = \frac{1}{2} \times 2 \times 3 = \boxed{3}$.


Question: As shown in the figure, a goalkeeper at point $O$ kicks a high ball, which starts its flight from point $A$, 1 meter above the ground ($A$ is on the y-axis). Player B notices the ball directly above his head at its highest point $M$, 6 meters away from $O$ and approximately 4 meters above the ground. After the ball lands, it bounces back. Experimental calculations show that the parabolic trajectory of the ball after bouncing on the grass has the same shape as its original trajectory, but its maximum height is reduced to half of the original maximum height. Determine the expression of the parabola for the ball's trajectory from its launch to its first landing.

Solution:
\textbf{Solution:} As shown in the figure, assume the expression of the parabola when it first hits the ground is $y=a(x-6)^2+4$.\\
Given: when $x=0$, $y=1$.\\
Thus, $1=36a+4$, $\therefore a=-\frac{1}{12}$.\\
$\therefore$ The expression is $\boxed{y=-\frac{1}{12}(x-6)^2+4}$ (or $\boxed{y=-\frac{1}{12}x^2+x+1}$).


Question: As shown in the figure, a goalkeeper kicks a high ball from point $O$ on the soccer field. The ball takes off from point $A$, which is 1 meter above the ground and located on the $y$-axis, and reaches its highest point $M$ directly above player B, who is 6 meters away from point $O$, at approximately 4 meters above the ground. After the ball lands, it bounces back into the air. Experimental measurements have shown that the trajectory of the ball after bouncing forms a parabola that is identical in shape to its initial trajectory, but its maximum height is reduced to half of the original maximum height. How many meters away from the goalkeeper is the first landing point $C$ of the ball?

Solution:
Solution: Let $y=0$, then $-\frac{1}{12}(x-6)^2+4=0$. \\
$\therefore (x-6)^2=48$. The solutions are ${x}_1=4\sqrt{3}+6\approx 13$ and ${x}_2=-4\sqrt{3}+6<0$ (discard the latter). \\
$\therefore$ The first landing point $C$ of the football is approximately $\boxed{13}$ meters away from the goalkeeper.


Question: As shown in the diagram, the goalkeeper kicks a high ball from point $O$, and the ball starts its flight from point $A$, 1 meter above the ground ($A$ is on the $y$-axis). The player Yi discovers that the ball reaches its highest point, $M$, right above his head at a height of approximately 4 meters from the ground, and this point is 6 meters away from $O$ at point $B$. After landing, the ball bounces again. According to experimental measurements, the parabolic trajectory of the ball after bouncing on the lawn retains the same shape as the original trajectory, but its maximum height is reduced to half of the original height. To reach the second landing point $D$, how much further does the player Yi need to run?

Solution:
\textbf{Solution:} As shown in the diagram, the distance after the soccer ball bounces for the second time is $CD$.\\
According to the problem: $CD=EF$ (which means the parabola $AEMFC$ is translated downward by $2$ units).\\
$\therefore 2=-\frac{1}{12}(x-6)^{2}+4$ solving this gives $x_{1}=6-2\sqrt{6}$, $x_{2}=6+2\sqrt{6}$.\\
$\therefore CD=|x_{1}-x_{2}|=4\sqrt{6}\approx 10$.\\
$\therefore BD=13-6+10=17$ (meters).\\
Answer: He should run another $\boxed{17}$ meters forward.


Question: As shown in the figure, there is a certain mass of carbon dioxide in a sealed container. When the volume $V$ (in $m^3$) of the container changes, the density $\rho$ (in $kg/m^3$) of the gas changes accordingly. It is known that the density $\rho$ and the volume $V$ have an inverse proportional function relationship. The graph of this relationship is shown in the figure. When $V=5m^3$, $\rho=1.98kg/m^3$. Find the analytic formula of the density $\rho$ as a function of the volume $V$?

Solution:
\textbf{Solution:} Since density $\rho$ is inversely proportional to volume $V$,\\
we set $\rho=\frac{k}{V}\left(V>0\right)$.\\
Given $V=5\mathrm{m}^3$ and $\rho=1.98\mathrm{kg/m}^3$,\\
hence $1.98=\frac{k}{5}$,\\
thus $k=1.98\times 5=9.9$.\\
Therefore, the formula relating density $\rho$ to volume $V$ is: \boxed{\rho=\frac{9.9}{V}\left(V>0\right)}


Question: As shown in the diagram, a sealed container holds a certain mass of carbon dioxide. When the volume $V$ (in cubic meters, $m^{3}$) of the container changes, the density $\rho$ (in kilograms per cubic meter, $kg/m^{3}$) of the gas changes accordingly. It is known that the relationship between density $\rho$ and volume $V$ is an inverse proportion, as depicted in the graph. When $V = 5m^{3}$, $\rho = 1.98kg/m^{3}$. For $3 \leq V \leq 9$, determine the range of the carbon dioxide density $\rho$.

