Question: As shown in the figure, the parabola $y=ax^2+bx+6$ passes through points $A(-2, 0)$ and $B(4, 0)$, and intersects the $y$-axis at point $C$. Point $D$ is a moving point on the parabola with the $x$-coordinate denoted as $m(1<m<4)$. Connect $AC$, $BC$, $DB$, and $DC$. When $m=2$, if point $M$ is a moving point on the $x$-axis, and point $N$ is a moving point on the parabola, determine whether there exists such a point $M$ that the quadrilateral with vertices $B$, $D$, $M$, and $N$ forms a parallelogram. If it exists, please write down the coordinates of point $M$ directly; if it does not exist, please explain the reason.

Solution:
\textbf{Solution:} When $m=2$, the point $D(2, 6)$, assume point $M(x, 0)$, point $N(t, n)$, then $n=-\frac{3}{4}t^2+\frac{3}{2}t+6$. When $BD$ is a side, moving point $B$ 1 unit to the left and 6 units up gets point $D$, similarly moving point $M(N)$ 1 unit to the left and 6 units up gets point $N(M)$, hence
\[
\left\{
\begin{array}{l}
x-1=t \\
0+6=n
\end{array}
\right.
\]
or
\[
\left\{
\begin{array}{l}
x+1=t \\
0-6=n
\end{array}
\right.
\]
Solving this system gives:
\[
\left\{
\begin{array}{l}
x=3 \\
t=2 \\
n=6
\end{array}
\right.
\]
or
\[
\left\{
\begin{array}{l}
x=\frac{\sqrt{33}-1}{2} \\
t=\frac{1+\sqrt{33}}{2} \\
n=-6
\end{array}
\right.
\]
or
\[
\left\{
\begin{array}{l}
x=\frac{-\sqrt{33}-1}{2} \\
t=\frac{1-\sqrt{33}}{2} \\
n=-6
\end{array}
\right.
\]
(Values that do not meet the premise have been omitted); hence the coordinates of point $M$ are $\left(3, 0\right)$ or $\left(\frac{\sqrt{33}-1}{2}, 0\right)$ or $\left(\frac{-\sqrt{33}-1}{2}, 0\right)$.

When $BD$ is a diagonal, using the midpoint formula we get:
\[
\left\{
\begin{array}{l}
\frac{1}{2}(3+4)=\frac{1}{2}(x+m) \\
\frac{1}{2}(6+0)=\frac{1}{2}(n+0)
\end{array}
\right.
\]
Solving this system gives
\[
\left\{
\begin{array}{l}
m=3 \\
n=6 \\
x=4
\end{array}
\right.
\]
Therefore, the coordinates of point $M$ are $\left(4, 0\right)$.

In conclusion, the coordinates of point $M$ are $\boxed{\left(3, 0\right)}$ or $\boxed{\left(\frac{\sqrt{33}-1}{2}, 0\right)}$ or $\boxed{\left(\frac{-\sqrt{33}-1}{2}, 0\right)}$ or $\boxed{\left(4, 0\right)}$.


Question: In a Cartesian coordinate system, consider the inverse proportion function $y_{1}=\frac{k_{1}}{x}\left(k_{1}\ne 0\right)$ and the linear function $y_{2}=k_{2}x+b$ $(k_{2}\neq 0)$, whose graphs both pass through points $A$ and $B$, with the coordinates of point $A$ being $(1,m)$ and those of point $B$ being $(-2,-2)$. Determine the value of $m$ and the expression for the linear function $y_{2}$.

Solution:
\textbf{Solution:} Given that the inverse proportion function $y_{1}=\frac{k_{1}}{x}$ $(k_{1}\ne 0)$ and the linear function $y_{2}=k_{2}x+b$ $(k_{2}\neq 0)$ both pass through points A and B, with point A having coordinates $(1,m)$, and point B having coordinates $(-2,-2)$.\\
Thus, $k_{1}=-2\times(-2)=m$. Solving for this, we get $k_{1}=m=\boxed{4}$\\
Therefore, the expression for the inverse proportion function is $y_{1}=\frac{4}{x}$, with point A being $(1,4)$;\\
Therefore, $\left\{\begin{array}{l}k_{2}+b=4\\ -2k_{2}+b=-2\end{array}\right.$ Solving this, we get $\left\{\begin{array}{l}k_{2}=2\\ b=2\end{array}\right.$\\
Therefore, the expression for the linear function is $y_{2}=\boxed{2x+2}$.


Question: In the Cartesian coordinate system, suppose the inverse proportion function $y_{1}=\frac{k_{1}}{x}\left(k_{1}\ne 0\right)$ and the linear function $y_{2}=k_{2}x+b$ ($k_{2}\neq 0$) have graphs passing through points $A$ and $B$, where the coordinate of point $A$ is $(1, m)$ and the coordinate of point $B$ is $(-2, -2)$. When $y_{1}>y_{2}$, directly write the range of $x$.

Solution:
\textbf{Solution:} From the given, we have $x > -2$ or $0 < x < 1$,

$\therefore$ the value range of $x$ is \boxed{(-2, \infty) \cup (0, 1)}.


Question: In a Cartesian coordinate system, suppose the inverse proportional function $y_{1}=\frac{k_{1}}{x}$ ($k_{1}\ne 0$) and the linear function $y_{2}=k_{2}x+b$ ($k_{2}\neq 0$) both pass through points $A$ and $B$, with coordinates of point $A$ being $(1, m)$ and coordinates of point $B$ being $(-2, -2)$. If the graph of the function $y_{2}$ is translated downwards by $n$ ($n>0$) units, and intersects with the graph of the function $y_{1}$ at points $(p_{1}, q_{1})$ and $(p_{2}, q_{2})$, given that $p_{1}=-1$, find the values of $n$ and $p_{2}\times q_{2}$ at this time.

Solution: \textbf{Solution:} The equation of the function after translating the graph of $y_{2}$ downwards by $n$ ($n>0$) units is $y=2x+2-n$; When $p_{1}=-1$ (at point $(-1, q_{1})$), since translating the graph of $y_{2}$ downwards by $n$ ($n>0$) units intersects with the graph of $y_{1}$ at points $(-1, q_{1})$ and $(p_{2}, q_{2})$, hence $-1q_{1}=4$, $p_{2}\times q_{2}=4$, solving these we find $q_{1}=-4$, thus point $(-1, -4)$, hence $-2+2-n=-4$; solving for this gives $n=\boxed{4}$; therefore, at this instance $n=4$, $p_{2}\times q_{2}=\boxed{4}$.


Question:
As shown in the figure, the vertices $A$ and $C$ of the rectangle $OABC$ are on the positive semi-axes of the $x$ and $y$ axes, respectively. Point $F$ is a moving point on side $BC$ (not coinciding with points $B$, $C$). The inverse proportion function $y=\frac{k}{x}$ ($x>0$) passing through point $F$ intersects side $AB$ at point $E(8, m)$, with $AB=4$. If $BE=3AE$,
- (1) Find the expression of the inverse proportion function;
- (2) Fold rectangle $OABC$ such that point $O$ coincides with point $F$. The fold intersects the $x$ and $y$ axes at points $H$ and $G$, respectively. Find the length of the line segment $OG$.

Solution:
\textbf{Solution:}

(1) Since $BE=3AE$ and $AB=4$,\\
then $AE=1$ and $BE=3$,\\
thus $E(8,1)$,\\
thus $k=8\times 1=8$,\\
therefore, the expression for the inverse proportion function is $y=\frac{8}{x}$;\\
(2) When $y=4$, then $x=2$,\\
thus $F(2,4)$,\\
thus $CF=2$,\\
let $OG=x$, then $CG=4-x$ and $FG=x$,\\
by the Pythagorean theorem, we get,\\
$(4-x)^2+2^2=x^2$,\\
solving for $x$, we find $x=\frac{5}{2}$,\\
thus $OG=\boxed{\frac{5}{2}}$;


Question: The vertex $A$ and $C$ of the rectangle $OABC$ are respectively on the positive half-axis of $x$ and $y$. Point $F$ is a moving point on side $BC$ (not coinciding with points $B$ and $C$). The inverse proportion function $y=\frac{k}{x}$ $(x>0)$ passing through point $F$ intersects with side $AB$ at point $E(8,m)$, where $AB=4$. As shown in the figure, connect $OF$ and $EF$. Please use an equation involving $m$ to express the area of quadrilateral $OAEF$, and find the maximum value of the area of $OAEF$.

Solution:
\textbf{Solution:} Since points E and F are on the graph of the inverse proportion function $y=\frac{k}{x}$ $(x>0)$,\\
thus $CF \times 4 = 8m$,\\
therefore $CF = 2m$,\\
therefore, the area of quadrilateral $OAEF$ is $8 \times 4 - \frac{1}{2} \times 4 \times 2m - \frac{1}{2} \times (8 - 2m) \times (4 - m)$\\
$= -m^{2} + 4m + 16 = -\left(m - 2\right)^{2} + 20$,\\
since $0 < m < 4$,\\
therefore, when $m = 2$, the maximum area of quadrilateral $OAEF$ is $\boxed{20}$.


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 $C$. Furthermore, it is symmetrical about the line $x=1$, and the coordinates of point $A$ are $(-1,0)$. What is the expression of the quadratic function?

Solution:
\textbf{Solution:} From the given, we have:

\[
\left\{\begin{array}{l}
-\frac{b}{2a}=1\\
1-b+c=0
\end{array}\right.
\]

Therefore,

\[
\left\{\begin{array}{l}
b=-2\\
c=-3
\end{array}\right.
\]

Hence, \(y=x^{2}-2x-3=\boxed{x^{2}-2x-3}\)


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 $C$. It is symmetric about the line $x=1$, and the coordinates of point $A$ are $(-1,0)$. When $y<0$, write down the range of values for $x$.

Solution: \textbf{Solution:} When $y<0$, the range of $x$ is \boxed{-1<x<3}.


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 $C$. It is also symmetric about the line $x=1$, and the coordinate of point $A$ is $(-1,0)$. When $a \leq x \leq a+1$, the minimum value of the quadratic function $y=x^2+bx+c$ is $2a$. Find the value of $a$.

Solution:
Solution:
(1) For $a\le 0$, when $x=a+1$, y reaches its minimum value, which is
\[\left(a+1\right)^{2}-2\times \left(a+1\right)-3=a^{2}-4=2a\]
\[\therefore a=1\pm \sqrt{5}\] (discard the positive value)

(2) For $0<a\le 1$, when $x=1$, y reaches its minimum value, which is
\[1^{2}-2\times 1-3=-4=2a\]
\[\therefore a=-1\] (discard)

(3) For $a>1$, when $x=a$, y reaches its minimum value, which is
\[a^{2}-2a-3=2a\]
\[\therefore a=2\pm \sqrt{7}\] (discard the negative value)

Hence, $a= \boxed{1-\sqrt{5}}$ or $a= \boxed{2+\sqrt{7}}$


Question: As shown in the figure, in the Cartesian coordinate system, the vertex $C$ of the parabola passing through points $A\left(0, 4\right)$ and $B\left(5, 9\right)$ is located on the positive half of the $x$-axis. Find the equation of the parabola.

Solution:
\textbf{Solution:} Since the vertex of the parabola is on the x-axis, we can assume the equation of the parabola to be $y=a(x-h)^{2}$. Therefore,
\[
\left\{
\begin{array}{l}
a(0-h)^{2}=4\\
a(5-h)^{2}=9
\end{array}
\right.
\]
Solving this gives $h_{1}=2$, $h_{2}=-10$ (which is not consistent with the problem statement). When $h=2$, $a=1$, therefore the equation of the parabola is $y=(x-2)^{2}=x^{2}-4x+4$. Thus, the equation of the parabola is $\boxed{y=x^{2}-4x+4}$.


Question: As shown in the figure, in the Cartesian coordinate system, the vertex $C$ of the parabola passing through points $A\left(0, 4\right)$ and $B\left(5, 9\right)$ is on the positive half-axis of the $x$-axis. Find the coordinates of point $C$.

Solution:
\textbf{Solution:}
Since the equation of the parabola is $y=(x-2)^2$, \\
therefore, the coordinates of point $C$ are $\boxed{(2,0)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the vertex $C$ of the parabola passing through points $A(0, 4)$ and $B(5, 9)$ is located on the positive half of the $x$-axis. Let $P(x, y)$ be a point on segment $AB$, with $1\leq x\leq 4$. Draw $PM\parallel y$-axis to intersect the parabola at point $M$. Find the maximum and minimum values of $PM$.

Solution:
\textbf{Solution:} Let the equation of AB be: $y=kx+b$. Substituting points A$(0,4)$ and B$(5,9)$ into $y=kx+b$, we obtain:
\[
\left\{
\begin{array}{l}
b=4\\
5k+b=9
\end{array}
\right.
\]
Solving this gives:
\[
\left\{
\begin{array}{l}
k=1\\
b=4
\end{array}
\right.
\]
$\therefore$The equation of AB is: $y=x+4$,

$\because$Point P is a point on line segment AB, and point M is a point on the parabola $y=(x-2)^2$, $PM\parallel y$-axis,

Let $P(t, t+4)$, then $M(t, t^2-4t+4)$,

$\therefore PM=(t+4)-(t^2-4t+4)=-t^2+5t=-(t-\frac{5}{2})^2+\frac{25}{4}$,

Additionally, $\because 1\leq t\leq 4$,

$\therefore$When $1\le t\le \frac{5}{2}$, the value of PM increases with increasing t; when $\frac{5}{2}\le t\le 4$, the value of PM decreases with increasing t;

$\therefore$When $t=\frac{5}{2}$, $PM_{\max}=\boxed{\frac{25}{4}}$;

Since $\because 4-\frac{5}{2}=\frac{5}{2}-1=\frac{3}{2}$,

$\therefore$When $t=1$ and $t=4$, the minimum value of PM is reached, $PM_{\min}=\boxed{4}$.

In conclusion, the maximum value of PM is $\boxed{\frac{25}{4}}$, and the minimum value of PM is $\boxed{4}$.


Question: As shown in Figure 1, in the Cartesian coordinate system $xOy$, the parabola $y=ax^2+bx$ passes through the point $(-2, 5)$, and intersects with the line $y=-\frac{1}{2}x$ at point A in the second quadrant. Through point A, draw $AB\perp x$ axis, with the foot of the perpendicular at point $B\left(-4, 0\right)$. Suppose P is a point on the parabola above the line OA, through point P draw $PC\perp x$ axis at point C, intersecting OA at point D. Connect $OP$ and $PA$. What is the analytical expression of the parabola?

Solution:
\textbf{Solution:} Let $AB$ be perpendicular to the $x$-axis, with the point $B(-4, 0)$,\\
$\therefore y=-\frac{1}{2}\times 4=-2$,\\
$\therefore A(-4, 2)$. Furthermore, since the parabola passes through $(-2, 5)$,\\
$\therefore \left\{\begin{array}{l}
16a-4b=2\\
4a-2b=5
\end{array}\right.$, solving this gives: $\left\{\begin{array}{l}
a=-1\\
b=-\frac{9}{2}
\end{array}\right.$,\\
$\therefore$ the equation of the parabola is $y=-x^{2}-\frac{9}{2}x=\boxed{-x^{2}-\frac{9}{2}x}$.


