Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$, respectively, and intersects the graph of the inverse proportion function $y= \frac{m}{x}$ at point $C$ in the first quadrant. The line $CD$ is perpendicular to the $x$ axis, and the foot of the perpendicular is $D$. If $OA=OB=OD=1$, find the expression of this inverse proportion function.

Solution:
\textbf{Solution:} Substitute $C(1,2)$ into $y=\frac{m}{x}$,

$\therefore m=2$,

$\therefore$ the equation of the inverse proportion function is $\boxed{y=\frac{2}{x}}$.


Question: As shown in the figure, it is known that the graph of the linear function $y=kx+b$ intersects the $x$-axis and the $y$-axis at points $A$ and $B$, respectively, and intersects the graph of the inverse proportion function $y= \frac{m}{x}$ at point $C$ in the first quadrant. The line $CD$ is perpendicular to the $x$-axis, with $D$ being the foot of the perpendicular. If $OA=OB=OD=1$, find the expression for the linear function.

Solution:
\textbf{Solution:} From equation (1), we find points A(-1, 0) and B(0, 1), therefore the equation of line AB is $y = x + 1$, \\
thus, the equation of the linear function is: \boxed{y = x + 1}.


Question: As shown in the diagram, the parabola $y=\frac{3}{4}x^{2}-\frac{15}{4}x+3$ 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. What are the coordinates of points A, B, and C?

Solution:
Solution: Substituting \(x=0\) into \(y=\frac{3}{4}x^{2}-\frac{15}{4}x+3\), we get \(y=3\).
\\
Therefore, the coordinates of point \(C\) are \((0, 3)\).
\\
Substituting \(y=0\) into \(y=\frac{3}{4}x^{2}-\frac{15}{4}x+3\), we get \(\frac{3}{4}x^{2}-\frac{15}{4}x+3=0\).
\\
Solving this, we find \(x_{1}=1\) and \(x_{2}=4\).
\\
Therefore, the coordinates of point \(A\) are \((1, 0)\), and the coordinates of point \(B\) are \((4, 0)\).
\\
Therefore, the coordinates of point \(A\) are \(\boxed{(1, 0)}\), the coordinates of point \(B\) are \(\boxed{(4, 0)}\), and the coordinates of point \(C\) are \(\boxed{(0, 3)}\).


Question: As illustrated in the figure, the parabola $y=\frac{3}{4}x^{2}-\frac{15}{4}x+3$ intersects the x-axis at points $A$ and $B$, with point $A$ located to the left of point $B$, and intersects the y-axis at point $C$. Point $E$ is a moving point on line segment $BC$. A perpendicular line to the x-axis is drawn through point $E$ and it intersects the parabola at point $F$. When $EF=OC$, find the x-coordinate of point $E$.

Solution:
\textbf{Solution:} Let the equation of the line $BC$ be $y=-\frac{3}{4}x+3$.\\
Since it passes through point $B(4, 0)$,\\
thus we find $k=-\frac{3}{4}$.\\
Therefore, the equation of the line $BC$ is $y=-\frac{3}{4}x+3$.\\
Since point $E$ is a moving point on line $BC$, and a perpendicular to the $x$ axis through point $E$ intersects the parabola at point $F$,\\
thus, let the coordinates of point $E$ be $\left(m, -\frac{3}{4}m+3\right)$, then the coordinates of point $F$ are $\left(m, \frac{3}{4}m^2-\frac{15}{4}m+3\right)$.\\
Since $EF=OC$, and the coordinates of point $C$ are $(0, 3)$,\\
therefore, $\left|\frac{3}{4}m^2-\frac{15}{4}m+3-\left(-\frac{3}{4}m+3\right)\right|=3$.\\
Therefore, $\frac{3}{4}m^2-3m=3$, or $\frac{3}{4}m^2-3m=-3$.\\
Solving gives $m_1=2+2\sqrt{2}$, $m_2=2-2\sqrt{2}$, $m_3=m_4=2$.\\
Therefore, the $x$-coordinates of point $E$ are $\boxed{2+2\sqrt{2}}$, $\boxed{2-2\sqrt{2}}$, and $\boxed{2}$.


Question:
Synthesis and Inquiry\\
As shown in the figure, the parabola $y=\frac{3}{4}x^{2}-\frac{15}{4}x+3$ intersects the x-axis at points $A$ and $B$, with point $A$ located to the left of point $B$, and intersects the y-axis at point $C$. Point $P$ starts from point $B$ and moves towards the endpoint $C$ along $BC$ at a speed of $1$ unit length per second. Simultaneously, point $Q$ starts from point $O$ and moves towards the endpoint $B$ at the same speed along the positive half of the x-axis. Once arrived, both points stop moving. Connect $PQ$. When $\triangle BPQ$ is an isosceles triangle, please directly write down the time of movement.

Solution:
\textbf{Solution:} Let the motion time be $t$. According to the problem statement, in order to form $\triangle BPQ$, points P and Q must not coincide with B, hence the range of $t$ is $0<t<4$,\\
$\therefore PB=OQ=t$, $BQ=4-t$,\\
As shown in the figure, draw $PD\perp x$ axis at point D, let the coordinates of P be $\left(a, -\frac{3}{4}a+3\right)$, then $BD=4-a$, $PD=-\frac{3}{4}a+3$,\\
According to the Pythagorean theorem, in $\triangle PDB$,\\
$PD^2+DB^2=PB^2$,\\
$\left(-\frac{3}{4}a+3\right)^2+\left(4-a\right)^2=t^2$,\\
We find $a_1=4-\frac{4}{5}t$, $a_2=4+\frac{4}{5}t$ (this does not meet the conditions and is excluded),\\
$\therefore$The coordinates of point P are $\left(4-\frac{4t}{5}, \frac{3t}{5}\right)$,\\
$\because$The coordinates of point Q are $\left(t, 0\right)$,\\
$\therefore PQ^2=\left(4-\frac{4t}{5}-t\right)^2+\left(\frac{3t}{5}\right)^2=\frac{90}{25}t^2-\frac{72t}{5}+16$,\\
(1) When $BP=BQ$,\\
that is, $t=4-t$,\\
We solve: $t=\boxed{2}$;\\
(2) When $BP=PQ$,\\
$t^2=\frac{90}{25}t^2-\frac{72t}{5}+16$,\\
We solve: $t_1=\boxed{\frac{20}{13}}$, $t_2=4$ (this does not meet the conditions and is excluded),\\
(3) When $BQ=PQ$,\\
$\left(4-t\right)^2=\frac{90}{25}t^2-\frac{72t}{5}+16$,\\
We solve: $t_1=\boxed{\frac{32}{13}}$, $t_2=0$ (this does not meet the conditions and is excluded),\\
In summary: when $\triangle BPQ$ is an isosceles triangle, the times are \boxed{2} seconds, \boxed{\frac{20}{13}} seconds, \boxed{\frac{32}{13}} seconds.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=2x+4$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. It intersects the graph of the hyperbolic function $y=\frac{k}{x}$ ($k\neq 0$) at points $C$ and $D$. The coordinates of point $C$ are ($n$, $6$). What is the expression of this hyperbolic function?

Solution:
\textbf{Solution:} Since point C $(n, 6)$ lies on the graph of the linear function $y=2x+4$,\\
it follows that $6=2n+4$, solving this, we get $n=1$,\\
therefore, the coordinates of point C are $(1, 6)$.\\
Substituting the coordinates of point C $(1,6)$ into $y=\frac{k}{x}$, we obtain $k=6$,\\
therefore, the expression for the inverse proportion function is $\boxed{y=\frac{6}{x}}$;


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=2x+4$ intersects the x-axis at point A and the y-axis at point B. It intersects the graph of the inverse proportion function $y=\frac{k}{x}$ ($k\neq 0$) at points C and D, where the coordinate of point C is $(n, 6)$. Determine the coordinate of point D.

Solution:
\textbf{Solution:} Solve by combining the two function expressions,
\[
\begin{cases}
y=2x+4\\
y=\frac{6}{x}
\end{cases}
\]
, and find $x_{1}=-3$, $x_{2}=1$ (discard $x_{2}$).\\
When $x=-3$, $y=2\times(-3)+4=-2$,\\
$\therefore$ the coordinates of point D are $\boxed{(-3, -2)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=2x+4$ intersects the $x$-axis at point $A$ and the $y$-axis at point $B$. It also intersects the graph of the reciprocal function $y=\frac{k}{x}$ (where $k\neq 0$) at points $C$ and $D$, with the coordinates of point $C$ being ($n$, $6$). Connect $OC$ and $OD$. What is the area of $\triangle COD$?

Solution:
\textbf{Solution:} Given the problem, we have
\[S_{\triangle COD}=S_{\triangle CBO}+S_{\triangle DBO}\]
\[=\frac{1}{2}OB\cdot x_C+\frac{1}{2}OB\cdot |x_D|\]
\[=\frac{1}{2}\times 4\times 1+\frac{1}{2}\times 4\times 3\]
\[=\boxed{8}\]


Question: In the Cartesian coordinate system, the parabola $y = -x^2 - 2x + 3$ intersects the x-axis at points A and B (A is to the left of B) and intersects the y-axis at point C, with the vertex at D. Please directly write the coordinates of points A, C, and D.