Solution:
\textbf{Solution:} By observing the function graph, it is known that $\rho$ decreases with the increase of $V$,\\
when $V=3\,m^{3}$, $\rho=\frac{9.9}{3}=3.3\,kg/m^{3}$,\\
when $V=9\,m^{3}$, $\rho=\frac{9.9}{9}=1.1\,kg/m^{3}$,\\
$\therefore$ when $3\le V\le 9$, $1.1\le \rho \le 3.3\,(kg/m^{3})$\\
That is, the range of change in the density of carbon dioxide $\rho$ is $\boxed{1.1\le \rho \le 3.3\,(kg/m^{3})}$.


Question: It is known that the parabola $y=\frac{1}{4}x^{2}+1$ has the following property: the distance from any point on the parabola to the fixed point $F(0, 2)$ is equal to its distance to the x-axis, as illustrated in the figure. The point $M$ has coordinates $(\sqrt{3}, 3)$. Let $P$ be a moving point on the parabola $y=\frac{1}{4}x^{2}+1$. When the area of $\triangle POF$ is 4, find the coordinates of point $P$?

Solution:
\textbf{Solution:} Let the coordinates of point $P$ be $\left(x, \frac{1}{4}x^{2}+1\right)$,

since the coordinates of point $F$ are $\left(0, 2\right)$,

therefore $OF=2$,

therefore when the area of $\triangle POF$ is 4, $\frac{1}{2}\times 2\times |x|=4$,

solving this, we get: $x=\pm 4$,

therefore $y=\frac{1}{4}\times (\pm 4)^{2}+1=5$,

hence, the coordinates of point $P$ are $\boxed{\left(-4, 5\right)}$ or $\boxed{\left(4, 5\right)}$.


Question: It is known that the parabola $y=\frac{1}{4}x^{2}+1$ has the following property: for any point on the parabola, its distance to the fixed point $F(0, 2)$ is equal to its distance to the x-axis, as shown in the figure. The coordinate of point $M$ is $(\sqrt{3}, 3)$, and $P$ is a moving point on the parabola $y=\frac{1}{4}x^{2}+1$. What is the minimum perimeter of $\triangle PMF$?

Solution:
\textbf{Solution:} Draw $ME\perp x$ axis at point $E$, and $ME$ intersects the parabola at point ${P}^{\prime}$.\\
Since the distance from any point on the parabola to the fixed point $F(0, 2)$ is equal to its distance to the x-axis,\\
it follows that ${P}^{\prime}F={P}^{\prime}E$,\\
and since $MF$ is constant,\\
it follows that when point $P$ moves to point ${P}^{\prime}$, the perimeter of $\triangle PMF$ takes its minimum value,\\
since $F(0, 2)$, $M(\sqrt{3}, 3)$,\\
it follows that $ME=3$, $FM=\sqrt{(\sqrt{3}-0)^{2}+(3-2)^{2}}=2$,\\
thus $MP^{\prime}+P^{\prime}F+MF=MP^{\prime}+P^{\prime}E+MF=ME+MF=3+2=\boxed{5}$,\\
therefore, the minimum perimeter of $\triangle PMF$ is $\boxed{5}$.


Question: As shown in Figure 1, the nozzle B of a certain irrigation equipment is $1.25m$ above the ground. The parabolic water flow reaches its maximum height of $2.25m$ from the ground at a distance of $1m$ from the bottom of the nozzle A, as shown in Figure 2, where a Cartesian coordinate system is established. What is the quadratic function equation corresponding to this parabolic water flow?

Solution:
\textbf{Solution:}

Since the parabolic water flow corresponds to a quadratic function graph passing through the origin,

we assume the quadratic function relationship of the parabolic water flow to be $y=ax^{2}$;

According to the problem statement, the distance from point B to the x-axis is $1m$, and the distance from point B to the y-axis is $2.25-1.25=1m$,

therefore, the coordinates of point B are $(-1,-1)$,

substituting $B(-1,-1)$ into $y=ax^{2}$, we get $-1=a(-1)^{2}$,

solving this gives $a=-1$,

therefore, the quadratic function relationship of the parabolic water flow is $\boxed{y=-x^{2}}$;


Question: As shown in Figure 1, the nozzle B of an irrigation equipment is $1.25m$ above the ground. The parabolic water flow emitted reaches its maximum height of $2.25m$ above the ground at a distance of $1m$ from the bottom of the nozzle A, as illustrated in Figure 2 by establishing a Cartesian coordinate system. Find the horizontal distance between the nozzle and the point where the water flow touches the ground.