Question: As shown in Figure 1, in the Cartesian coordinate system $xOy$, the parabola $y=ax^2+bx$ passes through the point $\left(-2, 5\right)$, and intersects with the line $y=-\frac{1}{2}x$ at point A in the second quadrant. A line perpendicular to the $x$-axis passing through point A meets the $x$-axis at point $B\left(-4, 0\right)$. Let P be a moving point on the parabola above line OA, draw $PC\perp x$-axis at point C, and intersect line OA at point D. Connect $OP$ and $PA$. What is the maximum value of the area $S$ of $\triangle AOP$?

Solution:
\textbf{Solution}: Let point $P\left(t, -{t}^{2}-\frac{9}{2}t\right)$, then point $D\left(t, -\frac{1}{2}t\right)$,\\
$\therefore PD=\left(-{t}^{2}-\frac{9}{2}t\right)-\left(-\frac{1}{2}t\right)=-{t}^{2}-4t$,\\
$\therefore S=\frac{1}{2}\cdot PD\cdot 4=\frac{1}{2}\cdot \left(-{t}^{2}-4t\right)\cdot 4=-2{\left(t+2\right)}^{2}+8$,\\
$\therefore$ when $t=-2$, $S_{\max}=\boxed{8}$.


Question: As shown in Figure 1, in the Cartesian coordinate system $xOy$, the parabola $y=ax^2+bx$ passes through the point $\left(-2, 5\right)$, and intersects with the line $y=-\frac{1}{2}x$ at point $A$ in the second quadrant. A perpendicular line $AB$ is drawn from point $A$ to the $x$-axis, with the foot of the perpendicular being point $B\left(-4, 0\right)$. Let $P$ be a moving point on the parabola above line $OA$, and draw a line $PC\perp x$-axis at point $C$, intersecting $OA$ at point $D$. Connect $OP$ and $PA$. Connect $PB$ intersecting $OA$ at point $E$, as shown in Figure 2. Can the line segments $PB$ and $AD$ bisect each other? If so, please find the coordinates of point $E$; if not, please explain why.

Solution: \textbf{Solution:} Solution: The line segment $PB$ and $AD$ bisect each other. The reason is as follows: Connect $BD$, $\because$ the line segments $PB$ and $AD$ bisect each other, $\therefore$ the quadrilateral $ABDP$ is a parallelogram, $\therefore$ $PD=AB$, $\because A(-4, 2)$, $PD = -t^2 - 4t$, $\therefore$ $-t^2 - 4t = 2$, $\therefore$ $t = -2 + \sqrt{2}$ or $t = -2 - \sqrt{2}$. When $t = -2 + \sqrt{2}$, then $D(-2 + \sqrt{2}, 1 - \frac{\sqrt{2}}{2})$, $\because E$ is the midpoint of $AD$, $\therefore$ the coordinates of point E are $\boxed{\left(\frac{-6 + \sqrt{2}}{2}, \frac{6 - \sqrt{2}}{4}\right)}$. When $t = -2 - \sqrt{2}$, then $D(-2 - \sqrt{2}, 1 + \frac{\sqrt{2}}{2})$, $\because E$ is the midpoint of $AD$, $\therefore$ the coordinates of point E are $\boxed{\left(\frac{-6 - \sqrt{2}}{2}, \frac{6 + \sqrt{2}}{4}\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the quadratic function $y=-x^2+bx+c$ intersects the coordinate axes at points $A$, $B$, and $C$. The coordinates of point $B$ are $(1,0)$, the coordinates of point $C$ are $(0,4)$, and the coordinates of point $D$ are $(0,2)$. Point $P$ is a moving point on the graph of the quadratic function. Find the analytical expressions for the quadratic function and the line $AD$.

Solution:
Solution: Substitute B(1, 0) and C(0, 4) into $y=-x^{2}+bx+c$, we get
\[
\left\{
\begin{array}{l}
0=-1+b+c\\
4=c
\end{array}
\right.,
\]
Solving this, we have
\[
\left\{
\begin{array}{l}
b=-3\\
c=4
\end{array}
\right.,
\]
$\therefore$ The quadratic function can be written as $y=-x^{2}-3x+4$.

Substitute $y=0$ into $y=-x^{2}-3x+4$, we have $-x^{2}-3x+4=0$, solving this we find $x_{1}=-4$, $x_{2}=1$,

$\therefore$ \boxed{A(-4, 0)}.

Let's assume the equation of line AD as $y=kx+b'$.

$\because$ D(0, 2),

$\therefore$
\[
\left\{
\begin{array}{l}
0=-4k+b' \\
2=b'
\end{array}
\right.,
\]
Solving this, we find:
\[
\left\{
\begin{array}{l}
k=\frac{1}{2}\\
b'=2
\end{array}
\right.,
\]
$\therefore$ The equation of line AD is \boxed{y=\frac{1}{2}x+2}.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the quadratic function $y=-x^2+bx+c$ intersects the coordinate axes at points $A$, $B$, and $C$, where point $B$ has the coordinates $(1,0)$, point $C$ has the coordinates $(0,4)$, and point $D$ has the coordinates $(0,2)$. Let point $P$ be a moving point on the graph of the quadratic function. When point $P$ is located in the second quadrant on the graph of the quadratic function, connect points $AD$ and $AP$, and construct the parallelogram $APED$ with $AD$ and $AP$ as adjacent sides. Let the area of parallelogram $APED$ be $S$. Find the maximum value of $S$.

Solution:
\textbf{Solution:} Draw PD and construct PG $\parallel$ y-axis intersecting AD at point G, as shown in the figure.\\
Let P$(t，-t^{2}-3t+4)$ (where $-4<t<0$), then G$(t，\frac{1}{2}t+2)$,\\
$\therefore PG=-t^{2}-3t+4-\frac{1}{2}t-2=-t^{2}-\frac{7}{2}t+2$,\\
$\therefore S=2S_{\triangle APD}=2\times \frac{1}{2}PG\cdot |x_{D}-x_{A}|=-4t^{2}-14t+8$,\\
$=-4\left(t+\frac{7}{4}\right)^{2}+\frac{81}{4}$.\\
Since $-4<0$ and $-4<t<0$,\\
$\therefore$ when $t=-\frac{7}{4}$, $S$ attains its maximum value $\boxed{\frac{81}{4}}$.


Question: As shown in the figure, the vertices of $\triangle OAB$ are located at $O(0, 0)$, $A(3, 4)$, and $B(6, 0)$, respectively. Moving points $P$ and $Q$ start simultaneously from point $O$, moving in the positive directions of the x-axis and y-axis at speeds of 3 units per second and 2 units per second, respectively. They stop moving when point $P$ reaches point $B$. Through point $Q$, draw $MN \parallel OB$, intersecting $AO$ and $AB$ at points $M$ and $N$, respectively, and connect $PM$ and $PN$. Let the movement time be $t$ (seconds). What are the coordinates of point $M$ (expressed in terms of $t$)?

Solution:
\textbf{Solution:} Let the equation of the line $OA$ be $y=ax$. Substituting the coordinates of point $A$, we get $4=3a$, which yields $a=\frac{4}{3}$.\\
$\therefore$ The equation of the line $OA$ is $y=\frac{4}{3}x$.\\
$\because$ $MN\parallel OB$ and $OQ=2t$,\\
$\therefore$ The ordinate of $M$ is $2t$.\\
Substituting $y=2t$ into $y=\frac{4}{3}x$ gives $x=\frac{3}{2}t$,\\
$\therefore$ The abscissa of $M$ is $\boxed{\left(\frac{3}{2}t, 2t\right)}$.


Question: As shown in the figure, the vertices of $\triangle OAB$ are $O(0, 0)$, $A(3, 4)$, and $B(6, 0)$. Moving points P and Q start from point O simultaneously, heading along the positive directions of the x-axis and y-axis at speeds of 3 units per second and 2 units per second, respectively. Point P and Q stop moving when P reaches point B. A line through point Q, $MN \parallel OB$, intersects $AO$ and $AB$ at points M and N, respectively. Lines $PM$ and $PN$ are then drawn. Let the movement time be t (seconds). What value of t will maximize the area of quadrilateral $MNPB$?

Solution:
\textbf{Solution:} As shown in the figure, draw $AD\perp OB$ through point $A$, intersecting $MN$ and $OB$ at points $C$ and $D$, respectively.\\
$\because MN\parallel OB$, $A(3, 4)$, $M\left(\frac{3}{2}t, 2t\right)$,\\
$\therefore D(3, 0)$, $C(3, 2t)$.\\
$\therefore OD=BD$.\\
$\because MN\parallel OB$,\\
$\therefore \angle ANC=\angle AMC$, $AC\perp MN$,\\
$\therefore AM=AN$,\\
$\therefore MC=CN=\frac{1}{2}MN$,\\
$\therefore MN=2MC=2\left(3-\frac{3}{2}t\right)=6-3t$.\\
$\because$ Point P moves at a speed of $3$,\\
$\therefore OP=3t$.\\
$\because B(6, 0)$,\\
$\therefore PB=6-3t$.\\
$\therefore MN=PB$,\\
$\therefore$ Quadrilateral $MNPB$ is a parallelogram.\\
$\therefore S_{\text{quadrilateral }MNBP}=\frac{1}{2}\left[(6-3t)+(6-3t)\right]\times 2t=-6(t-1)^2+6$.\\
$\because$ Point P stops moving when it reaches point B along with point Q,\\
$\therefore 0<3t<6$, which means $0<t<2$.\\
$\therefore$ When $t=\boxed{1}$, the area of quadrilateral $MNBP$ is maximal, with the maximum value being $6$.


Question: As shown in the figure, the vertices of triangle $ \triangle OAB$ are respectively $ O(0, 0)$, $ A(3, 4)$, and $ B(6, 0)$. Moving points P and Q start simultaneously from point O and move along the positive direction of the x-axis and the y-axis at speeds of 3 units per second and 2 units per second, respectively. The motion of points P and Q stops simultaneously when point P reaches point B. Through point Q, draw $ MN\parallel OB$ to intersect $AO$ and $AB$ at points M and N, respectively, and connect PM and PN. Let the motion time be t (seconds). Connect AP. Determine whether there exists a point P such that $ \angle OAP=\angle BPN$. If it exists, find the value of t at that time; if it does not exist, explain the reason.

Solution:
\textbf{Solution:} As shown in the figure, connect AP,\\
since $A(3, 4)$, $M\left(\frac{3}{2}t, 2t\right)$,\\
thus $OA=\sqrt{3^{2}+4^{2}}=5$, $OM=\sqrt{\left(\frac{3}{2}t\right)^{2}+\left(2t\right)^{2}}=\frac{5}{2}t$.\\
According to the given conditions, $AO=AB$, $AM=AN$,\\
thus $OM=BN=\frac{5}{2}t$.\\
(1) When $t=0$, points $M$, $P$ are both at the origin, at this time point $N$ coincides with point $B$,\\
thus $\angle BPN=\angle OAP=0^\circ$.\\
(2) When $0<t<2$,\\
since $OA=OB$,\\
thus $\angle AOB=\angle PBN$,\\
and $\angle OAP=\angle BPN$,\\
thus $\triangle AOP\sim \triangle PBN$,\\
thus $\frac{OA}{BP}=\frac{OP}{BN}$,\\
namely $\frac{5}{6-3t}=\frac{3t}{\frac{5}{2}t}$,\\
solving this gives $t_{1}=\frac{11}{18}$, $t_{2}=0$ (discard).\\
(3) When $t=2$, it contradicts the given conditions, discard it.\\
In summary, $t=0$ or $t=\boxed{\frac{11}{18}}$.


Question: As shown in the figure, in the Cartesian coordinate system, the axis of symmetry of the parabola is the line $x=1$, and it intersects the x-axis at points A and B, and intersects the y-axis at point $C\left(0, -3\right)$, with $OB=OC$. Find the equation of the parabola.

Solution:
\textbf{Solution:} Given $C\left(0, -3\right)$ and $OB=OC$, it follows that $OB=OC=3$ and thus $B\left(3, 0\right)$. Suppose the equation of the parabola is $y=ax^2+bx+c$, then we have:
\[
\begin{cases}
-\frac{b}{2a}=1\\
9a+3b+c=0\\
c=-3
\end{cases}
\]
Solving these, we obtain:
\[
\begin{cases}
a=1\\
b=-2\\
c=-3
\end{cases}
\]
Therefore, the equation of the parabola is \boxed{y=x^2-2x-3}.


Question: As shown in the figure, in the Cartesian coordinate system, the axis of symmetry of the parabola is the line $x=1$, and it intersects the x-axis at points A and B, and intersects the y-axis at point $C\left(0, -3\right)$, where $OB=OC$. Is there a point Q on the parabola such that $\triangle BCQ$ is a right-angled triangle with BC as its right angle side? If it exists, find the coordinates of point Q; if not, please explain why.

Solution:
\textbf{Solution:} There exists a point $Q$ such that $ \triangle BCQ$ is a right triangle with $BC$ as the right angle side, for the reasons below:\\
(1) When $ \angle BCQ=90^\circ$, as shown in the figure:\\
Draw a line from point Q to the y-axis at point H, perpendicular to it, since $ OB=OC$ and $ \angle BOC=90^\circ$, thus $\triangle BOC$ is an isosceles right triangle, hence $ \angle OCB=45^\circ$; since $ \angle BCQ=90^\circ$, thus $ \angle HCQ=45^\circ$, hence $\triangle HCQ$ is also an isosceles right triangle, thus $ HC=HQ$; consider point $ Q\left(t, {t}^{2}-2t-3\right)$, then $ HQ=t$ and $ CH=-3-{t}^{2}+2t+3=-{t}^{2}+2t$, thus $ t=-{t}^{2}+2t$, solving this yields: ${t}_{1}=1$, ${t}_{2}=0$ (discarded as it does not meet the question's requirement), hence point $ Q\left(1, -4\right)$;\\
(2) When $ \angle CBQ=90^\circ$, as shown in the figure:\\
Draw a perpendicular line from point B to the x-axis, then from points Q and C draw lines QE$\perp$BE at point E, and CF$\perp$BE at point F, thus $ CF=BF=3$, hence $\triangle BFC$ is an isosceles right triangle, thus $ \angle CBF=45^\circ$; since $ \angle CBQ=90^\circ$, thus $ \angle EBQ=45^\circ$, hence $\triangle QEB$ is also an isosceles right triangle, thus $ EQ=EB$; consider point $ Q\left(t, {t}^{2}-2t-3\right)$, then $ EQ=3-t$ and $ EB={t}^{2}-2t-3$, thus $ 3-t={t}^{2}-2t-3$, solving this yields: ${t}_{1}=-2$, ${t}_{2}=3$ (discarded as it does not meet the question's requirement), hence point $ Q\left(-2, 5\right)$;\\
In conclusion: When $ \triangle BCQ$ is a right triangle with $BC$ as the right angle side, point $\boxed{Q\left(1, -4\right)}$ or $\boxed{Q\left(-2, 5\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the axis of symmetry of the parabola is the line $x=1$, and it intersects the x-axis at points A and B, and intersects the y-axis at point $C\left(0, -3\right)$, with $OB=OC$. Let the x-coordinate of a point $P$ on the parabola be $m$, and $P$ is below the line BC. Find the coordinates of point P when the area of $\triangle BCP$ is the largest integer.