Solution:
Solution: When $y=-x^2-2x+3$ and $y=0$, we have $-x^2-2x+3=0$,\\
we get: $x_1=-3$, $x_2=1$,\\
Since A is on the left side of B,\\
Therefore, \boxed{A (-3, 0)}, \boxed{B (1, 0)},\\
When $y=-x^2-2x+3$ and $x=0$, then $y=3$,\\
Therefore, \boxed{C (0, 3)},\\
Since $y=-x^2-2x+3=-(x+1)^2+4$,\\
Therefore, the vertex \boxed{D (-1, 4)};


Question: In the Cartesian coordinate system, the parabola $y=-x^2-2x+3$ intersects the x-axis at points $A$ and $B$ (where $A$ is to the left of $B$) and intersects the y-axis at point $C$, with the vertex denoted as $D$. As shown in the diagram, find a point $E$ on the x-axis such that the perimeter of $\triangle CDE$ is minimized. What are the coordinates of point $E$?

Solution:
\textbf{Solution:} Let's construct point $C'$ symmetric to point C with respect to the x-axis, and connect $C'D$ intersecting the x-axis at point E. At this moment, the perimeter of $\triangle CDE$ is minimized, as shown in figure 1,\\
$\because C(0, 3),$\\
$\therefore C'(0, -3),$\\
Suppose the equation of line $C'D$ is $y=kx+b$,\\
then $\begin{cases}4=-k+b\\-3=b\end{cases}$, solving this we get: $\begin{cases}k=-7\\b=-3\end{cases}$,\\
$\therefore$ the equation of line $C'D$ is $y=-7x-3$,\\
when in $y=-7x-3$, $y=0$, then $x=-\frac{3}{7}$,\\
$\therefore$ when the perimeter of $\triangle CDE$ is minimized, the coordinates of point E are $\boxed{\left(-\frac{3}{7}, 0\right)}$;


Question: In the Cartesian coordinate system, the parabola $y=-x^2-2x+3$ intersects the x-axis at points $A$ and $B$ (with $A$ on the left side of $B$) and intersects the y-axis at point $C$, with $D$ being the vertex. As shown in the figure, point $P$ is a moving point on the axis of symmetry of the parabola such that $\triangle ACP$ is an isosceles triangle. If it exists, find the coordinates of point $P$; if not, please state the reason.

Solution:
\textbf{Solution:}

There exists a point P(-1, t).

Since $A(-3,0)$ and $C(0,3)$,

Therefore, $AC^2=3^2+3^2=18$, $AP^2=[-1-(-3)]^2+t^2=t^2+4$,

$CP^2=[0-(-1)]^2+(t-3)^2=(t-3)^2+1$,

(1) When $AC=AP$, then $18=t^2+4$,

Solving this gives $t=\sqrt{14}$ or $t=-\sqrt{14}$,

Hence, P(-1, $\sqrt{14}$) or (-1, $-\sqrt{14}$);

(2) When $AC=CP$, then $18=(t-3)^2+1$,

Solving this gives $t=3+\sqrt{17}$ or $t=3-\sqrt{17}$,

P(-1, $3+\sqrt{17}$) or (-1, $3-\sqrt{17}$);

(3) When $AP=CP$, then $t^2+4=(t-3)^2+1$,

Solving this gives $t=1$,

Therefore, P(-1, 1);

In conclusion, the coordinates of point P are: \boxed{(-1, \sqrt{14})} or \boxed{(-1, -\sqrt{14})} or \boxed{(-1, 3+\sqrt{17})} or \boxed{(-1, 3-\sqrt{17})} or \boxed{(-1, 1)}.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the direct proportion function $y_1=-\frac{3}{2}x$ and the graph of the inverse proportion function $y_2=\frac{k}{x}$ intersect at point A in the second quadrant, and the x-coordinate of point A is $-2$. What is the analytic expression of the inverse proportion function?

Solution:
\textbf{Solution:} Given that the graph of the direct proportion function $y=-\frac{3}{2}x$ passes through point A, and the x-coordinate of point A is $-2$,\\
\[\therefore\] the y-coordinate of point A is 3, thus the coordinates of point A are $\left(-2, 3\right)$.\\
\[\because\] The graph of the inverse proportion function $y=\frac{k}{x}$ passes through point $A\left(-2, 3\right)$,\\
\[\therefore\] $3=\frac{k}{-2}$.\\
\[\therefore\] $k=-6$.\\
\[\therefore\] The equation of the inverse proportion function is $\boxed{y=-\frac{6}{x}}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the graph of the direct proportion function $y_1=-\frac{3}{2}x$ intersects with the graph of the inverse proportion function $y_2=\frac{k}{x}$ at point A in the second quadrant, and the x-coordinate of point A is $-2$. Write directly the range of values for $x$ when $y_1>y_2$.

Solution: \textbf{Solution:} When $y_1>y_2$, the range of values for $x$ is $\boxed{x\leq -2 or 0<x\leq 2}$.


Question: As shown in the graph, in the Cartesian coordinate system $xOy$, the direct proportion function $y_{1}=-\frac{3}{2}x$ and the inverse proportion function $y_{2}=\frac{k}{x}$ intersect at point A in the second quadrant, and the x-coordinate of point A is $-2$. The coordinates of point B are $(-4, 0)$. If point P is on the y-axis, and the area of $\triangle AOP$ is equal to the area of $\triangle AOB$, find the coordinates of point P.

Solution:
\textbf{Solution:} Since ${S}_{\triangle AOB}=\frac{1}{2}\times 4\times 3=6$,\\
it follows that ${S}_{\triangle APO}=\frac{1}{2}\times 2OP=OP$,\\
therefore $OP=6$,\\
thus, the coordinates of point P are $\boxed{(0, 6)}$ or $\boxed{(0, -6)}$.


Question: As shown in the diagram, in the Cartesian coordinate system, A(-3, 0) and B(1, 0) are given points. Point A is rotated counterclockwise about the origin until it lies on point C on the negative half of the y-axis. The parabola $y=ax^2+bx+c$ passes through points A, B, and C. Find the equation of the parabola by connecting points A and C.

Solution:
\textbf{Solution:} \text{Given that, we know } C(0, -3) \\
\text{Substituting points } \[A(-3, 0), B(1, 0), \text{ and } C(0, -3) \text{ into the parabola equation } y=ax^{2}+bx+c \text{ yields}\] \\
\[
\begin{cases}
9a+3b+c=0 \\
a+b+c=0 \\
c=-3
\end{cases}
\text{Solving this, we get: }
\begin{cases}
a=1 \\
b=2 \\
c=-3
\end{cases}
\]
\text{Therefore, the quadratic function is given by } \boxed{y=x^{2}+2x-3}


Question: It is known that, in the Cartesian coordinate system, points $A(-3, 0)$ and $B(1, 0)$ are given. Rotate point $A$ counterclockwise around the origin to let it fall on point $C$ located on the negative half of the y-axis. A parabola, $y=ax^2+bx+c$, passes through points $A$, $B$, and $C$. The line segment $AC$ is connected. The line $AC$ is translated upward, the translated line $DM$ intersects the y-axis at point $D$ and intersects the parabola to the right of the y-axis at point $M$. The segments $AM$ and $CM$ are connected. Given that the area of $\triangle ACM$ is $15$, what are the coordinates of point $M$?

Solution:
\textbf{Solution:} Let line $DM$ be obtained by translating line $AC$,\\
$\therefore DM\parallel AC$,\\
$\therefore S_{\triangle ACD}=S_{\triangle ACM}=15$,\\
Since $S_{\triangle ACM}=\frac{1}{2}CD\cdot AO=15, AO=3$,\\
$\therefore CD=10$,\\
$\therefore D(0, 7)$,\\
$\therefore A(-3, 0), C(0, -3)$,\\
Let $AC: y=kx+b(k\ne 0)$,\\
Solving gives: $\left\{\begin{array}{l}k=-1 \\ b=-3\end{array}\right.$,\\
$\therefore AC: y=-x-3$,\\
Since $DM\parallel AC$,\\
Let $DM: y=-x+b$,\\
Substituting $D(0, 7)$ gives,\\
$DM: y=-x+7$,\\
Combining it with the parabola, we solve to get,\\
$\left\{\begin{array}{l}x=2 \\ y=5\end{array}\right.$ or $\left\{\begin{array}{l}x=-5 \\ y=12\end{array}\right.$ (which does not meet the conditions, hence disregarded),\\
$\therefore \boxed{M(2, 5)}$.


Question: Given that in the Cartesian coordinate system, $A(-3,0)$, $B(1,0)$, point $A$ is rotated counter-clockwise around the origin to fall on point $C$ on the negative half of the $y$-axis. The parabola $y=ax^2+bx+c$ passes through points $A$, $B$, and $C$, connecting $AC$. Let point $N$ be a moving point on line $BC$, with the $x$-coordinate of $N$ denoted as $n$. If the triangle formed by points $A$, $C$, and $N$ is an obtuse triangle, determine the range of values for $n$.

Solution:
\textbf{Solution:}

Draw $AQ\perp AC$ through point A, intersecting $BC$ at point Q. Draw $AP\perp BC$ through point A at point P.

Since $B(1, 0)$, $C(0, -3)$,

let the equation of $BC$ be: $y=kx+b(k\ne 0)$,

solving the system
\[\begin{cases}
k + b = 0 \\
b = -3
\end{cases},\]
we get
\[\begin{cases}
k = 3 \\
b = -3
\end{cases}.\]

Thus, the equation of $BC$ is $y = 3x - 3$.

Since the equation of $AC$ is $y = -x - 3$,

thus $\angle CAB=45^\circ$.

Since $AQ \perp AC$,

thus $\angle QAB=45^\circ$.

Let the equation of $AQ$ be $y = x + b$.