Solution:
\textbf{Solution:} Let $y=-2.25$, i.e., $-{x}^{2}=-2.25$,\\
Solving gives: $x_{1}=1.5$, $x_{2}=-1.5$ (discard),\\
$1+1.5=2.5m$,\\
The horizontal distance between the nozzle and the point where the stream of water hits the ground is $\boxed{2.5m}$.


Question: As shown in the figure, it is known that the parabola $y = -x^2 + bx + c$ passes through points A$(-1, 0)$ and B$(3, 0)$. What is the analytic equation of the parabola?

Solution:
\textbf{Solution:} Let's substitute points A ($-1$, $0$) and B ($3$, $0$) into $y=-x^2+bx+c$,\\
$\therefore \begin{cases}0=-1^2-b+c\\ 0=-9+3b+c\end{cases}$,\\
Solving, we get $\begin{cases}b=2\\ c=3\end{cases}$,\\
$\therefore$ the equation of the parabola is: $y=-x^2+2x+3=\boxed{-x^2+2x+3}$.


Question: As shown in the figure, it is known that the parabola $y=-x^2+bx+c$ passes through points $A(-1,0)$ and $B(3,0)$. When $0<x<3$, directly write out the range of $y$.

Solution: \textbf{Solution:} When $0<x<3$, the range of $y$ is $0<y\le \boxed{4}$.


Question: As shown in the figure, it is known that the parabola $y=-x^2+bx+c$ passes through points $A(-1,0)$ and $B(3,0)$. Let point $P$ be a point on the parabola. If the area of $\triangle PAB$ is $10$, find the coordinates of point $P$ at this time.

Solution:
\textbf{Solution:} Let $P(x, y)$,\\
thus, the height of $\triangle PAB$ is $|y|$,\\
since $A(-1, 0)$, $B(3, 0)$,\\
thus, $AB=4$,\\
therefore, the area of $\triangle ABP$ is $\frac{1}{2} \times 4 \times |y| = 10$,\\
solving gives $y=\pm 5$,\\
when $y=5$,\\
$5=-x^2+2x+3$,\\
no solution for this equation,\\
when $y=-5$,\\
$-5=-x^2+2x+3$,\\
solving gives $x_1=4, x_2=-2$,\\
thus, the coordinates of $P$ are $\boxed{P(4, -5)}$ or $\boxed{P(-2, -5)}$.


Question: As shown in the figure, it is known that the graph of the inverse proportion function $y_{1}=\frac{k}{x}$ intersects with the graph of the linear function $y_{2}=x+b$ at point $A(1, 4)$ and point $B(-4, n)$. What is the value of $k$ and the area of $\triangle ABO$?

Solution:
\textbf{Solution:}
Given point $A(1,4)$ lies on the graph of the inverse proportion function $y_1=\frac{k}{x}$,\\
$\therefore k=1\times 4=4$,\\
$\therefore$ the expression for the inverse proportion function is $y_1=\frac{4}{x}$,\\
$\because$ point $B(-4,-1)$ also lies on the graph of the inverse proportion function $y_1=\frac{4}{x}$,\\
$\therefore n=4-4=-1$, that is, $B(-4,-1)$,\\
Substituting point $A(1,4)$ into the linear function $y_2=x+b$,\\
we get $1+b=4$, solving for $b$ gives $b=3$,\\
$\therefore$ the expression for the linear function is $y_2=x+3$;\\
Thus, the analytical expression for the inverse proportion function is $y_1=\frac{4}{x}$, and the linear function is found to be $y_2=x+3$;\\
Let the intersection of the line with the x-axis be C,\\
In $y_2=x+3$, when $y=0$, we get $x=-3$,\\
$\therefore$ the intersection of line $y_2=x+3$ and the x-axis is $C(-3,0)$,\\
$\because$ the segment $OC$ divides $\triangle AOB$ into $\triangle AOC$ and $\triangle BOC$,\\
$\therefore S_{\triangle AOB}=S_{\triangle AOC}+S_{\triangle BOC}=\frac{1}{2}\times 3\times 4+\frac{1}{2}\times 3\times 1=\boxed{7.5}$;


Question: As shown in the figure, it is known that the graph of the inverse proportion function \(y_{1}=\frac{k}{x}\) and the graph of the linear function \(y_{2}=x+b\) intersect at point \(A(1, 4)\) and point \(B(-4, n)\). Directly write out the range of values of the independent variable \(x\) when \(y_{2}>y_{1}\).

Solution:
\textbf{Solution:} When $-4 < x < 0$ or $x > 1$, we have $y_{2} > y_{1}$.

We directly express the range of the independent variable $x$ as \boxed{x \in (-4, 0) \cup (1, +\infty)}.