Solution:
\textbf{Solution:} From equation (1), we know that $B(3, 0)$, $C(0, -3)$. Assume the equation of line BC is $y=kx+b$, then we have:
\[
\begin{cases}
3k+b=0\\
b=-3
\end{cases}
\]
Solving these, we obtain:
\[
\begin{cases}
k=1\\
b=-3
\end{cases}
\]
Thus, the equation of line BC is $y=x-3$. Draw PM from point P perpendicular to the x-axis, intersecting BC at point M, as shown in the diagram. Therefore, $P(m, m^{2}-2m-3)$, thus $M(m, m-3)$, therefore $PM=m-3-m^{2}+2m+3=-m^{2}+3m$. Then, the area of $\triangle BCP$, ${S}_{\triangle BCP}={S}_{\triangle CPM}+{S}_{\triangle BPM}=\frac{1}{2}PM\cdot (x_{B}-x_{C})=\frac{3}{2}(-m^{2}+3m)=-\frac{3}{2}\left(m-\frac{3}{2}\right)^{2}+\frac{27}{8}$. Since $-\frac{3}{2}<0$, the parabola opens downwards, therefore ${S}_{\triangle BCP}\le \frac{27}{8}$. Hence, when the area of $\triangle BCP$ is the largest integer, it equals $\boxed{3}$. Thus, $-\frac{3}{2}m^{2}+\frac{9}{2}m=3$, solving for $m$ yields $m_{1}=1$, $m_{2}=2$. Therefore, point $\boxed{P(1, -4)}$ or $\boxed{P(2, -3)}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=3$, $BC=4$. On segment $AB$, a moving point $M$ starts at point $A$ and moves towards point $B$ at a constant speed, while another moving point $N$ starts at point $B$ and moves towards point $A$ at a constant speed. When one of the points $M$ or $N$ stops moving, the other point stops moving as well. Perpendicular lines are drawn from points $M$ and $N$ to $AB$, intersecting the lines through the right-angle sides $AC$ and $BC$ at points $D$ and $E$, respectively. Line $DE$ is drawn. If the movement time is $t$ seconds, throughout the movement, the quadrilateral $DENM$ always forms a rectangle (except when points $M$ and $N$ coincide). As shown in the figure, let the length of $DM$ be $x$, and the area of rectangle $DENM$ be $S$. Find the functional relationship between $S$ and $x$; what value of $x$ maximizes the area of rectangle $DENM$? What is the maximum area?

Solution: \textbf{Solution:} Since in $\triangle ABC$, $\angle C=90^\circ$, $AC=3$, $BC=4$, we have $AB=\sqrt{AC^2+BC^2}=5$. Because $\triangle ADM\sim \triangle ABC$, we get $\frac{AM}{AC}=\frac{MD}{BC}$. Considering $DM=x$, we find $\frac{AM}{3}=\frac{x}{4}$, which means $AM=\frac{3}{4}x$. Therefore, $AD=\sqrt{AM^2+DM^2}=\frac{5}{4}x$. Thus, $CD=AC-AD=3-\frac{5}{4}x$. Given $\triangle ADM\sim \triangle ABC$ and $\triangle DEC\sim \triangle ABC$, it follows that $\triangle ADM\sim \triangle DEC$. Therefore, $\frac{DE}{AD}=\frac{CD}{AM}$, that is $\frac{DE}{\frac{5}{4}x}=\frac{3-\frac{5}{4}x}{\frac{3}{4}x}$, which leads to $DE=5-\frac{25}{12}x$. Thus, the area of quadrilateral $DENM$ is $DM\cdot DE=5x-\frac{25}{12}x^2=-\frac{25}{12}\left(x^2-\frac{12}{5}x+\frac{36}{25}\right)+3=-\frac{25}{12}\left(x-\frac{6}{5}\right)^2+3$. Since $-\frac{25}{12}<0$, the area of rectangle $DENM$ reaches its maximum when $x=\frac{6}{5}$, with the maximal area being \boxed{3}.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=3$, $BC=4$. On segment $AB$, a moving point $M$ starts from point $A$ moving towards point $B$ at a uniform speed, while another moving point $N$ starts from point $B$ moving towards point $A$ at a uniform speed. When one of the points $M$ or $N$ stops moving, the other point stops as well. Perpendiculars are drawn from points $M$ and $N$ to $AB$, intersecting the lines containing the right angle sides $AC$ and $BC$ at points $D$ and $E$ respectively. $DE$ is then connected. If the movement time is $t$ seconds, during the movement, the quadrilateral $DENM$ is always a rectangle (except when points $M$ and $N$ coincide). During the movement, if the speed of point $M$ is 1 unit length per second, find the speed of point $N$. For what value of $t$ seconds does the rectangle $DENM$ become a square?

Solution: \textbf{Solution:} Given that before points M and N meet, since the quadrilateral DENM is a rectangle, it follows that NE = MD. Since $\triangle AMD \sim \triangle ABC$, it follows that $\frac{AM}{AC} = \frac{MD}{BC}$. According to the problem, $AM = t$, thus $\frac{t}{3} = \frac{MD}{4}$, which implies $MD = \frac{4}{3}t$; Since $\triangle BEN \sim \triangle BAC$, it follows that $\frac{BN}{BC} = \frac{EN}{AC}$, which implies $\frac{BN}{4} = \frac{\frac{4}{3}t}{3}$, so $BN = \frac{16}{9}t$. Thus, the speed of point N is $\frac{16}{9}$ units per second. Since at the point where N and M meet, $AM + BM = AB$, it follows that $t + \frac{16}{9}t = 5$. Solving this equation yields $t = \frac{9}{5}$, meaning the meeting time for M and N is $\frac{9}{5}$. Continuing movement after N and M meet, when point N reaches point A, solving $\frac{16}{9}t = 5$ gives $t = \frac{45}{16}$, which means the time for point N to reach the bottom point A is $\frac{45}{16}$. Since the rectangle DENM is a square, therefore DM = MN = EN. Before N and M meet, i.e., when $0 < t < \frac{9}{5}$, we have $MD = \frac{4}{3}t$, $BN = \frac{16}{9}t$, $AM = t$. Thus, $MN = AB - AM - BN = 5 - t - \frac{16}{9}t = 5 - \frac{25}{9}t$, resulting in $5 - \frac{25}{9}t = \frac{4}{3}t$. Solving this equation yields $t = \boxed{\frac{45}{37}}$. After N and M meet, i.e., when $\frac{9}{5} < t < \frac{45}{16}$, with $BN = \frac{16}{9}t$, $AM = t$, $MD = \frac{4}{3}t$, we get $AN = AB - BN = 5 - \frac{16}{9}t$, $BM = AB - AM = 5 - t$. Thus, $MN = AB - AN - BM = 5 - \left( 5 - \frac{16}{9}t \right) - \left( 5 - t \right) = \frac{25}{9}t - 5$, leading to $\frac{25}{9}t - 5 = \frac{4}{3}t$. Solving this equation yields $t = \frac{45}{13} > \frac{45}{16}$, which is not reasonable. Therefore, summarizing the above discussion, the speed of point N is $\frac{16}{9}$ units per second and when $t = \boxed{\frac{45}{37}}$, the rectangle DENM forms a square.


Question: As shown in the figure, the parabola $y=ax^2-2ax-3a$ (where $a$ is a constant and $a<0$) intersects the x-axis at points A and B (point A is to the left of point B), and intersects the y-axis at point C, and $OB=OC$. Find the value of $a$.

Solution:
\textbf{Solution.} From the given information, we have
\[y = a{x}^{2}-2ax-3a \\
= a({x}^{2}-2x-3) \\
= a(x-3)(x+1)\]
Setting $y=0$, we find ${x}_{1}=-1$, ${x}_{2}=3$. Setting $x=0$, we have $y=-3a$.

The parabola $y=a{x}^{2}-2ax-3a$ (where $a$ is a constant, $a<0$) intersects the $x$-axis at points $A$ and $B$ (with $A$ to the left of $B$) and the $y$-axis at point $C$, making the intersection points with the $x$-axis $A(-1,0)$, $B(3,0)$, and with the $y$-axis $C(0,-3a)$.

Since $OB=OC$, we have $OC=3$, hence $C(0,3)$, which implies $-3a=3$.

Solving for $a$ gives $a=\boxed{-1}$.


Question: As shown in the figure, the parabola $y=ax^2-2ax-3a$ (where $a$ is a constant and $a<0$) 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, with OB=OC. Point D is the vertex of the parabola, and point P $(m, n)$ is a point on the parabola in the third quadrant. Lines BD, BC, CD, and BP are drawn. When $\angle PBA=\angle CBD$, find the value of $m$?

Solution:
\textbf{Solution:} As shown in the figure, draw $PE\perp x$ axis at point $E$,\\
$\because y=-x^2+2x+3=-(x-1)^2+4$\\
$\therefore D(1, 4)$\\
$\because B(3, 0), C(3, 0)$\\
$\therefore CD=\sqrt{1^2+1^2}=\sqrt{2}, BC=\sqrt{3^2+3^2}=3\sqrt{2}, BD=\sqrt{(3-1)^2+4^2}=2\sqrt{5}$\\
$\therefore CD^2+BC^2=20, BD^2=20$\\
$\therefore CD^2+BC^2=BD^2$\\
$\therefore \triangle BCD$ is a right triangle, with $\angle BCD=90^\circ$\\
$\because PE\perp AB$\\
$\therefore \angle PEB=\angle PCD=90^\circ$\\
Also, $\because \angle PBA=\angle CBD$\\
$\therefore \triangle BCD\sim \triangle BEP$\\
$\therefore \frac{CD}{PE}=\frac{BC}{BE}$\\
$\because P(m, n)$ lies on the parabola $y=-x^2+2x+3$,\\
$\therefore n=-m^2+2m+3$\\
$\therefore PE=-n=m^2-2m-3, BE=3-m$\\
$\therefore \frac{\sqrt{2}}{m^2-2m-3}=\frac{3\sqrt{2}}{3-m}$\\
After rearranging, we get $(3m+4)(m-3)=0$\\
Solving this, we find $m_1=-\frac{4}{3}, m_2=3$ (discard)\\
$\because P(m, n)$ is in the third quadrant,\\
$\therefore m=\boxed{-\frac{4}{3}}$


Question: As shown in the figure, the parabola $y=ax^2-2ax-3a$ (where \(a\) is a constant, $a<0$) 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, with OB=OC. Point K is a point in the coordinate plane, and DK=2, point M is the midpoint of segment BK. Connect AM. When AM is maximized, find the coordinates of point K.

Solution:
\textbf{Solution:} Connect $BD$ and take the midpoint $Q$ of $BD$. Connect $QM$. Therefore, $QM$ is the median of $\triangle BDK$, thus $QM=\frac{1}{2}DK=1$. According to the problem, point $K$ lies on the circle with center $D$ and radius 2, so $M$ moves on the circle with center $Q$ and radius 1. When points $A$, $Q$, $M$ are collinear, and $M$ lies on the extension of $AQ$, $AM$ is maximized. Since $B(3, 0)$ and $D(1, 4)$, thus $Q\left(\frac{1+3}{2}, \frac{4+0}{2}\right)$ i.e., $Q(2, 2)$. Since $A(-1, 0)$ and $Q(2, 2)$, let the equation of line $AM$ be $y=kx+d$. Substitute points $A(-1, 0)$ and $Q(2, 2)$ to get
\[
\begin{cases}
0=-k+d\\
2=2k+d
\end{cases}
\]
Solving these equations gives
\[
\begin{cases}
k=\frac{2}{3}\\
b=\frac{2}{3}
\end{cases}
\]
Therefore, the equation of line $AM$ is $y=\frac{2}{3}x+\frac{2}{3}$. Since $DK\parallel QM$, let the equation of line $DK$ be $y=\frac{2}{3}x+b$. Given $D(1, 4)$, thus $4=\frac{2}{3}+b$, solving this gives $b=\frac{10}{3}$, so the equation of $DK$ is $y=\frac{2}{3}x+\frac{10}{3}$. Let point $K(m, \frac{2}{3}m+\frac{10}{3})$ $(m>0)$, given $D(1, 4)$ and $DK=2$, thus $(m-1)^2+\left(\frac{2}{3}m+\frac{10}{3}-4\right)^2=2^2$, solving this gives $m_1=\frac{6\sqrt{13}+13}{13}$, $m_2=\frac{-6\sqrt{13}+13}{13}$ (discard), thus $m=\frac{6\sqrt{13}+13}{13}$, thus $\frac{2}{3}m+\frac{10}{3}=\frac{2}{3}\times \frac{6\sqrt{13}+13}{13}+\frac{10}{3}=\frac{4\sqrt{13}+52}{13}$, thus $K\boxed{\left(\frac{6\sqrt{13}+13}{13}, \frac{4\sqrt{13}+52}{13}\right)}$.


Question: As shown in the diagram, in the Cartesian coordinate system $xOy$, the parabola $y=ax^{2}+bx-4$ intersects the $x$-axis at points $A$ and $B$ (with $B$ to the right of $A$) and intersects the $y$-axis at point $C$. It is known that $OA=1$ and $OB=4OA$. Connect $BC$. What is the equation of the parabola?

Solution:
\[ \text{Solution: From } OA=1 \text{ we get } A(-1, 0). \]
\[ \because OB=4OA=4, \]
\[ \therefore B(4, 0). \]
Substituting $A(-1, 0)$ and $B(4, 0)$ into $y=ax^2+bx-4$ we get:
\[ \begin{cases} a-b-4=0 \\ 16a+4b-4=0 \end{cases}, \]
Solving this yields
\[ \begin{cases} a=1 \\ b=-3 \end{cases}, \]
\[ \therefore y=x^2-3x-4; \]
Thus, the equation of the parabola is $\boxed{y=x^2-3x-4}$.


Question: As shown in the figure, in the Cartesian coordinate system \(xOy\), the parabola \(y=ax^2+bx-4\) intersects the \(x\)-axis at points \(A\) and \(B\) (with \(B\) to the right of \(A\)) and intersects the \(y\)-axis at point \(C\). It is known that \(OA=1\), \(OB=4OA\), and segment \(BC\) is drawn. As shown in Figure 1, point \(P\) is a moving point on the parabola below \(BC\), and segments \(BP\) and \(CP\) are drawn. When the area of \(\triangle BCP\) equals the area of \(\triangle BOC\), what are the coordinates of point \(P\)?