Substituting $A(-3, 0)$ into it, we get $b = 3$,

thus the equation of $AQ$ is $y = x + 3$.

Solving the system of equations of lines $BC$ and $AQ$,
\[\begin{cases}
y = 3x - 3 \\
y = x + 3
\end{cases},\] we find $x = 3$.

Thus, the x-coordinate of point Q is 3.

Let $AP$ intersect the y-axis at point E.

Since $\angle BAP + \angle ABP = \angle OBC + \angle ABP = 90^\circ$,

thus $\angle BAP = \angle OBC$.

Since $\angle APB = \angle BOC = 90^\circ$ and $OA = OC$,

thus $\triangle AOE \cong \triangle COB$,

therefore $OE = OB = 1$,

thus E(-1, 0).

Therefore, the equation of $AP$ is $y = \frac{1}{3}x - 1$.

Solving the system of equations of lines $BC$ and $AP$,
\[\begin{cases}
y = 3x - 3 \\
y = \frac{1}{3}x - 1
\end{cases},\] we find $x = \frac{3}{5}$.

Thus, the x-coordinate of point P is $\frac{3}{5}$.

(1) When point N is above point Q, $\angle NAC$ is obtuse, $\triangle CAN$ is an obtuse triangle, in this case, $n > 3$.

(2) When point N coincides with point Q or point P, $\triangle CAN$ is a right triangle, which does not meet the criteria.

(3) When point N is on the line segment $PQ$ (excluding the endpoints), $\triangle CAN$ is an acute triangle, which does not meet the criteria.

(4) When point N is on the line segment $PC$ (excluding the endpoints), $\angle ANC$ is obtuse, $\triangle CAN$ is an obtuse triangle, in this case, $0 < n < \frac{3}{5}$.

(5) When point N coincides with point C, it cannot form a triangle, which does not meet the criteria.

(6) When point N is below point C, $\angle ACN$ is obtuse, $\triangle CAN$ is an obtuse triangle, in this case, $n < 0$.

Overall, the range of $n$ is $n > 3$ or $0 < n < \frac{3}{5}$ or $n < 0$ (also can be written as $n > 3$ or $n < \frac{3}{5}$ and $n \neq 0$).

Therefore, the range of $n$ is $\boxed{n > 3 \text{ or } 0 < n < \frac{3}{5} \text{ or } n < 0}$.


Question: A triangular piece of material is shown in the figure, where $ \angle A=30^\circ$, $ \angle C=90^\circ$, and $AB=12$. Cut out a rectangle CDEF from this material, where points D, E, and F are respectively on AC, AB, and BC. Let the length of AE be x, and the area of rectangle CDEF be S. Write down the expression of S as a function of x, and state the domain of x.

Solution:
\textbf{Solution:} We have $AB=12$, $AE=x$, and the point E does not coincide with points A or B,\\
therefore $0<x<12$,\\
the quadrilateral CDEF is a rectangle,\\
therefore $\angle AFE=90^\circ$,\\
since $\angle A=30^\circ$,\\
therefore $EF=\frac{1}{2}AE$,\\
in $\triangle ABC$, $\angle C=90^\circ$, $\angle A=30^\circ$, $AB=12$,\\
therefore $BC=\frac{1}{2}AB=6$,\\
according to the Pythagorean theorem: $AC=\sqrt{12^2-6^2}=6\sqrt{3}$,\\
therefore $CF=AC-AF=6\sqrt{3}-\frac{\sqrt{3}}{2}AE$,\\
therefore the area $S_{\text{rectangle CDEF}}=CF\cdot EF=\frac{1}{2}x\left(6\sqrt{3}-\frac{\sqrt{3}}{2}x\right)=-\frac{\sqrt{3}}{4}x^2+3\sqrt{3}x$;\\
The range of $x$ is \boxed{0<x<12}.


Question: As illustrated, a triangular piece of material is given, with $\angle A=30^\circ$, $\angle C=90^\circ$, and $AB=12$. From this material, a rectangle CDEF is to be cut out, where points D, E, and F are located on AC, AB, and BC, respectively. Let the length of AE be $x$, and the area of rectangle CDEF be $S$. When the area of rectangle CDEF is $8\sqrt{3}$, what is the length of AE?

Solution:
\textbf{Solution:} According to the problem statement, we have $S_{\text{rectangle }CDEF}=-\frac{\sqrt{3}}{4}x^2+3\sqrt{3}x=8\sqrt{3}$\\
$\therefore -\frac{\sqrt{3}}{4}\left[(x-6)^2-36\right]=8\sqrt{3}$\\
$-\frac{\sqrt{3}}{4}(x-6)^2+9\sqrt{3}=8\sqrt{3}$,\\
$(x-6)^2=4$\\
$\therefore x-6=\pm 2$\\
Solving gives: $x_1=4$, $x_2=8$,\\
$\therefore$ The length of AE is either $\boxed{4}$ or $\boxed{8}$.


Question: A triangular piece of material is shown in the figure, where $ \angle A=30^\circ$, $ \angle C=90^\circ$, and $ AB=12$. A rectangle CDEF is cut from this material, with points D, E, and F located on AC, AB, and BC, respectively. Let the length of AE be $x$, and the area of rectangle CDEF be $S$. What is the length of AE when the area of rectangle CDEF is maximized? What is the maximum area?

Solution:
\textbf{Solution:} Given: $S_{\text{rectangle } CDEF} = -\frac{\sqrt{3}}{4}x^2 + 3\sqrt{3}x\ (0 < x < 12)$\\
$= -\frac{\sqrt{3}}{4}(x^2 - 12x)$\\
$= -\frac{\sqrt{3}}{4}(x^2 - 12x + 6^2 - 6^2)$\\
$= -\frac{\sqrt{3}}{4}[(x-6)^2 - 36]$\\
$= -\frac{\sqrt{3}}{4}(x-6)^2 + 9\sqrt{3}$,\\
Therefore, when $x=\boxed{6}$, the area of rectangle CDEF is maximized,\\
meaning when point E is the midpoint of AB, the area of rectangle CDEF is maximized, with a maximum area of $\boxed{9\sqrt{3}}$.


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 inverse proportion function $y=-\frac{8}{x}$ at points $A$ and $B$, and both the $x$-coordinate of point $A$ and the $y$-coordinate of point $B$ are $-2$. Find the analytical expression of the linear function.

Solution:
\textbf{Solution:} Let's consider the inverse proportion function $y=-\frac{8}{x}$. When $x=-2$, we have $y=4$,

$\therefore$ the coordinates of point A are $\left(-2, 4\right)$;

For the inverse proportion function $y=-\frac{8}{x}$, when $y=-2$, we have $-2=-\frac{8}{x}$,

Solving this gives: $x=4$,

$\therefore$ the coordinates of point B are $\left(4, -2\right)$.

$\because$ A linear function passes through points A and B,

$\therefore$
$\begin{cases}
-2k+b=4\\
4k+b=-2
\end{cases}$, solving these equations yields:
$\begin{cases}
k=-1\\
b=2
\end{cases}$,

$\therefore$ the equation of the linear function is $\boxed{y=-x+2}$.


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 inverse proportional function $y = -\frac{8}{x}$ at points $A$ and $B$. Moreover, the x-coordinate of point $A$ and the y-coordinate of point $B$ are both $-2$. Find: What is the area of $\triangle AOB$?

Solution:
\textbf{Solution:} Let $y=-x+2$ and $x=0$, then $y=2$,\\
$\therefore$ the coordinates of point N are $\left(0, 2\right)$,\\
$\therefore$ the area of $\triangle AOB$ is $\frac{1}{2}ON\cdot \left({x}_{B}-{x}_{A}\right)=\frac{1}{2}\times 2\times \left[4-\left(-2\right)\right]=\boxed{6}$.


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 inverse proportion function $y=-\frac{8}{x}$ at points $A$ and $B$, and the $x$-coordinate of point $A$ and the $y$-coordinate of point $B$ are both $-2$. Find: For what value of $x$ is the function value of the linear function greater than that of the inverse proportion function?

Solution:
\textbf{Solution:} Observe the graph of the function and find: \\
When $x<-2$ or $0<x<4$, the graph of the linear function is above the graph of the inverse proportion function, \\
$\therefore$ The range of $x$ for which the function value of the linear function is greater than the function value of the inverse proportion function is $\boxed{x<-2}$ or $\boxed{0<x<4}$.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y_1 = kx + b$ $(k \neq 0)$ intersects with the graph of the inverse proportion function $y_2 = \frac{m}{x}$ $(m \neq 0)$ at points $A$ and $B$ within the second and fourth quadrants, and intersects with the $x$-axis at point $C$. The coordinates of point $B$ are $(n, -2)$. The length of segment $OA$ is 5. Point $E$ is located on the negative half of the x-axis, and $\sin\angle AOE = \frac{4}{5}$. Determine the analytic expressions of the inverse proportion function and the linear function.

Solution:
\textbf{Solution:} As shown in the figure, draw $AD\perp x$-axis at point D through point A,\\
$\because \sin\angle AOE=\frac{4}{5}$ and OA=5,\\
$\therefore \frac{AD}{OA}=\frac{4}{5}$,\\
$\therefore AD=4$,\\
By the Pythagorean theorem, we have $OD=\sqrt{OA^{2}-AD^{2}}=\sqrt{5^{2}-4^{2}}=3$,\\
$\therefore OD=3$,\\
$\therefore$The coordinates of point A are (-3, 4),\\
Substituting point A (-3, 4) into $y_{2}=\frac{m}{x}$,\\
We get $4=\frac{m}{-3}$, solving for $m$ yields $m=-12$,\\
$\therefore$The equation of the inverse proportion function is $y_{2}=-\frac{12}{x}$,\\
Substituting B (n, -2) into $y_{2}=-\frac{12}{x}$ gives $n=6$,\\
$\therefore$The coordinates of point B are (6, -2),\\
Substituting points A (-3, 4) and B (6, -2) into $y_{1}=kx+b$,\\
We get $\begin{cases}4=-3k+b \\ -2=6k+b\end{cases}$, solving for $\begin{cases}k=-\frac{2}{3} \\ b=2\end{cases}$,\\
$\therefore$The equation of the linear function is $y_{1}=-\frac{2}{3}x+2$,\\
$\therefore$The equation for the linear function is $\boxed{y_{1}=-\frac{2}{3}x+2}$, and the equation for the inverse proportion function is $\boxed{y_{2}=-\frac{12}{x}}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the linear function $y_{1}=kx+b$ ($k\ne 0$) intersects with the graph of the inverse proportion function $y_{2}=\frac{m}{x}$ ($m\ne 0$) at points A and B in the second and fourth quadrants, respectively, and intersects the x-axis at point C. The coordinates of point B are $(n, -2)$. The length of segment OA is $5$, and point E is located on the negative half of the x-axis, with $\sin\angle AOE=\frac{4}{5}$. What is the area of $\triangle AOB$?

Solution:
\textbf{Solution:}\\
As shown in the figure, draw line $BH\perp x$-axis through point B,\\
according to the coordinates of point A and point B, we know: AD=4, BH=2\\
From (1), we know the analytical expression of line AB is $y_{1}=-\frac{2}{3}x+2$\\
When y=0, we get $0=-\frac{2}{3}x+2$, and solve for x to find x=3\\
$\therefore$ the coordinates of point C are (3, 0)\\
$\therefore$ OC=3\\
$S_{\triangle AOB}=S_{\triangle AOC}+S_{\triangle BOC}$\\
$=\frac{1}{2}\times 4\times 3+\frac{1}{2}\times 2\times 3$\\
$=\boxed{9}$\\
$\therefore$ the area of $\triangle AOB$ is $\boxed{9}$


Question: As shown in the diagram, in the Cartesian coordinate system, the graph of the linear function $y_{1}=kx+b$ $(k\ne 0)$ intersects with the graph of the inverse proportion function $y_{2}=\frac{m}{x}$ $(m\ne 0)$ at points A and B within the second and fourth quadrants, respectively, and intersects with the x-axis at point C. Point B is located at $(n, -2)$. The segment OA has a length of 5, and point E is located on the negative half-axis of the x-axis, with $\sin\angle AOE=\frac{4}{5}$. On the graph of the inverse proportion function, find a point D such that the area of $\triangle ABD$ is equal to the area of $\triangle AOB$. Please write down the coordinates of point D directly.

Solution: \textbf{Solution}: The coordinates of point D are $\boxed{D_1(-3\sqrt{2}, 2\sqrt{2})}$, $\boxed{D_2(3\sqrt{2}, 2\sqrt{2})}$, $\boxed{D_3(3-3\sqrt{3}, 2+2\sqrt{3})}$, and $\boxed{D_4(3+3\sqrt{3}, 2-2\sqrt{3})}$.


Question: As shown in the figure, it is known that the parabola $y=ax^2+2x+c$ intersects the x-axis at point $A(-1,0)$ and point $B$, and intersects the y-axis at point $C(0,3)$. Find the equation of the parabola, and calculate the coordinates of point $B$.

Solution:
\textbf{Solution:} Since the parabola passes through points $A(-1, 0)$, $B$, and $C(0, 3)$, \\
we have $\begin{cases}a-2+c=0\\ c=3\end{cases}$,\\
thus $\begin{cases}a=-1\\ c=3\end{cases}$,\\
therefore, the equation of the parabola is: $y=-x^2+2x+3$;\\
Setting $y=0$ yields $0=-x^2+2x+3$, solving this gives: $x_1=-1, x_2=3$,\\
hence the point \boxed{B\left(3, 0\right)}.


Question: As shown in the figure, it is known that the parabola $y=ax^2+2x+c$ intersects the x-axis at points $A(-1,0)$ and $B$, and intersects the y-axis at point $C(0,3)$. Point $M$ is a point on line segment $BC$ (not coinciding with $B$ or $C$), and a line through $M$ is drawn parallel to the y-axis intersecting the parabola at $N$. If the abscissa of point $M$ is $m$, please express the length of $MN$ using an algebraic expression that includes $m$.

Solution:
\textbf{Solution}: From (1) it is known that $B(3, 0)$. Assume the equation of line $BC$ is: $y=kx+b$. Substituting $B(3, 0)$, $C(0, 3)$, we get: \[
\begin{cases}
3k+b=0 \\
b=3
\end{cases}
\], solving this system gives: \[
\begin{cases}
k=-1 \\
b=3
\end{cases}
\]. Therefore, the equation of line $BC$ is $y=-x+3$, thus $M(m, -m+3)$. Furthermore, since $MN \perp x$ axis, hence $N(m, -m^2+2m+3)$, therefore $MN=\boxed{-m^2+3m}$ $(0<m<3)$.


Question: As shown in the figure, it is known that the parabola $y=ax^2+2x+c$ intersects the x-axis at points $A(-1,0)$ and $B$, and intersects the y-axis at point $C(0,3)$. Connect $NB$, $NC$. Is there a point $M$ that maximizes the area of $\triangle BNC$? If it exists, find the coordinates of point $M$; if it does not exist, explain why.

Solution:
\textbf{Solution:} Yes, let's connect $CN$ and $BN$, as shown in the figure:\\
$S_{\triangle BNC}=S_{\triangle CMN}+S_{\triangle MNB}=\frac{1}{2}|MN|\cdot |OB|$\\
$\therefore$ When $|MN|$ is at its maximum, the area of $\triangle BNC$ is maximized,\\
$MN=-m^2+3m=-(m-\frac{3}{2})^2+\frac{9}{4}$,\\
When $m=\frac{3}{2}$, $MN$ reaches its maximum value of $\frac{9}{4}$,\\
Therefore, when $m=\frac{3}{2}$, the area of $\triangle BNC$ is maximized, and the coordinates of point $M$ are $\boxed{\left(\frac{3}{2}, \frac{3}{2}\right)}$.


Question: As shown in the figure, the parabola $y=ax^2+bx-3$ $(a \neq 0)$ intersects the x-axis at points $A(-3,0)$ and $B(1,0)$, and intersects the y-axis at point C. Find the equation of this parabola.

Solution:
\textbf{Solution:} Because the parabola $y=ax^{2}+bx-3\left(a\ne 0\right)$ intersects the x-axis at point $A(-3,0)$ and point $B(1,0)$, \\
therefore $\begin{cases}9a-3b-3=0\\ a+b-3=0\end{cases}$, \\
solving the system of equations, we get $\begin{cases}a=1\\ b=2\end{cases}$, \\
therefore, the expression of the parabola is $y=x^{2}+2x-3=\boxed{x^{2}+2x-3}$.


Question: Synthesis and Inquiry: As shown in the diagram, the parabola $y=ax^{2}+bx-3\ (a\neq 0)$ intersects the x-axis at points $A(-3,0)$ and $B(1,0)$, and intersects the y-axis at point C. If point D is a moving point on the parabola in the third quadrant, and lines $AD$ and $CD$ are drawn, find the maximum area of the quadrilateral $AOCD$ and the coordinates of point D at that time.

Solution:
\textbf{Solution:} Let's draw a figure, passing through point D draw $DF\perp x$-axis at point F, let $D\left(m, m^{2}+2m-3\right)$  $\left(-3<m<0\right)$.\\
$\therefore DF=-m^{2}-2m+3$, $AF=m+3$, $OF=-m$.\\
When $x=0$, $y=-3$.\\
$\therefore$ The coordinates of point C are $\left(0, -3\right)$.\\
$\therefore OC=3$.\\
$\therefore S_{\text{quadrilateral}AOCD}=\frac{1}{2}AF\cdot DF+\frac{1}{2}\left(OC+DF\right)\cdot OF$,\\
$=\frac{1}{2}\left(m+3\right)\left(-m^{2}-2m+3\right)+\frac{1}{2}\left(3-m^{2}-2m+3\right)\left(-m\right)$,\\
$=-\frac{3}{2}m^{2}-\frac{9}{2}m+\frac{9}{2}$,\\
$=-\frac{3}{2}\left(m+\frac{3}{2}\right)^{2}+\frac{63}{8}$.\\
$\therefore$ When $m=-\frac{3}{2}$, $S_{\text{quadrilateral}AOCD}$ is maximized, and the maximum value is $\boxed{\frac{63}{8}}$.\\
At this time, the coordinates of point D are $\boxed{\left(-\frac{3}{2}, -\frac{15}{4}\right)}$.


Question: Integrated and Investigative: As shown in the figure, the parabola $y = ax^{2} + bx - 3 \left( a \neq 0 \right)$ intersects the x-axis at points $A(-3,0)$ and $B(1,0)$, and intersects the y-axis at point C. If point E is on the axis of symmetry of the parabola, and the line segment $EB$ rotates $90^\circ$ clockwise around point E, the corresponding point of B, ${B}^{\prime}$, also lies on this parabola. Find the coordinates of point E.