Solution:
\textbf{Solution:} Given $y=ax^2+bx-4$ intersects at point $C$,\\
hence $C(0, -4)$,\\
Let $y=kx+b$, substituting $B(4, 0)$ and $C(0, -4)$, we get\\
\[\left\{\begin{array}{l}
0=4k+b \\
-4=b
\end{array}\right.,\]\\
from which we find\\
\[\left\{\begin{array}{l}
k=1 \\
b=-4
\end{array}\right.,\]\\
therefore, $y=x-4$,\\
Draw $PH\perp x$-axis from point $P$, intersecting the $x$-axis at $H$, and denote the intersection with $BC$ as point $D$,\\
Let $P(m, m^2-3m-4)$,\\
substituting $x=m$ into $y=x-4$, we obtain $y=m-4$,\\
thus $D(m, m-4)$,\\
therefore, $DP=m-4-(m^2-3m-4)=-m^2+4m$,\\
since $S_{\triangle BOC}=\frac{1}{2}OB\cdot OC=\frac{1}{2}\times 4\times 4=8$,\\
therefore, $S_{\triangle BCP}=8$,\\
thus, $S_{\triangle PDC}+S_{\triangle PDB}=8$,\\
which leads to $\frac{1}{2}PD\cdot OH+\frac{1}{2}PD\cdot BH=8$, that is $\frac{1}{2}PD\cdot OB=8$,\\
hence $(-m^2+4m)\times 4\times \frac{1}{2}=8$,\\
solving this gives $m=2$,\\
therefore, \boxed{P\left(2, -6\right)}.


Question: In the cartesian coordinate system $xOy$, the parabola $y=ax^2+bx-4$ intersects the $x$-axis at points $A$ and $B$ (with $B$ to the right of $A$) and intersects the $y$-axis at point $C$. It is given that $OA=1$ and $OB=4OA$. Connect $BC$. As shown in figure 2, let point $N$ be a point on segment $OC$. Find the minimum value of $AN+\frac{\sqrt{2}}{2}CN$?

Solution:
\textbf{Solution:} Let point $N$ be constructed such that $NM\perp BC$ at point $M$, and draw $AH\perp BC$ at $H$,\\
$\because OB=OC$, $\angle BOC=90^\circ$\\
$\therefore \triangle BOC$ is an isosceles right triangle, $\angle BCO=\angle OBC=45^\circ$\\
$\because NM\perp BC$\\
$\therefore \angle NMC=90^\circ$,\\
$\therefore \angle CNM=90^\circ-\angle NCM=90^\circ-\angle OCB=45^\circ$,\\
$\therefore \triangle NCM$ is an isosceles right triangle,\\
$\therefore NM=NC\cdot \sin45^\circ=\frac{\sqrt{2}}{2}NC$,\\
$\therefore AN+\frac{\sqrt{2}}{2}CN=AN+NE \geq AH$,\\
When points A, $N$, and $M$ are collinear, the minimum value of $AN+NM$ $=AH$,\\
$\because AB=OA+OB=5$,\\
$\because \angle ABH=45^\circ$, $AH\perp BC$,\\
$\therefore \angle BAH=90^\circ-\angle ABH=45^\circ=\angle ABH$,\\
$\therefore \triangle AEB$ is an isosceles right triangle, $\angle ABE=45^\circ$,\\
$\therefore AH=AB\cdot \sin45^\circ=\boxed{\frac{5\sqrt{2}}{2}}$\\
$\therefore$ The minimum value of $AN+\frac{\sqrt{2}}{2}CN$ $\boxed{\frac{5\sqrt{2}}{2}}$


Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\frac{k_2}{x}$ at points A and B. The coordinates of point A are $(-1, 4)$, and the coordinates of point B are $(4, n)$. What are the expressions of these two functions?

Solution:
\textbf{Solution}: According to the problem, the graph of the function $y=\frac{k_2}{x}$ passes through points $A(-1, 4)$ and $B(4, n)$,\\
$\therefore k_2=-1\times 4=4n$,\\
$\therefore k_2=-4$, $n=-1$,\\
$\therefore$ point $B(4, -1)$,\\
Furthermore, since the linear function $y=k_1x+b$ passes through points $A$ and $B$,\\
$\therefore \left\{\begin{array}{l}
-k_1+b=4 \\
4k_1+b=-1
\end{array}\right.$, yielding: $\left\{\begin{array}{l}
k_1=-1 \\
b=3
\end{array}\right.$;\\
$\therefore$ the expression for the linear function is $\boxed{y=-x+3}$, and for the inverse proportion function, $\boxed{y=-\frac{4}{x}}$.


Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\frac{k_2}{x}$ at points A and B, where the coordinates of point A are $(-1, 4)$, and the coordinates of point B are $(4, n)$. Point P is on line segment AB, and $\frac{S_{\triangle AOP}}{S_{\triangle BOP}}=1:2$. Find the coordinates of point P.

Solution:
\textbf{Solution:} Let the intersection of line $AB$ and the $y$-axis be point $C$, \\
since the analytical expression of the linear function is $y=-x+3$, \\
thus $C(0, 3)$, \\
since the area of $\triangle AOC = \frac{1}{2} \times 3 \times 1 = \frac{3}{2}$, \\
thus the area of $\triangle AOB = \text{area of} \ \triangle AOC + \text{area of} \ \triangle BOC = \frac{1}{2} \times 3 \times 1 + \frac{1}{2} \times 3 \times 4 = \frac{15}{2}$, \\
since the ratio of the areas $S_{\triangle AOP} : S_{\triangle BOP} = 1:2$, \\
thus $S_{\triangle AOP} = \frac{15}{2} \times \frac{1}{3} = \frac{5}{2}$, \\
therefore, $S_{\triangle AOC} < S_{\triangle AOP}$, the area of $\triangle COP = \frac{5}{2} - \frac{3}{2} = 1$, \\
therefore, $\frac{1}{2} \times 3 \times x_{P} = 1$, \\
thus $x_{P} = \frac{2}{3}$, \\
since point $P$ is on the segment $AB$, \\
thus $y = -\frac{2}{3} + 3 = \frac{7}{3}$, \\
therefore, \boxed{P\left(\frac{2}{3}, \frac{7}{3}\right)}.


Question: As shown in the figure, the parabola $y=ax^2+bx-3$ intersects the x-axis at points $A(-1,0)$ and $B(3,0)$. The line $l$ passing through point $A$ intersects the parabola at point $C(2,n)$. Find the analytical expression of the parabola.

Solution:
Solution: Substituting A ($-1$, $0$) and B ($3$, $0$) into $y=ax^2+bx-3$, we get
\[
\left\{
\begin{array}{l}
a-b-3=0 \\
9a+3b-3=0
\end{array}
\right.
\]
Solving this, we obtain
\[
\left\{
\begin{array}{l}
a=1 \\
b=-2
\end{array}
\right.
\]
$\therefore$The equation of the parabola is \boxed{y=x^2-2x-3}.


Question: As shown in the figure, the parabola $y=ax^2+bx-3$ intersects the $x$-axis at points $A(-1,0)$ and $B(3,0)$. A line $l$ passing through point $A$ intersects the parabola at point $C(2,n)$. Point $P$ is a moving point on segment $AC$, and a perpendicular line to the $x$-axis passing through point $P$ intersects the parabola at point $E$. Find the coordinates of point $P$ when the length of segment $PE$ is maximal.

Solution:
Solution: To solve this problem, substitute the x-coordinate $x=2$ into $y=x^2-2x-3$,\\
we get $y=-3$,\\
$\therefore C(2, -3)$,\\
Substituting $A(-1, 0)$ and $C(2, -3)$ into $y=kx+b$ yields $\left\{\begin{array}{l}-k+b=0 \\ 2k+b=-3\end{array}\right.$,\\
from which we solve $\left\{\begin{array}{l}k=-1 \\ b=-1\end{array}\right.$,\\
$\therefore$ the equation of line AC is $y=-x-1$,\\
Let the x-coordinate of point P be $m(-1\leq m\leq 2)$, then, the coordinates of P and E are: $P\left(m, -m-1\right)$, $E\left(m, m^2-2m-3\right)$;\\
$\therefore PE=\left(-m-1\right)-\left(m^2-2m-3\right)$\\
$=-m^2+m+2$\\
$=-\left(m-\frac{1}{2}\right)^2+\frac{9}{4}$,\\
$\because -1<0$,\\
$\therefore$ when $m=\frac{1}{2}$, the maximum value of PE $=\frac{9}{4}$,\\
$\therefore -m-1=-\frac{1}{2}-1=-\frac{3}{2}$,\\
at this time $P\left(\frac{1}{2}, -\frac{3}{2}\right)$,\\
$\therefore$ when the segment PE is at its maximum, the coordinates of point P are $\boxed{\left(\frac{1}{2}, -\frac{3}{2}\right)}$.


Question: As shown in the figure, the parabola $y=ax^2+bx-3$ intersects the x-axis at points $A(-1,0)$ and $B(3,0)$. The line $l$ passing through point $A$ intersects the parabola at point $C(2,n)$. Given a point $F$ moving on the parabola, is there a point $D$ on the x-axis such that the quadrilateral with vertices $A$, $C$, $D$, $F$ is a parallelogram? If such a point $D$ exists, please write down the coordinates of all points $D$ that satisfy the condition; if not, please explain why.

Solution: \textbf{Solution:} Exist; The coordinates of point $D$ meeting the conditions are $\boxed{(-3,0)}$ or $\boxed{(1,0)}$ or $\boxed{\left(4 -\sqrt{7},0\right)}$ or $\boxed{\left(4 +\sqrt{7},0\right)}$.


Question: As shown in the figure, in the rectangle $AOBC$, using the lines through $OB$ and $OA$ as the x-axis and y-axis respectively, a Cartesian coordinate system is established as shown in the figure. The coordinates of point A are $(0, 3)$, and the coordinates of point B are $(4, 0)$. F is a moving point on line segment $BC$ (not coinciding with B or C). The graph of the inverse proportion function $y=\frac{k}{x} (x > 0)$ passing through point F intersects the edge $AC$ at point E. Lines $OE$ and $OF$ are then drawn, and line EF is constructed. If $CF=2$，find the new equation of the inverse proportion function.

Solution:
\textbf{Solution:} Since quadrilateral AOBC is a rectangle, and the coordinates of point A are (0, 3), and the coordinates of point B are (4, 0), \\
thus $AC=OB=4$, and $BC=OA=3$, \\
since $CF=2$, \\
therefore, the coordinates of F are (4, 1), \\
Substituting the coordinates of F (4, 1) into $y=\frac{k}{x}$ gives $1=\frac{k}{4}$, \\
Solving this yields $k=4$, \\
therefore, the equation of the inverse proportion function is $\boxed{y=\frac{4}{x}}$.


Question: As shown in the figure, in the rectangle $AOBC$, taking the lines containing $OB$ and $OA$ as the x-axis and y-axis respectively, a Cartesian coordinate system is established as shown. The coordinates of point A are $(0, 3)$ and the coordinates of point B are $(4, 0)$. Let F be a moving point on $BC$ (not coinciding with B or C). The graph of the inverse proportion function $y=\frac{4}{x}$ ($x>0$) passing through point F intersects side $AC$ at point E. Connect $OE$ and $OF$, and draw line EF. Given $CF=2$, calculate the area of $\triangle EOF$.

Solution:
\textbf{Solution:}

Substituting $y=3$ into $y=\frac{4}{x}$, we get $3=\frac{4}{x}$.

Solving, we find $x=\frac{4}{3}$.

$\therefore$ The coordinates of point E are $\left(\frac{4}{3}, 3\right)$.

$\therefore AE=\frac{4}{3}$, and $CE=AC-AE=\frac{8}{3}$.

Since $BF=1$,

$\therefore CF=2$.

Given that $S_{\triangle EOF}=S_{\text{rectangle} AOBC}-S_{\triangle CEF}-S_{\triangle AOE}-S_{\triangle BOF}$,

$=OA \times OB - \frac{1}{2} \times CF \times CE - \frac{1}{2} \times OA \times AE - \frac{1}{2} \times OB \times BF$,

$=3 \times 4 - \frac{1}{2} \times 2 \times \frac{8}{3} - \frac{1}{2} \times 3 \times \frac{4}{3} - \frac{1}{2} \times 4 \times 1$,

$=12 - \frac{8}{3} - 2 - 2$,

$=\boxed{\frac{16}{3}}$.

$\therefore$ The area of $\triangle EOF$ is $\frac{16}{3}\text{cm}^2$.


Question: As shown in the figure, pass point C(8, 6) to draw $CB\perp x$-axis and $CA\perp y$-axis respectively, with perpendicular feet at points B and A. Point F is a mobile point on segment BC but does not coincide with points B and C. The graph of the inverse proportion function $y=\frac{k}{x}$ ($k>0$) passes through point F and intersects segment AC at point E. Connect EF. When point E is the midpoint of segment AC, what are the coordinates of point F?

Solution:
\textbf{Solution:} Given that point E is the midpoint of line segment AC,\\
thus E(4, 6),\\
since point E lies on the graph of the inverse proportion function $y=\frac{24}{x}$,\\
thus $k=xy=4\times 6=24$,\\
therefore $y=\frac{24}{x}$,\\
when $x=8$, $y=3$,\\
thus \boxed{F(8,3)}.


Question: As shown in the figure, from point $C(8, 6)$, construct $CB\perp x$-axis and $CA\perp y$-axis, with the feet of the perpendiculars being points $B$ and $A$, respectively. Point $F$ is a moving point on line segment $BC$, but does not coincide with points $B$ or $C$. The graph of the inverse proportion function $y=\frac{k}{x}$ $(k>0)$ passes through point $F$ and intersects line segment $AC$ at point $E$. Connect $EF$. If the area of $\triangle CEF$ is $6$, find the expression of the inverse proportion function?

Solution:
\textbf{Solution:} Since the area of $\triangle CEF$ is 6,\\
we have $\frac{1}{2} \times CE \times CF=6$,\\
thus $\frac{1}{2} \times \frac{48-k}{6} \times \frac{48-k}{8}=6$,\\
solving this gives $k_{1}=24$, $k_{2}=72>48$ (which we discard),\\
therefore, the equation of the inverse proportion function is $\boxed{y=\frac{24}{x}}$.


Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\frac{k_2}{x}$ at points A and B. The coordinates of point A are $(-1, 4)$, and the coordinates of point B are $(4, n)$. What are the expressions for these two functions?

Solution:
\textbf{Solution:} To solve, substitute $A(-1, 4)$ into $y=\frac{k_2}{x}$, obtaining $k_2=-4$, \\
$\therefore y=-\frac{4}{x}$,\\
$\because$ point $B(4, n)$ is on $y=-\frac{4}{x}$,\\
$\therefore n=-1$,\\
$\therefore \boxed{B(4, -1)}$,\\
Substituting $A(-1, 4)$, $B(4, -1)$ into $y=k_1x+b_1$ yields\\
$\begin{cases}
-k_1+b=4\\
4k_1+b=-1
\end{cases}$, solving gives $\begin{cases}
k_1=-1\\
b=3
\end{cases}$,\\
$\therefore \boxed{y=-x+3}$;


Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\frac{k_2}{x}$ at points A and B, where the coordinates of point A are $(-1, 4)$, and the coordinates of point B are $(4, n)$. Based on the graph, directly write down the range of values of $x$ for which $k_1x+b>\frac{k_2}{x}$.