Solution:
\textbf{Solution:} From the given condition, we have $y=x^{2}+2x-3=(x+1)^{2}-4$, \\
$\therefore$ the axis of symmetry of the parabola $y=x^{2}+2x-3$ is $x=-1$, \\
$\because$ point E is on the axis of symmetry of the parabola, \\
$\therefore$ let $E(-1, n)$. \\
$\because$ when segment $EB$ rotates clockwise $90^\circ$ around point E, the corresponding point of B, ${B}^{\prime}$, also falls exactly on this parabola, \\
$\therefore$ consider two cases: \\
(1) When $n\ge 0$, for $EB=E{B}^{\prime}$, from the graph it is known that point ${B}^{\prime}$ coincides with point A. \\
Let the point where the axis of symmetry intersects the x-axis be point M. \\
$\because$ $\angle BEA=90^\circ$, \\
$\therefore$ $ME=MA=2$. \\
$\therefore$ $E(-1, 2)$. \\
(2) When $n<0$, \\
From the given condition, we have $EB=E{B}^{\prime}$, $\angle BE{B}^{\prime}=90^\circ$. \\
As shown in the figure, draw ${B}^{\prime}N\perp$ to the axis of symmetry at point N. \\
$\therefore$ $\angle NE{B}^{\prime}+\angle MEB=\angle NE{B}^{\prime}+\angle N{B}^{\prime}E=90^\circ$. \\
$\therefore$ $\angle N{B}^{\prime}E=\angle MEB$. \\
In $\triangle {B}^{\prime}NE$ and $\triangle EMB$, \\
$\left\{\begin{array}{l}
\angle {B}^{\prime}NE=\angle EMB=90^\circ \\
\angle N{B}^{\prime}E=\angle MEB \\
E{B}^{\prime}=EB
\end{array}\right.$, \\
$\therefore$ $\triangle {B}^{\prime}NE\cong \triangle EMB$. \\
$\therefore$ ${B}^{\prime}N=EM=-n$, $EN=BM=2$. \\
$\therefore$ ${B}^{\prime}(-n-1, n-2)$. \\
Substitute into $y=x^{2}+2x-3$, we get $n-2=(-n-1)^{2}+2(-n-1)-3$, \\
Solving gives $n=-1$, $n=2$ (discard). \\
$\therefore$ The coordinates of point E that meet the conditions are $\boxed{(-1, 2)}$ or $\boxed{(-1, -1)}$.


Question: As shown in the figure, the parabola $y=\frac{4}{3}x^{2}+bx+c$ intersects the x-axis at points $A(-1,0)$ and $B$ (point $B$ is to the right of point $A$), and intersects the y-axis at point $C(0,-4)$. Find the analytical expression of the parabola and the coordinates of the vertex $E$.

Solution:
\textbf{Solution:} Given that the parabola $y=\frac{4}{3}x^{2}+bx+c$ intersects the $y$ axis at point $C(0, -4)$,\\
therefore, the equation of the parabola is $y=\frac{4}{3}x^{2}+bx-4$.\\
Since the parabola $y=\frac{4}{3}x^{2}+bx-4$ intersects the $x$ axis at $A(-1, 0)$,\\
thus, $0=\frac{4}{3} \times (-1)^{2}-b-4$.\\
Solving this, we get: $b=-\frac{8}{3}$,\\
therefore, the equation of the parabola is $y=\frac{4}{3}x^{2}-\frac{8}{3}x-4=\frac{4}{3}(x-1)^{2}-\frac{16}{3}$.\\
Hence, the vertex $E$ has the coordinates $\boxed{(1, -\frac{16}{3})}$.


Question: Given the parabola $y=\frac{4}{3}x^{2}+bx+c$ intersects the x-axis at points $A(-1,0)$ and $B$ (with point $B$ to the right of point $A$) and intersects the y-axis at point $C(0,-4)$. As shown in the figure, point $H$ is a moving point on the parabola, and the area of $\triangle OCH$ is $8$. What are the coordinates of point $H$?

Solution:
\textbf{Solution:} Let the x-coordinate of point H be $m$,

$\therefore S_{\triangle OCH}=\frac{1}{2}\text{OC} \cdot |x_H|$,

$\therefore 2|m|=8$,

Solving this gives: $m=4$ or $m=-4$,

When $m=4$, $y=\frac{4}{3}(4-1)^2-\frac{16}{3}=\frac{20}{3}$,

When $m=-4$, $y=\frac{4}{3}(-4-1)^2-\frac{16}{3}=\frac{84}{3}$,

$\therefore$ The coordinates of point H are $\boxed{(4, \frac{20}{3})}$ or $\boxed{(-4, \frac{84}{3})}$;


Question:
Comprehensive and Practice

As shown in the figure, the parabola $y=\frac{4}{3}x^{2}+bx+c$ intersects the x-axis at points $A(-1,0)$ and $B$ (with point $B$ to the right of point $A$) and intersects the y-axis at point $C(0,-4)$. Point $D$ is on the parabola. Line $ED$ intersects line $BC$ at point $G$. When $DG=2GE$, directly write down the coordinates of point $D$.

Solution:
\textbf{Solution:} From the construction, it is found that the parabola intersects the x-axis at $A(-1,0)$, and the vertex $E$ has coordinates $(1,-\frac{16}{3})$,

$\therefore$ the other intersection point of the parabola with the x-axis is $B(3,0)$,

Assume the equation of line $BC$ is $y=kx-4$,

thus $0=3k-4$,

Solving gives $k=\frac{4}{3}$,

$\therefore$ the equation of line $BC$ is $y=\frac{4}{3}x-4$,

$\because$ the coordinates of vertex $E$ are $(1,-\frac{16}{3})$,

$\therefore$ the coordinates of point $M$ are $(-1,-\frac{16}{3})$,

Assume the coordinates of point $D$ are $(a, \frac{4}{3}a^{2}-\frac{8}{3}a-4)$, then the coordinates of point $N$ are $(a^{2}-2a, \frac{4}{3}a^{2}-\frac{8}{3}a-4)$,

From the construction, it is known $ME\parallel DN$ and $DG=2GE$,

$\therefore\triangle MEG\cong \triangle NDG$,

From the properties of similar triangles, it follows that $DN=2ME$,

$ME=1-(-1)=2$, $DN= a^{2}-2a-a=a^{2}-3a$,

$\therefore a^{2}-3a=4$, i.e., $a^{2}-3a-4=0$,

Solving gives: $a=4$ or $a=-1$,

$\therefore$ the coordinates of point $D$ are $\boxed{(4,\frac{20}{3})}$ or $\boxed{(-1,\frac{84}{3})}$.


Question: As shown in the figure, the graph of the inverse proportion function $y=\frac{k}{x} (k \neq 0)$ passes through the point $(-2, 4)$ and the point $A(a, -2)$. Find the analytical expression of this inverse proportion function and the value of $a$.

Solution:
\textbf{Solution}: Since the graph of the inverse proportion function $y=\frac{k}{x} (k\ne 0)$ passes through the points $(-2, 4)$ and point $A(a, -2)$, \\
$\therefore k=-2\times 4=-8$, \\
$\therefore$ the inverse proportion function is $y=-\frac{8}{x}$, \\
$\therefore -2a=-8$, \\
Solving for $a$ gives: $a=\boxed{4}$.


Question: As shown in the figure, the graph of the inverse proportion function $y=\frac{k}{x} (k\ne 0)$ passes through the points $(-2, 4)$ and point $A(a, -2)$. If the point $C(x, y)$ is also on the graph of the inverse proportion function $y=\frac{k}{x} (k\ne 0)$, when $2<x<8$, find the range of values for the function $y$.

Solution:
\textbf{Solution:} Given that point $C(x, y)$ is also on the graph of the inverse proportion function $y=-\frac{8}{x}$, \\
$\therefore$ when $2<x<8$, $y$ increases as $x$ increases, \\
when $x=2$, we have $y=-4,$ \\
when $x=8$, we have $y=-1,$ \\
$\therefore -4<y<-1.$ \\
Therefore, the range of function $y$ is $\boxed{-4<y<-1}$.


Question: As shown in the figure, points $A$ and $B$ are two points in the Cartesian coordinate system. Connect $OA$ and $OB$, with $OA=5$, $OB=10$, and $OA \perp OB$. If the x-coordinate of point $A$ is $-4$, and the graph of the inverse proportional function $y=\frac{k_2}{x}$ passes through point $B$, while the graph of the inverse proportional function $y=\frac{k_1}{x}$ passes through point $A$. What are the values of $k_1$ and $k_2$?