Solution: \textbf{Solution:} By observing the graph, it is known that when $x<-1$ or $0<x<4$, it satisfies \boxed{k_1x+b>\frac{k_2}{x}}.


Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the inverse proportion function $y=\frac{k_2}{x}$ at points A and B, where the coordinates of point A are $(-1, 4)$, and the coordinates of point B are $(4, n)$. Connect OA and OB, what is the area of $\triangle AOB$?

Solution:
\[
\text{{Solution}}: \text{{As shown in the figure,}} \\
\text{{let }} AB \text{{ intersect the }} y \text{{-axis at point }} C, \\
\text{{since }} C \text{{ lies on the line }} y = -x + 3, \\
\text{{it follows that }} C(0, 3), \\
\text{{therefore, }} {S}_{\triangle AOB} = \frac{1}{2} OC \cdot (\left|{x}_{A}\right| + \left|{x}_{B}\right|) = \frac{1}{2} \times 3 \times (1+4) = \boxed{7.5};
\]


Question: As shown in the figure, the graph of the linear function $y=k_1x+b$ intersects with the graph of the hyperbolic function $y=\frac{k_2}{x}$ at points $A$ and $B$, where the coordinates of point $A$ are $(-1, 4)$, and the coordinates of point $B$ are $(4, n)$. Point $P$ is on segment $AB$, and $\frac{S_{\triangle AOP}}{S_{\triangle BOP}}=1\colon 2$. Find the coordinates of point $P$.

Solution:
\textbf{Solution:} As shown in the diagram, \\
$\because S_{\triangle AOP} : S_{\triangle BOP} = 1 : 2$, \\
$\therefore S_{\triangle AOP} = \frac{1}{3} \times 7.5 = 2.5$, $S_{\triangle BOP} = 5$, \\
Moreover, $S_{\triangle AOC} = \frac{1}{2} \times 3 \times 1 = 1.5$, \\
$\therefore$ point $P$ is in the first quadrant, \\
$\therefore S_{\triangle COP} = 2.5 - 1.5 = 1$, \\
Moreover, $OC = 3$, \\
$\therefore \frac{1}{2} \times 3 \times x_{P} = 1$, thus solving gives $x_{P} = \frac{2}{3}$, \\
Substituting $x_{P} = \frac{2}{3}$ into $y = -x + 3$, we get $y_{P} = \frac{7}{3}$, \\
$\therefore \boxed{P\left(\frac{2}{3}, \frac{7}{3}\right)}$.


Question: As shown in the figure, the linear function $y=kx+2$ intersects the $y$-axis at point A and intersects the graph of the inverse proportion function $y=\frac{m}{x}$ at points B and C. Line $BD\perp y$-axis intersects the $y$-axis at point D, $OA=OD$, and the area of triangle $\triangle ABD$ is $8$. Find the expressions of the linear function and the inverse proportion function.

Solution:
Solution: From the given information, point $A$ is the intersection point of the linear function $y=kx+2$ with the $y$-axis,\\
$\therefore$ setting $x=0$, we get $y=k\times 0+2=2$,\\
which means $A(0,2)$,\\
$\therefore OA=2$,\\
and since $OD=OA$,\\
$\therefore OD=2$,\\
$\therefore D(0,-2)$,\\
$\therefore AD=2OD=4$,\\
since $BD \perp y$-axis,\\
$\therefore$ the ordinate of point $B$ is $-2$,\\
since $S_{\triangle ABD}=8$,\\
$\therefore \frac{1}{2}AD\times BD=8$,\\
$\therefore \frac{1}{2}\times 4\times BD=8$,\\
$\therefore BD=4$,\\
$\therefore$ the coordinates of point $B$ are $(-4,-2)$,\\
Substituting point $B(-4,-2)$ into the linear function $y=kx+2$ and the inverse proportion function $y=\frac{m}{x}$ respectively,\\
we find: $-4=-2k+2$, $-2=\frac{m}{-4}$,\\
$\therefore k=1$, $m=8$,\\
$\therefore$ the equation of the linear function is: \boxed{y=x+2}, and the equation of the inverse proportion function is: \boxed{y=\frac{8}{x}};


Question: As shown in the figure, a linear function $y=kx+2$ intersects the $y$-axis at point A and intersects the graph of the hyperbolic function $y=\frac{m}{x}$ at points B and C. Line BD is perpendicular to the $y$-axis and intersects the $y$-axis at point D. It is given that OA＝OD and the area of $\triangle ABD$ is $8$. Determine the coordinates of point C, and directly write the solution set of the inequality $kx+2>\frac{m}{x}$.

Solution:
\textbf{Solution:} From equation (1), we can set up the system of equations
\[
\begin{cases}
y=x+2\\
y=\frac{8}{x}
\end{cases}
\]
Solving this system, we get:
\[
\begin{cases}
x=-4\\
y=-2
\end{cases}
\]
or
\[
\begin{cases}
x=2\\
y=4
\end{cases}
\]
$\because$ Point $C$ is in the first quadrant,\\
thus the coordinates of point $C$ are $\boxed{(2, 4)}$,\\
From the graph, it is obtained that when $-4<x<0$ or $x>2$, $kx+2>\frac{m}{x}$;


Question: As illustrated in the figure, the linear function \(y=kx+2\) intersects the \(y\)-axis at point A and intersects the graph of the inverse proportion function \(y=\frac{m}{x}\) at points B and C. Line BD is perpendicular to the \(y\)-axis and intersects it at point D. OA is equal to OD, and the area of triangle \(ABD\) is 8. In the given plane, there exists a point E such that the quadrilateral formed by points B, C, D, and E is a parallelogram. Please directly provide the coordinates of all possible points E that satisfy these conditions.

Solution: \textbf{Solution:} The coordinates of the point $E$ that meet the conditions are: $\boxed{(6,4)}$, $\boxed{(-6,-8)}$, $\boxed{(-2,4)}$.


Question: As shown in the figure, the line $y=k_1x+b$ intersects with the hyperbola $y=\frac{k_2}{x}$ at points $A$ and $B$. The line $AB$ intersects the $x$-axis at point $C$ and the $y$-axis at point $D(0, -5)$. The length of $OA$ is $5$, and $\tan\angle AOC=\frac{3}{4}$. The $y$-coordinate of point $B$ is $-8$. What is the analytical expression of the inverse proportion function?

Solution:
Solution: As shown in the diagram, construct line $AE$ perpendicular to the $x$-axis at point $E$, \\
$\because \tan\angle AOC=\frac{3}{4}$, let $AE=3a$, then $EO=4a$, \\
$\therefore OA=5a$ \\
$\because OA=5$, \\
$\therefore a=1$ \\
$\therefore AE=3$, $EO=4$, \\
$\therefore$ the coordinates of point $A$ are $\left(-4, 3\right)$, \\
$\therefore$ the equation of the hyperbola is $\boxed{y=\frac{-12}{x}}$;


Question: As illustrated, the line $y=k_1x+b$ intersects the hyperbola $y=\frac{k_2}{x}$ at points $A$ and $B$. The line $AB$ intersects the $x$-axis at point $C$ and the $y$-axis at point $D(0, -5)$. Given that $OA=5$ and $\tan\angle AOC=\frac{3}{4}$, and the $y$-coordinate of point $B$ is $-8$, what is the area of $\triangle AOD$?

Solution:
\textbf{Solution:} Substituting $A(-4, 3)$ and $D(0, -5)$ into $y=k_1x+b$, we obtain $b=-5$ and $-4k_1+b=3$. Solving these equations yields $k_1=-2$ and $b=-5$, \\
$\therefore$ the equation of line $AB$ is $y=-2x-5$, \\
Substituting $y=0$ into $y=-2x-5$, we get $x=-\frac{5}{2}$, \\
$\therefore$ the coordinates of point $C$ are $\left(-\frac{5}{2}, 0\right)$, and $OC=\frac{5}{2}$, \\
$\therefore$ the area of $\triangle AOD$ is $\frac{1}{2}\times \frac{5}{2}\times 3+\frac{1}{2}\times \frac{5}{2}\times 5=\boxed{10}$;


Question: As shown in the figure, the line $y=k_1x+b$ intersects with the hyperbola $y=\frac{k_2}{x}$ at points $A$ and $B$. The line $AB$ intersects the $x$-axis at point $C$ and the $y$-axis at point $D(0, -5)$. It is given that $OA=5$, $\tan\angle AOC=\frac{3}{4}$, and the $y$-coordinate of point $B$ is $-8$. Directly write the solution set for the inequality $k_1x+b\le \frac{k_2}{x}$.

Solution:
\textbf{Solution:} The solution set to the inequality ${k}_{1}x+b\le \frac{{k}_{2}}{x}$ is: $-4\le x<0$ or $x\ge \frac{3}{2}$.

Therefore, the solution set to the inequality is: $-4\le x<0$ or $x\ge \frac{3}{2}$.

Hence, the answer is: \boxed{-4\le x<0} or \boxed{x\ge \frac{3}{2}}.


Question: A car passes through a certain section of the road at a constant speed. The required time $t$ (h) and the driving speed $v$ (km/h) meet the functional relationship: $t=\frac{k}{v}$, whose graph is a curve as shown, with endpoints $A(40, 1)$ and $B(m, 0.5)$. What are the values of $k$ and $m$?

Solution:
\textbf{Solution:} From the given, the function $t=\frac{k}{v}$ passes through the point $(40,1)$, \\
$1=\frac{k}{40}$, then we get $k=\boxed{40}$, \\
Hence, the function equation is: $t=\frac{40}{v}$ \\
Substituting $(m,0.5)$ into $t=\frac{40}{v}$, we find $m=\boxed{80}$;


Question: A car passes a certain section of the road at a constant speed. The time $t$ (h) required and the driving speed $v$ (km/h) satisfy the function relationship: $t=\frac{k}{v}$, and its graph is a curve as shown, with endpoints at $A(40, 1)$ and $B(m, 0.5)$. If the driving speed cannot exceed $50\text{km/h}$, what is the minimum amount of time needed for the car to pass this section of the road?

Solution:
\textbf{Solution:} Substituting $v=50$ into $t=\frac{40}{v}$, we get $t=\frac{4}{5}$,\\
since $t$ decreases as $v$ increases,\\
therefore, the minimum time required for the car to pass through the section is $\boxed{\frac{4}{5}}$ hours.


Question: As shown in the figure, the graph of the linear function $y=k_1x-4$ intersects with the graph of the hyperbolic function $y=\frac{k_2}{x}$ (where $x>0$) at point $A(3, -6)$, and intersects with the $x$-axis at point $B$. Point $D$ is a point on segment $AB$, and lines $OD$ and $OA$ are connected, with $\frac{S_{\triangle BOD}}{S_{\triangle BOA}}=1\colon 3$. Find the analytic expressions of the linear and hyperbolic functions?

Solution:
\textbf{Solution:} Substituting point A (3, −6) into $y=k_1x-4$,\\
we get $-6=3k_1-4$,\\
from which we find $k_1=-\frac{2}{3}$,\\
Substituting point A (3, −6) into $y=\frac{k_2}{x}$ (where $x>0$), we get $-6=\frac{k_2}{3}$,\\
$\therefore k_2=-18$,\\
$\therefore$ the equation of the linear function is $\boxed{y=-\frac{2}{3}x-4}$, and the equation of the inverse proportion function is $\boxed{y=-\frac{18}{x}}$.


Question: Given the diagram, the graph of the linear function $y=k_1x-4$ intersects with the graph of the inverse proportion function $y=\frac{k_2}{x}$ ($x>0$) at point $A(3, -6)$, and intersects with the x-axis at point $B$. Point $D$ lies on line segment $AB$. Connecting $OD$ and $OA$, and given that the area ratio $S_{\triangle BOD}: S_{\triangle BOA}=1:3$, what are the coordinates of point $D$?

Solution:
\textbf{Solution:} As shown in the figure, draw $DM\perp x$-axis through point D, with M being the foot of the perpendicular, and draw $AN\perp x$-axis through point A, with N being the foot of the perpendicular, \\
$\because \frac{S_{\triangle BOD}}{S_{\triangle BOA}}=\frac{\frac{1}{2}OB\cdot DM}{\frac{1}{2}OB\cdot AN}=\frac{1}{3}$, \\
$\therefore \frac{DM}{AN}=\frac{1}{3}$, \\
$\because$ the coordinates of point A are (3, -6), \\
$\therefore AN=6$, \\
$\therefore DM=2$, i.e., the ordinate of point D is -2, \\
Substituting $y=-2$ into $y=-\frac{2}{3}x-4$, \\
we get $x=-3$, \\
$\therefore$ the coordinates of point D are $\boxed{(-3, -2)}$;


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

Solution:
Solution: Since the parabola $y=x^{2}+bx+c$ intersects the x-axis at points $A(-3,0)$ and $B(1,0)$, \\
it follows that the roots of $x^{2}+bx+c=0$ are $-3$ and $1$. \\
Hence, $c=-3\times 1=-3$, and $b=-(-3+1)=2$. \\
Therefore, the equation of the parabola is \boxed{y=x^{2}+2x-3};


Question: As shown in the figure, the cross-section of a tunnel is composed of a parabola and a rectangle. The rectangle has a length of 8m and a width of 2m. The highest point of the tunnel, point P, is located at the center of AB and is 6m above the ground. Establish the coordinate system as shown in the figure. Find the expression of the parabola.

Solution: \textbf{Solution:} Given the problem, we know that the vertex of the parabola is at $(4,6)$. Assume the equation of the parabola is $y=a(x-4)^2+6$. Since point A$(0,2)$ lies on the parabola, it follows that $2=a(0-4)^2+6$. Solving this gives $a=-\frac{1}{4}$. Therefore, the equation of the parabola is: $y=-\frac{1}{4}(x-4)^2+6=\boxed{-\frac{1}{4}(x-4)^2+6}$.


Question: As shown in the figure, water is sprayed out from window A of a building, forming a parabola (the plane of the parabola is perpendicular to the wall). The height of point A from the ground is 6 meters, the vertical distance of the highest point P from the wall is 2 meters, and its vertical distance from the ground is 8 meters. A Cartesian coordinate system is established as shown in the figure. What is the analytical equation of the parabola?

Solution:
Solution: From the given information, we know that the vertex of the parabola is at $(2,8)$, and it passes through point $A(0,6)$,\\
$\therefore$let the equation of the parabola be $y=a(x-2)^2+8$,\\
Substituting $A(0,6)$ into the equation, we get $4a+8=6$,\\
Solving for $a$, we find $a=-\frac{1}{2}$,\\
$\therefore$the equation of the parabola is $y=-\frac{1}{2}(x-2)^2+8$,\\
Hence, the equation of the parabola is $\boxed{y=-\frac{1}{2}(x-2)^2+8}$.