Solution:
\textbf{Solution:} Construct perpendicular lines from points A and B to the x-axis, with feet at M and N, respectively, as shown in the figure. \\
According to the problem, we know: OM = 4, OA = 5 \\
By the Pythagorean theorem, we have: $AM = \sqrt{OA^{2} - OM^{2}} = \sqrt{5^{2} - 4^{2}} = 3$ \\
$\therefore$ The coordinates of point A are (-4, 3) \\
Substituting the coordinates of point A into $y = \frac{k_1}{x}$ gives: $\frac{k_1}{-4} = 3$ \\
$\therefore k_1 = \boxed{-12}$ \\
$\because$ OA$\perp$OB, AM$\perp$OM, BN$\perp$ON \\
$\therefore \angle AOB = \angle AMO = \angle BNO = 90^{\circ}$ \\
$\therefore \angle AOM + \angle BON = 90^{\circ}$, $\angle AOM + \angle MAO = 90^{\circ}$ \\
$\therefore \angle MAO = \angle BON$ \\
$\because \angle AMO = \angle BNO = 90^{\circ}$ \\
$\therefore \triangle AMO \sim \triangle ONB$ \\
$\therefore \frac{AM}{ON} = \frac{OA}{OB} = \frac{5}{10} = \frac{1}{2}$ \\
$\therefore ON = 2AM = 6$ \\
In $\triangle BON$, by the Pythagorean theorem, we get: $BN = \sqrt{OB^{2} - ON^{2}} = \sqrt{100 - 36} = 8$ \\
$\therefore$ The coordinates of point B are (6, 8) \\
Substituting the coordinates of point B into $y = \frac{k_2}{x}$ gives: $\frac{k_2}{6} = 8$ \\
$\therefore k_2 = \boxed{48}$


Question: As shown in the figure, points $A$ and $B$ are two points in the Cartesian coordinate system. Connect $OA$ and $OB$, where $OA=5$, $OB=10$, and $OA \perp OB$. If the abscissa of point $A$ is $-4$, and the graph of the inverse proportion function $y=\frac{k_2}{x}$ passes through point $B$, while the graph of the inverse proportion function $y=\frac{k_1}{x}$ passes through point $A$. If point $C$ is on the line segment $AB$, and the area of $\triangle OBC$ is $\frac{1}{2}$ the area of $\triangle OAB$, find the coordinates of point $C$.

Solution:
\textbf{Solution:} Since $S_{\triangle OBC} = \frac{1}{2}S_{\triangle OAB}$,\\
it follows that point C is the midpoint of AB.\\
Draw AD$\perp$BN at point D, and draw CE$\parallel$y-axis intersecting AD at point E, as shown in the figure.\\
Since BN$\perp$x-axis,\\
it follows that BN$\parallel$CE,\\
thus $\triangle ACE \sim \triangle ABD$,\\
which implies $\frac{AE}{AD} = \frac{CE}{BD} = \frac{AC}{AB} = \frac{1}{2}$,\\
therefore, AD=2AE, BD=2CE,\\
meaning point E is the midpoint of AD.\\
Since quadrilateral AMND is a rectangle,\\
it follows that DN=AM=3, AD=MN=10,\\
hence BD=BN−DN=5,\\
Let the coordinates of point C be $(m, n)$,\\
then $6−m=\frac{1}{2}\times 10$, $n−3=\frac{1}{2}\times 5$,\\
thus $m=1$, $n=5.5$,\\
hence $\boxed{C(1, 5.5)}$.


Question: As shown in the figure, the graph of the linear function $y_{1}=kx+b$ intersects the graph of the inverse proportion function $y_{2}=\frac{m}{x}$ at points $A(-2, -5)$ and $C(5, n)$, intersects the y-axis at point B, and intersects the x-axis at point D. What are the expressions for the inverse proportion function and the linear function?

Solution:
\textbf{Solution:} Solution: Since substituting point A ($-2$, $-5$) into $y_{2}=\frac{m}{x}$ gives: $-5=\frac{m}{-2}$,\\
thus $m=10$,\\
therefore, the expression for the inverse proportion function is $y_{2}=\frac{10}{x}$,\\
since substituting point C ($5$, $n$) into the inverse proportion function $y_{2}=\frac{10}{x}$ yields: $n=\frac{10}{5}=2$,\\
therefore, the coordinates of point C are ($5$, $2$),\\
since substituting the coordinates of A and C into $y_{1}=kx+b$ gives: $\left\{\begin{array}{l} -5=-2k+b \\ 2=5k+b \end{array}\right.$,\\
thus $\left\{\begin{array}{l} k=1 \\ b=-3 \end{array}\right.$\\
therefore, the expression for the linear function is $y_{1}=x-3$;\\
Hence, the expression for the inverse proportion function is $y_{2}=\boxed{\frac{10}{x}}$, and the expression for the linear function is $y_{1}=\boxed{x-3}$.


Question: As shown in the figure, the graph of the linear function $y_{1}=kx+b$ intersects with the graph of the inverse proportion function $y_{2}=\frac{m}{x}$ at points $A\left(-2, -5\right)$ and $C\left(5, n\right)$. It intersects the y-axis at point B and the x-axis at point D. Connect $OA$ and $OC$. What is the area of $\triangle AOC$?

Solution:
Solution: Substituting $y=0$ into $y_1=x-3$ yields: $x=3$,\\
therefore, the coordinates of point D are $(3,0)$,\\
therefore, $OD=3$,\\
therefore, $S_{\triangle AOC}=S_{\triangle DOC}+S_{\triangle AOD}=\frac{1}{2}\times 3\times 2+\frac{1}{2}\times 3\times |{-5}|=\boxed{10.5}$;


Question: As shown in the figure, the graph of the linear function $y_1=kx+b$ intersects with the graph of the inverse proportional function $y_2=\frac{m}{x}$ at points $A(-2, -5)$ and $C(5, n)$, intersects the y-axis at point B, and intersects the x-axis at point D. Directly write the range of values of $x$ when $y_1<y_2$.

Solution: \textbf{Solution:} When $y_1<y_2$, the range of $x$ is \boxed{x<-2} or \boxed{0<x<5}.


Question: As shown in the figure, the parabola $y=ax^2+bx+c$ ($a \neq 0$) has its axis of symmetry as the line $x=-1$, intersects the x-axis at points A and B, and intersects the y-axis at point C, where OA = OC. The coordinates of point A are ($-3$, $0$). What is the equation of the parabola?

Solution:
\textbf{Solution:} Let $x=0$, then $y=c$, \\
$\therefore OC=-c$, \\
$\because OA=OC$, \\
$\therefore 3=-c$, i.e., $c=-3$. \\
$\because$ The axis of symmetry is the line $x=-1$, and the coordinates of point A are $(-3,0)$, \\
according to the problem, we get: $\begin{cases}-\frac{b}{2a}=-1\\ 9a-3b+c=0\end{cases}$, \\
Solving this, we get: $\begin{cases}a=1\\ b=2\end{cases}$. \\
$\therefore$ The equation of the parabola is $y=x^{2}+2x-3$. Therefore, the answer is $y=x^{2}+2x-3=\boxed{x^{2}+2x-3}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+c$ ($a\ne 0$) has its axis of symmetry as the line $x=-1$, intersects with the $x$-axis at points $A$ and $B$, and intersects with the $y$-axis at point $C$. It is given that $OA=OC$, and the coordinates of point $A$ are $(-3,0)$. If point $P$ lies on the parabola, and the area of $\triangle POC$ is 4 times the area of $\triangle BOC$, what are the coordinates of point $P$?

Solution:
\textbf{Solution:} When $x=0$, $y=-3$,\\
$\therefore$ point $C(0, -3)$, i.e., $OC=3$,\\
$\because$ A and B are symmetric about the axis of symmetry,\\
$\therefore$ B$(1,0)$, i.e., $OB=1$,\\
$\therefore$ $S_{\triangle BOC}=\frac{1}{2}OB\times OC=\frac{3}{2}$,\\
Let $P(x,{x}^{2}+2x-3)$,\\
$\therefore$ $S_{\triangle POC}=\frac{1}{2}\times3\times|x|$,\\
$\because$ $S_{\triangle POC}=4S_{\triangle BOC}$,\\
$\therefore$ $\frac{3}{2}|x|=4\times \frac{3}{2}$,\\
$\therefore$ $x=\pm 4$,\\
$\therefore$ the coordinates of point $P$ are $\boxed{P(4,21)}$, $\boxed{P(-4,5)}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+c$ ($a\neq 0$) has its axis of symmetry as the line $x=-1$, intersects the $x$-axis at points $A$ and $B$, and intersects the $y$-axis at point $C$, with $OA=OC$. The coordinate of point $A$ is $(-3,0)$. Suppose point $Q$ is a moving point on segment $AC$, and a line is drawn from $Q$ perpendicular to the $x$-axis, intersecting the parabola at point $D$. What is the maximum length of segment $QD$?

Solution:
\textbf{Solution:} Given point A ($-3$, $0$) and point C ($0$, $-3$),\\
hence, the equation of line AC is $y=-x-3$,\\
Let point Q be ($m$, $-m-3$) where $-3\leq m\leq 0$,\\
then point $D(m, m^2+2m-3)$,\\
thus $QD=-m-3-(m^2+2m-3)=-\left(m+\frac{3}{2}\right)^2+\frac{9}{4}$,\\
Therefore, when $m=-\frac{3}{2}$, the maximum value of $QD$ is $\boxed{\frac{9}{4}}$.


Question: As shown in the figure, the parabola $y = -x^2 + bx + c$ intersects the $x$-axis at points $A(2, 0)$ and $B(-4, 0)$. What is the equation of this parabola?

Solution:
\textbf{Solution:} Substituting A(2,0) and B(-4,0) into $y=-x^{2}+bx+c$, we get:
\[
\begin{cases}
-4+2b+c=0\\
-16-4b+c=0
\end{cases}
\]
Solving this, we get:
\[
\begin{cases}
b=-2\\
c=8
\end{cases}
\]
$\therefore$ The equation of the parabola is $y=-x^{2}-2x+8$. Therefore, the answer is $\boxed{y=-x^{2}-2x+8}$.


Question: As shown in the figure, the parabola $y=-x^{2}+bx+c$ intersects the x-axis at points $A(2, 0)$ and $B(-4, 0)$. Determine the axis of symmetry and the coordinates of the vertex of the parabola.