Question: As shown in the figure, water is sprayed outwards from the window $A$ of a building, forming a parabolic trajectory (the plane of the parabola is perpendicular to the wall). The height of point $A$ from the ground is $6$ meters. The highest point of the parabola, $P$, is $2$ meters away from the wall in the perpendicular direction, and $8$ meters away from the ground in the vertical direction. A Cartesian coordinate system is established as shown in the figure. Determine the maximum distance from the wall where the water hits the ground, in meters.

Solution:
\textbf{Solution:} Let $y=0$, we get $-\frac{1}{2}(x-2)^2+8=0$,\\
Solving this gives: $x_1=6$, $x_2=-2$ (discard),\\
$\therefore$ Hence, the furthest distance the water falls from the wall is $\boxed{6}$ meters.


Question: As shown in Figure 1, the vertices A and C of the rectangle $OABC$ are located on the positive semi-axes of the x-axis and y-axis, respectively, and point B (4, 3). The graph of the inverse proportion function $y=\frac{k}{x}$ ($x>0$) intersects with $AB$ and $BC$ at points $D$ and $E$, respectively, with $BD=1$. Let point $P$ be a moving point on the line segment $OA$. Determine the equation of the inverse proportion function and the coordinates of point $E$?

Solution:
\textbf{Solution:} Because the coordinates of point B are (4, 3),\\
thus OC=AB=3, OA=BC=4.\\
Because BD=1,\\
thus AD=2,\\
thus the coordinates of point D are (4, 2).\\
Because the graph of the inverse proportion function $y= \frac{k}{x}$ (where $x>0$) passes through point D,\\
thus $k=4\times 2=8$,\\
thus the equation of the inverse proportion function is $y= \frac{8}{x}$.\\
When $y=3$, $3= \frac{8}{x}$, solving this gives: $x= \frac{8}{3}$,\\
thus the coordinates of point E are $\boxed{\left( \frac{8}{3}, 3 \right)}$.


Question: As shown in Figure 1, the vertices $A$ and $C$ of rectangle $OABC$ are located on the positive half-axes of the $x$-axis and $y$-axis, respectively, with point $B(4,3)$. The graph of the inverse proportion function $y=\frac{k}{x}$ ($x>0$) intersects $AB$ and $BC$ at points $D$ and $E$ respectively, with $BD=1$. Point $P$ is a moving point on segment $OA$. As shown in Figure 2, when $PE$ and $PD$ are connected, what is the minimum value of $PD+PE$?

Solution:
\textbf{Solution:} In Figure 2, construct the symmetric point $D'$ of point D about the x-axis. Draw a line from $D'$ to E, intersecting the x-axis at point P. Connect PD. At this point, $PD+PE$ reaches its minimum value, which is equal to $D'E$. \\
$\because$ The coordinates of point D are (4, 2), \\
$\therefore$ The coordinates of point $D'$ are (4, −2). \\
Also, $\because$ The coordinates of point E are $\left(\frac{8}{3}, 3\right)$, \\
$\therefore D'E=\sqrt{\left(4-\frac{8}{3}\right)^{2}+\left(-2-3\right)^{2}}=\frac{\sqrt{241}}{3}$. \\
$\therefore$ The minimum value of $PD+PE$ is $\boxed{\frac{\sqrt{241}}{3}}$.


Question: As shown in the figure, vertices $A$ and $C$ of rectangle $OABC$ lie on the positive semiaxis of the $x$-axis and $y$-axis, respectively, with point $B(4,3)$. The graph of the inverse proportion function $y=\frac{k}{x}$ ($x>0$) intersects lines $AB$ and $BC$ at points $D$ and $E$, respectively, with $BD=1$. Point $P$ is a moving point on line segment $OA$. When $\angle PDO=45^\circ$, find the length of line segment $OP$.

Solution:
\textbf{Solution}: In the figure, draw $PF\perp OD$ at point F, then $\triangle PDF$ is an isosceles right triangle.\\
$\because$OA=4, AD=2,\\
$\therefore$OD=$\sqrt{OA^2+AD^2}=2\sqrt{5}$.\\
Let AP=m, then OP=4−m,\\
$\therefore$PD=$\sqrt{AD^2+AP^2}=\sqrt{4+m^2}$.\\
$\because$$\triangle PDF$ is an isosceles right triangle,\\
$\therefore$DF=PF=$\frac{\sqrt{2}}{2}PD=\frac{\sqrt{8+2m^2}}{2}$,\\
$\therefore$OF=OD−DF=$2\sqrt{5}-\frac{\sqrt{8+2m^2}}{2}$.\\
$\because$OF$^2$+PF$^2$=OP$^2$, that is, $(2\sqrt{5}-\frac{\sqrt{8+2m^2}}{2})^2+(\frac{\sqrt{8+2m^2}}{2})^2=(4−m)^2$,\\
After simplification, we get: $3m^2+16m-12=0$,\\
Solving, we find: $m_1=\frac{2}{3}$, $m_2=-6$ (discard as it is not meaningful for this problem),\\
$\therefore$OP=4−m=$\boxed{\frac{10}{3}}$.


Question: As shown in the figure, in the Cartesian coordinate system \(xOy\), the graph of the linear function \(y=x\) intersects the graph of the quadratic function \(y=-x^2+bx\) (where \(b\) is a constant) at points \(O\) and \(A\). The coordinates of point \(A\) are \((3, m)\). What is the value of \(m\) and the expression of the quadratic function?

Solution:
\textbf{Solution:}

Substitute the coordinates of point A, $(3,m)$, into the linear function $y=x$ to get:\\
$m=3$,\\
$\therefore A(3,3)$,\\
Substituting the coordinates of point A, $(3,3)$, into the quadratic function $y=-x^{2}+bx$ gives:\\
$3=-9+3b$,\\
Solving for $b$ yields: $b=4$,\\
$\therefore y=-x^{2}+4x$,\\
Answer: The value of $m$ is $\boxed{3}$, and the expression for the quadratic function is: $y=-x^{2}+4x$;


Question: As shown in the Cartesian coordinate system $xOy$, the graph of the linear function $y=x$ intersects with the graph of the quadratic function $y=-x^2+bx$ (where $b$ is a constant) at points $O$ and $A$. The coordinates of point $A$ are $(3, m)$. If point $P$ is the vertex of the parabola, and lines $OP$ and $AP$ are drawn, what is the area of $\triangle POA$?

Solution:
\textbf{Solution:} Construct $PC\perp x$-axis through point P, with the foot at C, and intersecting OA at point D. Construct $AE\perp PC$ through point A, with the foot at E,\\
$\because y=-x^{2}+4x=-(x-2)^{2}+4$,\\
$\therefore$ vertex P$(2,4)$,\\
Substituting $x=2$ into $y=x$ yields:\\
$y=2$,\\
$\therefore D(2,2)$,\\
$\therefore PD=4-2=2$,\\
$\because$ the area of $\triangle POA$ equals the sum of the areas of $\triangle OPD$ and $\triangle APD$,\\
$\therefore$ the area of $\triangle POA$ $=\frac{1}{2}PD\cdot OC+\frac{1}{2}PD\cdot AE$\\
$=\frac{1}{2}PD(OC+AE)$\\
$=\frac{1}{2}\times 2\times 3$\\
$=\boxed{3}$,\\


Question: As shown in Figure 1, the graph of the linear function $y=kx-3$ ($k\neq 0$) intersects the y-axis at point B and intersects the graph of the inverse proportion function $y=\frac{m}{x}$ ($x>0$) at point A (8,1). What are the analytical expressions of the linear function and the inverse proportion function?

Solution:
\textbf{Solution:} Given that point A$(8,1)$ lies on the line $y=kx-3$,\\
therefore $1=8k-3$,\\
solving for $k$, we find $k=\frac{1}{2}$,\\
thus, the linear function can be expressed as $\boxed{y=\frac{1}{2}x-3}$,\\
since A$(8,1)$ also lies on the graph of $y=\frac{m}{x}$ (for $x>0$),\\
therefore $1=\frac{m}{8}$,\\
solving for $m$, we obtain $m=8$,\\
hence, the inverse proportionality function is expressed as $\boxed{y=\frac{8}{x}}$;


Question: As shown in Figure 1, the graph of the linear function $y=kx-3$ (where $k\neq 0$) intersects the y-axis at point B and intersects the graph of the inverse proportion function $y=\frac{m}{x}$ (where $x>0$) at point A (8, 1). Point C is located on the line segment AB (distinct from points A and B). A line parallel to the y-axis passing through point C intersects the graph of the inverse proportion function at point D. Lines OC, OD, and AD are connected. When CD equals 6, what are the coordinates of point C and the area of $\triangle ACD$?

Solution:
\textbf{Solution:} Let C$(a, \frac{1}{2}a-3)$ (where $0<a<8$), then we have D$(a, \frac{8}{a})$, \\
$\therefore$CD$=\frac{8}{a}-\left(\frac{1}{2}a-3\right)=\frac{8}{a}-\frac{1}{2}a+3$, \\
$\because$CD$=6$, \\
$\therefore \frac{8}{a}-\frac{1}{2}a+3=6$, \\
Solving this, we get: $a=-8$ (discard) or $a=2$, \\
$\therefore \frac{1}{2}a-3=1-3=-2$, \\
$\therefore$C$(2, -2)$, \\
Draw AE$\perp$CD at point E from A, then AE$=8-2=6$, \\
$\therefore$Area of $\triangle ACD=\frac{1}{2}\times \text{CD}\times \text{AE}=\frac{1}{2}\times 6\times 6=\boxed{18}$;


Question: As shown in Figure 1, the graph of the linear function $y=kx-3$ ($k\neq 0$) intersects the y-axis at point B and intersects the graph of the inverse proportion function $y=\frac{m}{x}$ ($x>0$) at point A (8, 1). Assuming condition (2), by translating $\triangle OCD$ a certain distance in the direction of ray BA, we obtain $\triangle O'CD'$. If the corresponding point of point O, O', falls exactly on the graph of the inverse proportion function (as shown in Figure 2), determine the coordinates of points O' and D'.

Solution:
\textbf{Solution:} Connect $OO'$, due to the translation we have: $OO' \parallel AC$, \\
therefore, the equation of line $OO'$ is $y= \frac{1}{2}x$, \\
solving simultaneously, we get: $\left\{\begin{array}{l}y=\frac{8}{x}\\ y=\frac{1}{2}x\end{array}\right.$, \\
this yields: $\left\{\begin{array}{l}x=4\\ y=2\end{array}\right.$ or $\left\{\begin{array}{l}x=-4\\ y=-2\end{array}\right.$ (discard as it is irrelevant to the question), \\
therefore, the coordinates of $O'$ are $\boxed{(4,2)}$, \\
which means point $O$ (0,0) is translated 4 units to the right and 2 units up to obtain $O'$ (4,2), \\
furthermore, from (2) we know the coordinates of D are (2,4), \\
therefore, point D (2,4) translated 4 units to the right and 2 units up gives the coordinates of $D'$ as $\boxed{(6,6)}$.


Question: As shown in Figure 1, it is known that the line $l$: $y=-x+6$ intersects with the $x$ axis at point $A$, and with the $y$ axis at point $B$. The line $m$ intersects with the $y$ axis at point $C(0, -2)$, and intersects with the line $l$ at point $D(t, 1)$. What is the analytical expression of line $m$?

Solution:
\textbf{Solution:} Let $\because D(5, 1)$ lies on the line $l: y = -x + 6$,\\
$\therefore 1 = -5 + 6$,\\
$\therefore t = 5$,\\
$\therefore D(5, 1)$,\\
Let the equation of line $m$ be $y = kx + b$,\\
Substituting points C and D, we get,\\
$\begin{cases}
5k + b = 1\\
b = -2
\end{cases}$,\\
Solving this, we find,\\
$\begin{cases}
k = \frac{3}{5}\\
b = -2
\end{cases}$,\\
Therefore, the equation of line $m$ is $y = \boxed{\frac{3}{5}x - 2}$.


Question: As illustrated in Figure 1, it is known that line $l$: $y=-x+6$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$. Line $m$ intersects the $y$ axis at point $C(0, -2)$ and intersects line $l$ at point $D(t, 1)$. As shown in Figure 2, point $P$ is on line $l$ and to the left of the $y$ axis. A line is drawn from point $P$ parallel to the $y$ axis intersecting line $m$ at point $Q$ and the $x$ axis at point $G$. When ${S}_{\triangle PCG}=2{S}_{\triangle QCG}$, determine the coordinates of points $P$ and $Q$.

Solution:
\textbf{Solution:} Let $P(a, 6-a)$. Since point $P$ is on the left side of the x-axis, we have $a<0$. Given $PQ\parallel$ to the y-axis, let $G(a,0)$, and $Q(a, \frac{3}{5}a-2)$. As illustrated, points $P$ and $Q$ are on opposite sides of the x-axis.

Since the area of $\triangle PCG$ is $S_{\triangle PCG}=\frac{1}{2}PG\cdot(-a)$, and the area of $\triangle QCG$ is $S_{\triangle QCG}=\frac{1}{2}GQ\cdot(-a)$, and given that $S_{\triangle PCG}=2S_{\triangle QCG}$,

it follows that $PG=2QG$,

leading to $6-a=2\left(2-\frac{3}{5}a\right)$.

Solving for $a$, we find $a=\boxed{-10}$,

Therefore, $6-a=6-(-10)=\boxed{16}$, and $\frac{3}{5}a-2=\frac{3}{5}\times (-10)-2=\boxed{-8}$.

Thus, the coordinates of point $P$ are $\boxed{(-10, 16)}$, and the coordinates of point $Q$ are $\boxed{(-10, -8)}$.


Question: Given the graph of a linear function $y=kx+b$ intersects with the graph of the inverse proportional function $y=\frac{m}{x}$ $(x>0)$ at point A, and intersects with the x-axis at point B $(5,0)$. If $OB=AB$ and the area of $\triangle OAB$ is $\frac{15}{2}$, find the expressions of the inverse proportional function and the linear function.

Solution:
\textbf{Solution:} Let's consider the diagram, draw $AM\perp x$-axis at M,

$\because B(5,0)$,

$\therefore OB=5$,

$\because S_{\triangle OAB}=\frac{15}{2}$,

$\therefore \frac{1}{2} \cdot OB \cdot AM=\frac{15}{2}$,

$\therefore AM=3$,

$\because OB=AB$,

$\therefore AB=5$,

Within $\triangle AMB$, by applying the Pythagorean theorem, $BM=\sqrt{AB^2-AM^2}=4$,

$\therefore OM=OB+BM=9$,

$\therefore A(9,3)$,

$\because$ point A is on the graph of the inverse proportion function $y=\frac{m}{x}$,

$\therefore m=9\times 3=27$,

$\therefore$ the expression for the inverse proportion function is $\boxed{y=\frac{27}{x}}$;

$\because$ points A$(9,3)$, B$(5,0)$ are on the graph of the linear function $y=kx+b$,

$\therefore \left\{\begin{array}{l}9k+b=3\\ 5k+b=0\end{array}\right.$,

$\therefore \left\{\begin{array}{c}k=\frac{3}{4}\\ b=-\frac{15}{4}\end{array}\right.$,

$\therefore$ the expression for the linear function is $\boxed{y=\frac{3}{4}x-\frac{15}{4}}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ intersects with the graph of the hyperbolic function $y=\frac{m}{x}$ $(x>0)$ at point A and with the x-axis at point B$(5,0)$. If $OB=AB$ and the area of $\triangle OAB$ is $\frac{15}{2}$, directly state the solution set for the inequality $\frac{m}{x}>kx+b$ when $x>0$.

Solution: \textbf{Solution:} Since point A is (9, 3), from the graph, the solution set of $\frac{m}{x} > kx + b$ is $\boxed{0<x<9}$.


Question: Given a linear function $y=kx+b$ whose graph intersects with the graph of the inverse proportion function $y=\frac{m}{x}$ $(x>0)$ at point A, and intersects with the x-axis at point B$(5,0)$. If $OB=AB$, and the area of $\triangle OAB$ is $\frac{15}{2}$. If point P is on the y-axis, and point Q is on the graph of the inverse proportion function $y=\frac{m}{x}$ $(x>0)$, and the quadrilateral ABPQ is precisely a parallelogram, directly write down the coordinates of point P.

Solution:
\textbf{Solution:} According to the given problem, PQ is located to the top left of AB. Let P be $(0,t)$, then PQ can be seen as moving AB 5 units to the left, and then t units up.\\
$\therefore$ The coordinates of point Q are $(4,3+t)$\\
$\because$ Point Q is on the graph of $y= \frac{27}{x}$,\\
$\therefore$ $3+t=\frac{27}{4}$\\
$\therefore$ $t=\frac{15}{4}$\\
$\therefore$ $P\boxed{\left(0,\frac{15}{4}\right)}$


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=-\frac{1}{2}x$ intersects the graph of the inverse proportion function $y=\frac{k}{x}$ at two points $A$ and $B$ that are symmetric about the origin. Given that the y-coordinate of point $A$ is 3, what is the expression for the inverse proportion function?

Solution:
\textbf{Solution:} Let the linear function be $y=-\frac{1}{2}x$ and set $y=3$, then $3=-\frac{1}{2}x$. Solving this, we get: $x=-6$, thus the coordinates of point $A$ are $\left(-6, 3\right)$.\\
Since point $A\left(-6, 3\right)$ lies on the graph of the inverse proportion function $y=\frac{k}{x}$,\\
it follows that $k=-6\times 3=-18$,\\
hence, the expression for the inverse proportion function is $\boxed{y=-\frac{18}{x}}$.


Question: In the figure, in the Cartesian coordinate system, the line $y=-\frac{1}{2}x$ intersects with the graph of the inverse proportion function $y=\frac{k}{x}$ at points $A$ and $B$, which are symmetric about the origin. It is known that the ordinate of point $A$ is 3. Based on the graph, directly write the solution set of $-\frac{1}{2}x<\frac{k}{x}$.

Solution:
\textbf{Solution:} Solve the system of equations:
\[
\begin{cases}
y=-\frac{1}{2}x \\
y=\frac{-18}{x}
\end{cases}
\]
The solutions are:
\[
\begin{cases}
x=-6 \\
y=3
\end{cases}
\]
or
\[
\begin{cases}
x=6 \\
y=-3
\end{cases}
\]
Therefore, for point $B(6, -3)$, when $-6<x<0$ or $x>6$, we have $-\frac{1}{2}x<\frac{k}{x}$. Hence, the solution set for $-\frac{1}{2}x<\frac{k}{x}$ is \boxed{-6<x<0} or \boxed{x>6}.


Question: As shown in the Cartesian coordinate system, the line $y=-\frac{1}{2}x$ intersects with the inverse proportion function $y=\frac{k}{x}$ at two points, $A$ and $B$, which are symmetrical about the origin. It is known that the ordinate of point $A$ is 3. After the line $y=-\frac{1}{2}x$ is translated upwards, it intersects with the inverse proportion function in the second quadrant at point $C$. If the area of $\triangle ABC$ is 36, what is the equation of the translated line?

Solution:
\textbf{Solution:} Set the line after translation intersects the $y$-axis at point $F$, and connect $AF$, $BF$ as shown in the figure. Assume the equation of the line after translation is $y=-\frac{1}{2}x+b$,\\
$\because$ this line is parallel to line $AB$,\\
$\therefore S_{\triangle ABC}=S_{\triangle ABF}$,\\
$\because$ the area of $\triangle ABC$ is 36,\\
$\therefore S_{\triangle ABF}=\frac{1}{2}OF\cdot (x_{B}-x_{A})=36$,\\
By symmetry, it is known: $x_{B}=-x_{A}$,\\
$\because x_{A}=-6$,\\
$\therefore x_{B}=6$,\\
$\therefore \frac{1}{2}b\times 12=36$,\\
$\therefore b=6$.\\
$\therefore$ The equation of the line after translation is $\boxed{y=-\frac{1}{2}x+6}$.


Question: As shown in the diagram, 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 and B, and intersects with the x-axis at point C. The coordinates of point A are (2, 3), and the coordinates of point B are (-6, n). What are the analytic expressions for the linear function and the inverse proportion function?

Solution:
\textbf{Solution:} Given that the graph of the inverse proportion function $y=\frac{k}{x}$ passes through point A(2, 3),\\
we have $k=6$,\\
thus, the expression for the inverse proportion function is $y=\frac{6}{x}$.\\
Since point B(-6, n) lies on the graph of the inverse proportion function $y=\frac{6}{x}$,\\
it follows that $n=-1$.\\
Substituting A(2, 3) and B(-6, -1) into $y=kx+b$, we get
\[
\left\{
\begin{array}{l}
2k+b=3, \\
-6k+b=-1,
\end{array}
\right.
\]
From which we derive
\[
\left\{
\begin{array}{l}
k=\frac{1}{2}, \\
b=2,
\end{array}
\right.
\]
Therefore, the expression for the linear function is $\boxed{y=\frac{1}{2}x+2}$ and the expression for the inverse proportion function is $\boxed{y=\frac{6}{x}}$.


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 and B, and intersects with the x-axis at point C. The coordinates of point A are $(2, 3)$, and the coordinates of point B are $(-6, n)$. Connect AO and OB, then find the area of $\triangle AOB$.

Solution:
\textbf{Solution:} When $y=0$, $x=4$, \\
$\therefore C(-4,0)$, \\
$\therefore OC=4$. Construct $AD\perp x$ axis at point $D$, and $BE\perp x$ axis at point $E$. \\
Then, $AD=3$, $BE=1$, \\
$\therefore$ the area of $\triangle AOB$ is $S_{\triangle AOB} = S_{\triangle BOC} + S_{\triangle AOC} = \frac{1}{2} \times 4 \times 1 + \frac{1}{2} \times 4 \times 3 = \boxed{8}$.


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 and B, and intersects with the x-axis at point C. The coordinate of point A is $(2, 3)$, and the coordinate of point B is $(-6, n)$. Based on the graph, directly write down the solution set of the system of inequalities $\frac{m}{x}\le kx+b<0$.

Solution: \textbf{Solution:} From the graph, it is known that the graph of the inverse proportion function $y=\frac{m}{x}$ is below that of the linear function $y=kx+b$. The range of the variable $x$, which is less than 0, is: $-6\le x<-4$. Therefore, the solution set is $\boxed{-6\le x<-4}$.


Question: As shown in the figure, the line $AB$ intersects the x-axis at point A and the y-axis at point B, where $\angle OAB=30^\circ$, and the coordinates of point B are $(0, 2\sqrt{3})$. The line $y=kx+b$ passes through point A and intersects the y-axis at point C, and $OC=OA$. What is the equation of the line $AC$?

Solution: \textbf{Solution:} Given $B(0,2\sqrt{3})$, we have $OB=2\sqrt{3}$. In $\triangle AOB$, $\angle OAB=30^\circ$, hence $AB=2OB=4\sqrt{3}$. Therefore, $AO=\sqrt{AB^{2}-OB^{2}}=6$, resulting in $A(6,0)$. Since $OA=6$ and $OA=OC$, we get $OC=6$ and consequently $C(0,-6)$. Assuming the equation of line AC is $y=kx-6$ and substituting point A into it gives $k=1$. Therefore, the equation of line AC is $y=x-6$. Hence, the answer is $\boxed{y=x-6}$.


Question: As illustrated, the line $AB$ intersects the x-axis at point A and the y-axis at point B, with $ \angle OAB=30^\circ$. The coordinates of point B are $(0, 2\sqrt{3})$. The line $y=kx+b$ passes through point A and intersects the y-axis at point C, and $OC=OA$. Point D is on the perpendicular bisector $l$ of segment $AB$, located in the first quadrant. By folding $\triangle ABD$ along $BD$, $\triangle A'BD$ is obtained. If point $A'$ precisely falls on the line $l$, find the coordinates of points D and $A'$.

Solution:
\textbf{Solution:} As shown in Figure 1, connect $A$ to $A'$,

$\because$ Folding $\triangle ABD$ along $BD$ results in $\triangle A'BD$,

$\therefore$ $AB = A'B$, and $BD$ bisects $A$ to $A'$ perpendicularly, $\angle ABD = \angle A'BD$,

Again, $\because$ $l$ is the perpendicular bisector of $AB$, and $A'$ is a point on line $l$,

$\therefore$ $A'B = AA'$,

$\therefore$ $A'B = AB = AA'$,

$\therefore$ $\triangle AA'B$ is an equilateral triangle,

$\therefore$ $\angle ABD = \angle OAB = 30^\circ$,

$\therefore$ $BD \parallel OA$,

$\therefore$ The ordinate of $D$ is $2\sqrt{3}$,

$\because$ $l$ is the perpendicular bisector of $AB$, and $D$ is a point on line $l$,

$\therefore$ $DB = DA = A'D$,

$\therefore$ $\angle DBA = \angle DAB = 30^\circ$,

$\therefore$ $\angle DAO = 60^\circ$,

Construct $DM \perp OA$ at $M$,

$\therefore$ $\angle ADM = 90^\circ - \angle DAO = 30^\circ$,

Let $AM = a$, then $AD = 2a$,

$\because$ $AD^2 - AM^2 = DM^2$,

$\therefore$ $3a^2 = 12$,

$\therefore$ $a = 2$,

$\therefore$ $AM = 2$,

$\therefore$ $OM = 6 - AM = 4$,

$\therefore$ The coordinates of $D$ are $\boxed{(4, 2\sqrt{3})}$,

$\because$ $\angle A'AB = 60^\circ$,

$\therefore$ $\angle A'AO = \angle A'AB + \angle BAO = 90^\circ$,

$\therefore$ $AA' \perp OA$,

Also, $AA' = AB = 4\sqrt{3}$,

$\therefore$ The coordinates of $A'$ are $\boxed{(6, 4\sqrt{3})}$.


Question: As shown in the figure, line $AB$ intersects the x-axis at point A and the y-axis at point B, where $ \angle OAB=30^\circ$, and the coordinates of point B are $(0, 2\sqrt{3})$. The line $y=kx+b$ passes through point A and intersects the y-axis at point C, with $OC=OA$. Let P be a point on line $AC$, and Q be a point on line $l$. When $\triangle APQ$ is an equilateral triangle, write down the side length of $\triangle APQ$.

Solution:
\textbf{Solution:}

(1) As shown in Figure 2, when P is on the extension line of segment CA, connect $A'$P, and $A'$ with BQ,\\
$\because \triangle APQ$ is an equilateral triangle,\\
$\therefore AP = AQ = PQ$,\\
$\angle BAA' = \angle PAQ = 60^\circ$,\\
$\therefore \angle BAQ = \angle A'AP$,\\
In $\triangle ABQ$ and $\triangle AA'P$,\\
$\left\{\begin{array}{l}
AB = AA' \\
\angle BAQ = \angle A'AP \\
AQ = AP
\end{array}\right.$,\\
$\therefore \triangle ABQ \cong \triangle AA'P$ (SAS),\\
$\therefore BQ = A'P$,\\
Also, $\because Q$ is a point on the perpendicular bisector of AB,\\
$\therefore BQ = AQ$,\\
$\therefore AQ = A'P$,\\
Also, $\because AP = AQ$,\\
$\therefore AP = A'P$,\\
$\therefore P$ is on the perpendicular bisector of $AA'$,\\
$\because AA' \perp OA$, and $AA' = 4\sqrt{3}$,\\
$\therefore$ The ordinate of P is $2\sqrt{3}$,\\
Let $y = 2\sqrt{3}$, then $x-6 = 2\sqrt{3}$,\\
$\therefore x = 6 + 2\sqrt{3}$,\\
$\therefore P(6+2\sqrt{3}, 2\sqrt{3})$,\\
$\therefore AP = \sqrt{(6+2\sqrt{3}-6)^2+(2\sqrt{3})^2} = 2\sqrt{6}$,\\
(2) As shown in Figure 3, when P is on the ray AC, connect BP,\\
Similarly, we can get $\triangle ABP \cong AA'Q$,\\
$\therefore PB = QA'$, $\angle PBA = \angle QA'A$,\\
$\because \angle BAA' = 60^\circ$, and $A'Q$ vertically bisects AB,\\
$\therefore \angle QA'A = 30^\circ$,\\
$\therefore \angle PBA = 30^\circ$,\\
Also, $\angle ABO = 90^\circ - \angle BAO = 60^\circ$,\\
$\therefore \angle PBO = \angle ABO - \angle PBA = 30^\circ$,\\
Draw $PG \perp y$-axis at point G from P, let P$(m, m-6)$,\\
In $\triangle PBG$, $\angle PBO = 30^\circ$,\\
$\therefore PB = 2PG = 2m$,\\
$\therefore BG = \sqrt{3}m$,\\
$\therefore \sqrt{3}m = 2\sqrt{3}-(m-6)$,\\
$\therefore m = 2\sqrt{3}$,\\
$\therefore P(2\sqrt{3}, 2\sqrt{3}-6)$,\\
$\therefore AP = \sqrt{(2\sqrt{3}-6)^2+(2\sqrt{3}-6)^2} = 6\sqrt{2}-2\sqrt{6}$.\\
In summary: The side length of $\triangle APQ$ is either \boxed{2\sqrt{6}} or \boxed{6\sqrt{2}-2\sqrt{6}}.


Question: As shown in the figure, it is known that the graph of the function $y=x+1$ intersects the $y$ axis at point $A$. The graph of the linear function $y=kx+b$ passes through point $B(0, -1)$ and intersects the $x$ axis and the graph of $y=x+1$ at points $C$ and $D$ respectively. If the $x$-coordinate of point $D$ is 1, determine the analytical expression of the linear function $y=kx+b$.