Solution:
\textbf{Solution:} Since $y=-x^2-2x+8=-\left(x+1\right)^2+9$,\\
therefore the axis of symmetry is the line $x=-1$, and the coordinates of the vertex are $\boxed{(-1,9)}$;


Question: As shown in the figure, the parabola $y= -x^{2} + bx + c$ intersects the x-axis at points $A(2,0)$ and $B(-4,0)$. If the parabola intersects the y-axis at point $C$, is there a point $Q$ on the axis of symmetry of the parabola such that the perimeter of $\triangle QAC$ is minimized? If such a point exists, find the coordinates of point $Q$; if it does not exist, please explain why.

Solution:
\textbf{Solution:} Exist, Reason:\\
$\because$ the perimeter of $\triangle QAC = AC + QA + QC$,\\
$\therefore$ to minimize the perimeter of $\triangle QAC$, we only need to minimize $QA + QC$,\\
Based on the given information, points A and C are symmetric with respect to the axis $x = -1$,\\
$\therefore$ the intersection of line $AC$ and line $x = -1$ is point $Q$, at this time $QA + QC$ is minimized, i.e., the perimeter of $\triangle AQC$ is minimized,\\
For $y = -x^2 - 2x + 8$, let $x = 0$, then $y = 8$,\\
$\therefore$ C$(0,8)$,\\
Let the equation of line $AC$ be $y = kx + 8$ ($k \neq 0$),\\
Substituting point C$(0,8)$, we get: $-4k + 8 = 0$, solving gives: $k = 2$,\\
$\therefore$ the equation of line $AC$ is $y = 2x + 8$,\\
When $x = -1$, $y = 2 \times (-1) + 8 = 6$,\\
$\therefore$ the coordinates of point Q are $\boxed{(-1,6)}$.


Question: As shown in the figure, the graph of the linear function \(y=ax+b\) intersects with the graph of the inverse proportion function \(y=\dfrac{k}{x}\) (\(k \neq 0\)) at points \(A(1, -3)\) and \(B(m, -1)\). Connect \(OA\) and \(OB\). What is the expression of the linear function?

Solution:
\textbf{Solution:} Since the inverse proportion function is $y=\frac{k}{x}$ (where $k \neq 0$) and its graph passes through the point A(1, −3), \\
we have $k=1\times(-3)=-3$, \\
thus, the expression for the inverse proportion function is $y=-\frac{3}{x}$. \\
Given that the graph of the inverse proportion function $y=-\frac{3}{x}$ passes through point B(m, −1), \\
it follows that $m=3$, \\
therefore, point B is (3, −1). \\
Since points A(1, −3) and B(3, −1) lie on the graph of the linear function $y=ax+b$, \\
we obtain $\left\{\begin{array}{l}a+b=-3 \\ 3a+b=-1\end{array}\right.$, from which we solve $\left\{\begin{array}{l}a=1 \\ b=-4\end{array}\right.$, \\
thus, the expression for the linear function is $\boxed{y=x-4}$;


Question: As shown in the figure, the graph of the linear function $y=ax+b$ intersects with the graph of the reciprocal function $y=\frac{k}{x}$ ($k\ne 0$) at points $A(1, -3)$ and $B(m, -1)$. Connect $OA$ and $OB$. What is the area of $\triangle OAB$?

Solution:
\textbf{Solution:} Let the intersection point of line AB with the y-axis be C. Then, the coordinates of point C are $(0, -4)$. \\
The area of triangle $OAB$, $S_{\triangle OAB}=S_{\triangle BOC}-S_{\triangle AOC}=\frac{1}{2}\times 4\times 3- \frac{1}{2}\times 4\times 1=\boxed{4}.$


Question: As shown in the figure, the line $l: y=k_{1}x+b$ intersects with the hyperbola $y=\frac{k_{2}}{x}(x>0)$ at points A and B, and intersects with the x-axis at point C. If the x-coordinates of points A and B are 1 and 2 respectively, please directly provide the solution set of $k_{1}x+b>\frac{k_{2}}{x}$.

Solution:
\textbf{Solution:} From the given problem, we have $k_{1}x+b>\frac{k_{2}}{x}$,\\
the solution set is $x\in(1,2)$,\\
$\therefore$ the solution set is $x\in\boxed{(1,2)}$.


Question: As shown in the figure, the line $l: y=k_{1}x+b$ intersects the hyperbola $y=\frac{k_{2}}{x}$ (for $x>0$) at points $A$ and $B$, and intersects the $x$-axis at point $C$. If the abscissas of points $A$ and $B$ are $1$ and $2$, respectively, and the area of $\triangle AOB$ is $3$, what is the value of $k_{2}$?

Solution:
\textbf{Solution:} Let $AM\perp x$-axis at $M$, and $BN\perp x$-axis at $N$. Hence, the area of $\triangle AOM$ and the area of $\triangle BON$ are both equal to $\frac{1}{2}|k_{2}|$, \\
Let $A(1,k_{2})$, and $B(2,\frac{k_{2}}{2})$, \\
Since the area of $\triangle AOB$ is $3$, \\
Therefore, $S_{\triangle AOB}=S_{\triangle AOM}+S_{\text{trapezoid} AMNB}-S_{\triangle BON}=S_{\text{trapezoid} AMNB}=\frac{1}{2}\left(k_{2}+\frac{k_{2}}{2}\right)\times(2-1)=3$, \\
Hence, $k_{2}=\boxed{4}$.


Question: As shown in the figure, the points $A(-6, 2)$ and $B(n, -4)$ are two intersection points of the linear function $y=kx+b$ and the inverse proportion function $y=\frac{m}{x}$. Find the expressions for the inverse proportion function and the linear function.

Solution:
\textbf{Solution:} Substitute $A(-6, 2)$ into $y=\frac{m}{x}$, thus we get $m=2\times(-6)=-12$, \\
$\therefore$ the inverse proportion function equation is $y=-\frac{12}{x}$, \\
Substituting $B(n, -4)$ into $y=-\frac{12}{x}$, it yields $-4=-\frac{12}{n}$, \\
Solving that gives $n=3$, \\
Substituting $A(-6, 2)$ and $B(3, -4)$ into $y=kx+b$, we get \\
$\begin{cases}
-6k+b=2\\
3k+b=-4
\end{cases}$, resulting in: $\begin{cases}
k=-\frac{2}{3}\\
b=-2
\end{cases}$, \\
$\therefore$ the linear function equation is $\boxed{y=-\frac{2}{3}x-2}$;


Question: As shown in the figure, points $A\left(-6, 2\right)$ and $B\left(n, -4\right)$ are two intersections of the linear function $y=kx+b$ and the inverse proportion function $y=\frac{m}{x}$. What is the area of $\triangle AOB$?

Solution:
\textbf{Solution:} Solving by substituting $y=0$ into $y=-\frac{2}{3}x-2$, we get $0=-\frac{2}{3}x-2$,\\
which yields: $x=-3$,\\
hence, the coordinates of point C are ($-3$, $0$),\\
thus, $S_{\triangle AOB}=S_{\triangle AOC}+S_{\triangle BOC}=\frac{1}{2} \times OC \times y_A + \frac{1}{2} \times OC \times (-y_B) = \frac{1}{2} \times OC \times (y_A-y_B) = \frac{1}{2} \times 3 \times (2+4) = \boxed{9}$.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y_1=kx+b$ intersects with the x-axis at point $A$ and with the y-axis at point $B$. It intersects with the graph of the inverse proportion function $y_2=\frac{m}{x}$ at points $C$ and $D$, where point $C(-2,3)$, and the ordinate of point $D$ is $-1$. Find the analytic expressions of the inverse proportion function and the linear function?

Solution:
\textbf{Solution:} Given the inverse proportional function $y_2=\frac{m}{x}$ passes through point $C(-2, 3)$, substituting gives:\\
$\therefore m=-2\times 3=-6$,\\
$\therefore$ the equation of the inverse proportional function is $y=-\frac{6}{x}$;\\
Given the y-coordinate of point $D$ is $-1$,\\
$\therefore D(6, -1)$,\\
Substituting points $C$ and $D$ into $y_1=kx+b$,\\
we get $\begin{cases}-2k+b=3\\ 6k+b=-1\end{cases}$, solving this gives $\begin{cases}k=-\frac{1}{2}\\ b=2\end{cases}$,\\
$\therefore$ the equation of the inverse proportional function is $\boxed{y=-\frac{6}{x}}$,\\
$\therefore$ the equation of the linear function is $\boxed{y=-\frac{1}{2}x+2}$;


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y_1=kx+b$ intersects the x-axis at point A and the y-axis at point B. It intersects with the graph of the hyperbolic function $y_2=\frac{m}{x}$ at points C and D, with point C being (-2, 3) and the ordinate of point D being -1. If point E is a point on the graph of the hyperbolic function in the fourth quadrant, and a line is drawn from point E perpendicular to the y-axis to point F, and lines OE and AF are connected, if the area of $\triangle BAF$ is 4 times that of $\triangle EFO$, determine the coordinates of point E.