Solution:
\textbf{Solution}: From the problem, we know that the x-coordinate of point $D$ is 1, and point $D$ lies on the line, thus $y=x+1$.

$\therefore$ Substituting $x=1$ into $y=x+1$, we get $y=2$.

$\therefore D(1, 2)$.

Substituting points $B(0, -1)$ and $D(1, 2)$ into $y=kx+b$, we obtain $\begin{cases}b=-1 \\ k+b=2\end{cases}$. Solving these equations, we find $\begin{cases}k=3 \\ b=-1\end{cases}$.

$\therefore$ The equation of line $BD$ is $\boxed{y=3x-1}$.


Question: As shown in the diagram, it is known that the graph of the function $y=x+1$ intersects the $y$-axis at point $A$. The graph of the linear function $y=kx+b$ passes through point $B(0, -1)$ and intersects the $x$-axis and the graph of $y=x+1$ at points $C$ and $D$, respectively. Find the area of quadrilateral $AOCD$ (i.e., the area of the shaded region in the diagram).

Solution:
\textbf{Solution:} From equation (1), substituting $x=0$ into $y=x+1$, we get $y=1$,\\
$\therefore A(0, 1)$. Substituting $y=0$ into $y=3x-1$ gives $x=\frac{1}{3}$,\\
$\therefore C\left(\frac{1}{3}, 0\right)$,\\
$\therefore$ the area of quadrilateral $AOCD=S_{\triangle AOD}+S_{\triangle COD}$\\
$=\frac{1}{2}\times 1\times 1+\frac{1}{2}\times \frac{1}{3}\times 2=\boxed{\frac{5}{6}}$


Question: As shown in the figure, the line $y=-x+m$ intersects the $x$ axis at point $B(4,0)$ and the $y$ axis at point $A$. Point $C$ is a point on the $x$ axis, and it is known that the area of $\triangle ABC$ is $4$. What is the equation of line $AB$?

Solution:
The solution is as follows: Substitute point B$(4,0)$ into $y=-x+m$ to obtain $-4+m=0$,\\
solving for $m$ gives $m=4$,\\
$\therefore$ the equation of line $AB$ is $\boxed{y=-x+4}$.


Question: As shown in the figure, the line $y = -x + m$ intersects the $x$-axis at point $B(4, 0)$ and the $y$-axis at point $A$. Point $C$ is on the $x$-axis, and the area of $\triangle ABC$ is known to be 4. Find the coordinates of point $C$.

Solution:
\textbf{Solution:} Given $x=0$, then $y=-x+4=4$, thus point $A(0,4)$,\\
since $S_{\triangle ABC}=4$,\\
thus $\frac{1}{2} BC \cdot 4=4$, solving gives $BC=2$,\\
since point $B(4,0)$,\\
thus $OB=4$,\\
thus $OC=OB-BC=4-2=2$ or $OC=OB+BC=4+2=6$,\\
thus point $\boxed{C(2,0)}$ or $\boxed{(6,0)}$.


Question: As shown in Figure 1, the graph of the linear function \(y=-\frac{1}{2}x+4\) intersects the \(x\)-axis and \(y\)-axis at point \(A\) and point \(B\), respectively, and intersects the graph of the direct proportion function \(y=\frac{3}{2}x\) at point \(C\). Point \(C\) is translated 1 unit to the right and 6 units downward to obtain point \(D\). Find the perimeter of \(\triangle OAB\) and the coordinates of point \(D\).

Solution:
\textbf{Solution:} Solve: In the equation $y=-\frac{1}{2}x+4$, when $x=0$, we get $y=4$, and when $y=0$, from $0=-\frac{1}{2}x+4$ we have: $x=8$,\\
$\therefore$ Points A(8,0) and B(0,4).\\
$\therefore$ $OA=8$, $OB=4$.\\
$\therefore$ $AB=\sqrt{OA^{2}+OB^{2}}=\sqrt{8^{2}+4^{2}}=4\sqrt{5}$.\\
$\therefore$ The perimeter of $\triangle OAB$ $=OA+OB+AB=12+4\sqrt{5}=\boxed{12+4\sqrt{5}}$.\\
By solving the system of equations $y=-\frac{1}{2}x+4$ and $y=\frac{3}{2}x$, we get:\\
$\left\{\begin{array}{l}
x=2\\
y=3
\end{array}\right.$.\\
$\therefore$ Point D is (3,−3); The coordinates of point D are $\boxed{(3,-3)}$.


Question: As shown in Figure 1, the graph of the linear function $y=-\frac{1}{2}x+4$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, and intersects the graph of the direct proportion function $y=\frac{3}{2}x$ at point $C$. By moving point $C$ 1 unit to the right and then 6 units downward, point $D$ is obtained. As shown in Figure 2, point $P$ is a moving point on the $y$-axis. When the sum of $CP+PD$ is minimized, what are the coordinates of point $P$?

Solution:
\textbf{Solution:} Let point $D'$ be the symmetric point of point $D$ about the $y$-axis, then $D'(-3, -3)$. Connect $CD'$ and let it intersect the $y$-axis at point $P'$. Connect $P'D$. At this time, $CP'+PD$ is minimal. Let the equation of line $CD'$ be $y=kx+b$. Substituting points $C(2,3)$ and $D'(-3,-3)$, we obtain:
\[
\left\{
\begin{array}{l}
2k+b=3 \\
-3k+b=-3
\end{array}
\right.
\]
Solving this yields $k=\frac{6}{5}$ and $b=\frac{3}{5}$. Therefore, the equation of line $CD'$ is $y=\frac{6}{5}x+\frac{3}{5}$. When $x=0$, we have $y=\frac{3}{5}$. Therefore, the coordinates of point $P'$ are $\boxed{(0, \frac{3}{5})}$, which represent the coordinates of point $P$ when $CP+PD$ is minimized.


Question: As shown in Figure 1, the graph of the linear function $y=-\frac{1}{2}x+4$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and intersects the graph of the direct proportionality function $y=\frac{3}{2}x$ at point $C$. Moving point $C$ 1 unit to the right and then 6 units downward yields point $D$. If point $Q$ is a moving point on the $x$-axis, and $\triangle OQD$ forms an isosceles triangle, directly write the coordinates of point $Q$.

Solution:
\textbf{Solution:} Let point Q be $(x,0)$. Since D is $(3,-3)$ and O is $(0,0)$, it follows that $OD^2=(3-0)^2+(-3-0)^2=18$, and $OQ^2=(x-0)^2+(0-0)^2=x^2$. The distance $DQ^2=(x-3)^2+(0+3)^2=x^2-6x+18$. For $\triangle OQD$ to be an isosceles triangle, there are three cases:
- When $OD=OQ$, from $18=x^2$, we have $x=\pm 3\sqrt{2}$. Hence, $Q(-3\sqrt{2},0)$ or $Q(3\sqrt{2},0)$,
- When $OD=DQ$, from $18=x^2-6x+18$, we get $x=6$ or $x=0$ (coinciding with O, which is disregarded). Therefore, $Q(6,0)$,
- When $OQ=DQ$, from $x^2=x^2-6x+18$, we obtain $x=3$. Therefore, $Q(3,0)$,

In summary, when $\triangle OQD$ is an isosceles triangle, the coordinates of point Q are $\boxed{Q(-3\sqrt{2},0)}$, or $\boxed{Q(3\sqrt{2},0)}$, or $\boxed{Q(3,0)}$, or $\boxed{Q(6,0)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, line $l_1$ intersects the x-axis at point $A$ and the y-axis at point $B$. The coordinates of point $B$ are $(0, 3)$. Line $l_2: y=2x$ intersects line $l_1$ at point $C$, and the x-coordinate of point $C$ is 1. What is the equation of line $l_1$?

Solution: \textbf{Solution:} Since line $l_{2}\colon y=2x$ intersects with line $l_{1}$ at point $C$, and the x-coordinate of point $C$ is 1, therefore $y=2$, thus point $C(1, 2)$. Let the equation of line $l_{1}$ be $y=kx+b$. Substituting the coordinates of point $C$ yields: $\begin{cases}k+b=2\\ b=3\end{cases}$, solving this gives: $\begin{cases}k=-1\\ b=3\end{cases}$, therefore the equation of line $l_{1}$ is $y=-x+3$; hence, the equation of line $l_{1}$ is $\boxed{y=-x+3}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $l_1$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. The coordinates of point $B$ are $(0, 3)$. The line $l_2: y=2x$ intersects line $l_1$ at point $C$, and the x-coordinate of point $C$ is 1. If point $D$ is on the $y$-axis and the area of $\triangle OCD$ is $\frac{2}{3}$ of the area of $\triangle AOC$, what are the coordinates of point $D$?

Solution:
\textbf{Solution:} From (1), we can get the equation of line $l_1$ as $y=-x+3$, and point $C(1, 2)$,\\
$\therefore$ when $y=0$, then $-x+3=0$, solving this we get: $x=3$,\\
$\therefore$ point $A(3, 0)$,\\
$\therefore$ the area of $\triangle AOC$ with OA as the base and the distance from point C to the x-axis as the height, i.e., $S_{\triangle AOC}=\frac{1}{2}\times 3\times 2=3$,\\
$\therefore$ $S_{\triangle OCD}=\frac{2}{3}S_{\triangle AOC}=2$,\\
Let point $D(0, a)$, then the area of $\triangle OCD$ has OD as the base and the distance from point C to the y-axis as the height,\\
$\therefore$ $S_{\triangle OCD}=\frac{1}{2}\times 1\cdot |a|=2$,\\
Solving this, we find: $a=\pm 4$,\\
$\therefore$ $\boxed{D\left(0, 4\right)$ or $D\left(0, -4\right)}$;


Question: As shown in the figure, the graph of the linear function $y_1=x+1$ intersects with the graph of the inverse proportion function $y_2= \frac{2}{x}$ at points $A$ and $B$. A line is drawn from point $A$ perpendicular to the $x$-axis, and the foot of the perpendicular is point $H$. Then, line $BH$ is drawn. What are the coordinates of points $A$ and $B$ and the area of $\triangle ABH$?

Solution: \textbf{Solution:} From the given problem, we have $x+1= \frac{2}{x}$, therefore $x_{1}=1$, $x_{2}=-2$, then $A(1,2)$, $B(-2,-1)$, $H(1,0)$. Let the intersection point of the line $y=x+1$ and the $x$-axis be $C$, thus $C(-1, 0)$. Therefore, $CH=2$, which leads to the area of $\triangle ABH = S_{\triangle ACH} + S_{\triangle BCH} = \frac{1}{2} \times 2 \times 2 + \frac{1}{2} \times 2 \times 1 = \boxed{3}$.


Question: As shown in the figure, the graph of the linear function $y_1=x+1$ intersects with the graph of the inverse proportion function $y_2= \frac{2}{x}$ at points $A$ and $B$. A line is drawn from point $A$ perpendicular to the $x$-axis, forming a perpendicular foot at point $H$, and then connecting $BH$. When $y_1>y_2$, please directly state the range of the independent variable $x$ using the graph.

Solution: Solution: As can be seen from the figure, when $y_1>y_2$, we have \boxed{x>1} or \boxed{-2<x<0}.


Question: As shown in the Cartesian coordinate system, the line $y=\frac{5}{2}x+5$ intersects the x-axis at point A and the y-axis at point B. Another line passing through point B intersects the positive half of the x-axis at point C, and the area of $\triangle ABC$ is 15. Find the coordinates of point C and the equation of line BC.

Solution:
\textbf{Solution:} Solve: The line $y=\frac{5}{2}x+5$ intersects the x-axis at point A and the y-axis at point B, \\
$\therefore A(-2,0), B(0,5)$, \\
that is $OA=2, OB=5$, \\
since the area of $\triangle ABC$ is $15$, \\
$\therefore \frac{1}{2}(OA+OC) \cdot OB=15$, \\
$\therefore OC=4$, \\
$\therefore C(4,0)$, \\
Let the expression of line BC be $y=kx+b$, \\
Substituting the coordinates of points B and C into the linear equation yields: $\left\{\begin{array}{l}4k+b=0 \\ b=5\end{array}\right.$ \\
Solving this, we get: $\left\{\begin{array}{l}k=-\frac{5}{4} \\ b=5\end{array}\right.$ \\
$\therefore$ The expression for line BC is: $y=-\frac{5}{4}x+5$; \\
Hence, the coordinates of point C are $C=\boxed{(4,0)}$, and the expression for line BC is $y=-\frac{5}{4}x+5$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=\frac{5}{2}x+5$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. Another line passing through point $B$ intersects the positive $x$-axis at point $C$, and the area of $\triangle ABC$ is $15$. If $M$ is a point on segment $BC$ and the area of $\triangle ABM$ is equal to the area of $\triangle AOB$, what are the coordinates of $M$?

Solution:
\textbf{Solution:} From the given problem, we have $S_{\triangle ACM} = S_{\triangle ABC} - S_{\triangle ABM} = S_{\triangle ABC} - S_{\triangle ABO} = 15 - \frac{1}{2} \times 2 \times 5 = 10$,\\
$\therefore S_{\triangle ACM} = \frac{1}{2} \times 6 \times y_m = 10$. Solving this, we find: $y_m = \frac{10}{3}$,\\
$\therefore \frac{10}{3} = -\frac{5}{4}x + 5$,\\
Solving for $x_m$, we obtain: $x_m = \frac{4}{3}$,\\
$\therefore M\boxed{\left(\frac{4}{3}, \frac{10}{3}\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=\frac{5}{2}x+5$ intersects the x-axis at point $A$ and the y-axis at point $B$. Another line passing through point $B$ intersects the positive half of the x-axis at point $C$, and the area of $\triangle ABC$ is $15$. Under the condition of (2), point $E$ is a moving point on the line $AM$. Is there a point $D$ on the x-axis such that the quadrilateral with vertices $D$, $E$, $B$, and $C$ is a parallelogram? If it exists, write down the coordinates of point $D$ directly; if not, explain why.

Solution: \textbf{Solution:} There exist points satisfying the condition, and the coordinates of point $D$ are $\boxed{(7,0)}$ or $\boxed{(-11,0)}$ or $\boxed{(1,0)}$.


Question: As shown in the graph, in the cartesian coordinate system, the line $L_1\colon y=-\frac{1}{2}x+8$ intersects the $x$-axis and $y$-axis at points $B$ and $C$, respectively, and intersects the line $L_2\colon y=x$ at point $A$. There is a moving point $P$ on line $L_1$. A perpendicular line to the $x$-axis passing through point $P$ intersects line $L_2$ at point $N$. Find the range of values for $x$ when point $P$ is above point $N$.

Solution:
\textbf{Solution:} Solve: The line $L_{1}\colon y=-\frac{1}{2}x+8$ intersects with the line $L_{2}\colon y=x$ at point $A$,\\
A perpendicular line from point $P$ to the $x$-axis intersects line $L_{2}$ at point $N$. When point $P$ is above point $N$,\\
That is: $-\frac{1}{2}x+8>x$,\\
We find: $\boxed{x<\frac{16}{3}}$;