Solution:
\textbf{Solution:} From the problem, since the graph of the linear function $y_{1}=-\frac{1}{2}x+2$ intersects the x-axis at point A and the y-axis at point B, \\
$\therefore A(4, 0)$, $B(0, 2)$, that is $OA=4$, $OB=2$, \\
Let $E(x, -\frac{6}{x})$, \\
since $E$ is in the fourth quadrant, \\
$\therefore EF=x$, $OF=\frac{6}{x}$, \\
$\therefore$ The area of $\triangle EFO=\frac{1}{2}EF\cdot OF=\frac{1}{2}x\cdot \frac{6}{x}=3$, \\
since the area of $\triangle BAF=4$ times the area of $\triangle EFO$, \\
$\therefore$ The area of $\triangle BAF=12$, \\
$\therefore$ The area of $\triangle BAF=\frac{1}{2}BF\cdot OA=\frac{1}{2}BF\cdot 4=12$, \\
$\therefore BF=6$, \\
$\therefore OF=6-2=4$, \\
$\therefore F(0, -4)$, \\
When $y=-4$, substitute into $y=-\frac{6}{x}$, and solve to obtain $x=\frac{3}{2}$, \\
$\therefore E\left(\frac{3}{2}, -4\right)$．Therefore, the coordinates of point E are $\boxed{\left(\frac{3}{2}, -4\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the parabola passes through points $A(-1,0)$, $B(4,0)$, and $C(0,-2)$. Determine the equation of the parabola and its axis of symmetry.

Solution:
\textbf{Solution}: Let the equation of the parabola be $y=ax^{2}+bx+c$, \\
since the parabola passes through points A($-1, 0$), B($4, 0$), and C($0, -2$), substituting, we get: \\
\[\begin{cases}
a-b+c=0\\
16a+4b+c=0\\
c=-2
\end{cases}\] \\
Solving, we find: \\
\[\begin{cases}
a=\frac{1}{2}\\
b=-\frac{3}{2}\\
c=-2
\end{cases}\] \\
Therefore, the equation of this parabola is $y=\frac{1}{2}x^{2}-\frac{3}{2}x-2$. \\
Since the equation of the parabola is $y=\frac{1}{2}x^{2}-\frac{3}{2}x-2=\frac{1}{2}(x-\frac{3}{2})^{2}-\frac{25}{8}$ \\
Therefore, the axis of symmetry of the parabola is $x=\boxed{\frac{3}{2}}$.


Question: As shown in the diagram, in the Cartesian coordinate system, the parabola passes through points A $(-1, 0)$, B $(4, 0)$, and C $(0, -2)$. Is there a point P on the axis of symmetry of this parabola such that the perimeter of $\triangle PAC$ is minimized? If such a point exists, find the coordinates of point P; if not, explain why.

Solution:
\textbf{Solution}: Exist, the reason is as follows:\\
Connect PB\\
Due to the symmetry of the parabola: $PA=PB$\\
$\therefore$ the perimeter of $\triangle PAC$ $PA+PC+AC=PB+PC+AC$,\\
$\therefore$ when points B, P, C are collinear, $PB+PC$ is minimized,\\
that is, when points B, P, C are collinear, the perimeter of $\triangle PAC$ is minimized,\\
Suppose the equation of line BC is $y=kx+m$,\\
Substitute points B$(4,0)$ and C$(0,-2)$ into it,\\
we get $\begin{cases}0=4k+m\\ -2=m\end{cases}$, and solve to get: $\begin{cases}k=\frac{1}{2}\\ m=-2\end{cases}$,\\
thus, the equation of line BC is $y=\frac{1}{2}x-2$.\\
Let $x=\frac{3}{2}$, then $y=\frac{1}{2}\times \frac{3}{2}-2=-\frac{5}{4}$,\\
thus, the coordinates of point P are $\left(\frac{3}{2},-\frac{5}{4}\right)$.\\
$\therefore$ There exists a point P on the axis of symmetry of this parabola that minimizes the perimeter of $\triangle PAC$, at which the coordinates of point P are $\boxed{\left(\frac{3}{2},-\frac{5}{4}\right)}$.


Question: Synthesis and Inquiry: As shown in the figure, in the Cartesian coordinate system, the graph of the quadratic function $y=x^2+bx+c$ passes through point $A\left(0, -\frac{7}{4}\right)$ and point $B\left(1, \frac{1}{4}\right)$. What is the analytical expression of this quadratic function?

Solution:
\textbf{Solution:} Substitute $A\left(0, -\frac{7}{4}\right)$ and point $B\left(1, \frac{1}{4}\right)$ into $y=x^{2}+bx+c$ to get: \\
\[
\begin{cases}
-\frac{7}{4}=c\\
\frac{1}{4}=1+b+c
\end{cases}
,\] solving this system of equations gives
\[
\begin{cases}
b=1\\
c=-\frac{7}{4}
\end{cases},
\]
$\therefore y=x^{2}+x-\frac{7}{4}$. Therefore, the quadratic function is $\boxed{y=x^{2}+x-\frac{7}{4}}$.


Question: Synthesis and Exploration: As shown in the diagram, in the Cartesian coordinate system, the graph of the quadratic function $y=x^2+bx+c$ passes through point $A\left(0, -\frac{7}{4}\right)$ and point $B\left(1, \frac{1}{4}\right)$. When $-2\le x\le 2$, what are the maximum and minimum values of the quadratic function $y=x^2+bx+c$?

Solution:
\textbf{Solution:} Given the equation $y=x^2+bx+c$ can be rewritten as $y=\left(x+\frac{1}{2}\right)^2-\frac{9}{4}$,\\
$\therefore$ the parabola opens upwards, and its axis of symmetry is the line $x=-\frac{1}{2}$.\\
$\therefore$ when $x=-\frac{1}{2}$, $y$ attains its minimum value of $-\frac{9}{4}$.\\
$\because$ $2-\left(-\frac{1}{2}\right) > -\frac{1}{2}-(-2)$,\\
$\therefore$ when $x=2$, $y$ reaches its maximum value $2^2+2\cdot2+c=\boxed{\frac{17}{4}}$.


Question: Synthesis and Exploration: As shown in the figure, in the Cartesian coordinate system, the graph of the quadratic function $y=x^2+bx+c$ passes through point $A\left(0, -\frac{7}{4}\right)$ and point $B\left(1, \frac{1}{4}\right)$. Point $P$ is an arbitrary point on the graph of this function, with the x-coordinate $m$. A line segment $PQ$ is drawn parallel to the x-axis from point $P$, and the x-coordinate of point $Q$ is $-2m+1$. It is known that point $P$ and point $Q$ do not coincide, and the length of segment $PQ$ decreases as $m$ increases. What is the range of values for $m$?

Solution:
Solution: $PQ=\left|-3m+1\right|$, \\
when $-3m+1>0$, $PQ=-3m+1$, the length of $PQ$ decreases as $m$ increases, \\
when $-3m+1<0$, $PQ=3m-1$, the length of $PQ$ increases as $m$ increases, \\
$\therefore$ $-3m+1>0$ satisfies the condition, \\
solving gives $\boxed{m<\frac{1}{3}}$.


Question: As shown in the figure, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\frac{k_2}{x} \left(x>0\right)$ at points $A$ and $B$ within the first quadrant. Given $A(1, m)$ and $B(2, 1)$, find the equations of the line and the hyperbola.

Solution:
\textbf{Solution:} Given point $B(2, 1)$ is on the hyperbola $y_{2}=\frac{k_{2}}{x}$, \\
$\therefore k_{2}=2\times 1=2$,\\
$\therefore$ the equation of the hyperbola is $y_{2}=\frac{2}{x}$\\
Since $A(1, m)$ is on the hyperbola $y_{2}=\frac{2}{x}$, \\
$\therefore m=\frac{2}{1}=2$,\\
$\therefore A(1, 2)$\\
Since line $y_{1}=k_{1}x+b$ passes through points $A(1, 2)$ and $B(2, 1)$,\\
$\therefore \left\{\begin{array}{l}k_{1}+b=2\\ 2k_{1}+b=1\end{array}\right.$, solving this gives $\left\{\begin{array}{l}k_{1}=-1\\ b=3\end{array}\right.$,\\
$\therefore$ the equation for line $AB$ is $\boxed{y_{1}=-x+3}$.


Question: As shown in the figure, the line $y_1=k_1x+b$ intersects with the hyperbola $y_2=\frac{k_2}{x}\ (x>0)$ at points $A$ and $B$ within the first quadrant, where $A(1, m)$ and $B(2, 1)$. Based on the graph, directly write down the solution set for the inequality $y_1<y_2$.

Solution: \textbf{Solution:} The solution set for the inequality $y_1<y_2$ according to the graph is $0<x<1$ or $x>2$. Therefore, the answer is $x\in\boxed{(0,1)\cup(2,+\infty)}$.


Question: As shown in the figure, the graph of the linear function $y=-\frac{1}{2}x+1$ intersects with the graph of the inverse proportion function $y=\frac{k}{x}$ at points $A$ and $B$. The x-coordinate of point $A$ is $-2$, and the x-coordinate of point $B$ is $4$. The graph of the linear function intersects with the y-axis at point $C$. What is the expression of the inverse proportion function?

Solution:
\textbf{Solution:} When $x=-2$, $y=-\frac{1}{2}\times (-2)+1=2$,\\
$\therefore \boxed{A(-2,2)}$,\\
$\because$ the graph of the inverse proportion function $y=\frac{k}{x}$ passes through point A,\\
$\therefore k=2\times (-2)=-4$,\\
$\therefore$ the expression for the inverse proportion function is $\boxed{y=-\frac{4}{x}}$.


Question: As shown in the figure, the graph of the linear function $y=-\frac{1}{2}x+1$ intersects the graph of the hyperbolic function $y=\frac{k}{x}$ at points $A$ and $B$, with the $x$-coordinate of point $A$ being $-2$ and the $x$-coordinate of point $B$ being $4$. The graph of the linear function intersects the $y$-axis at point $C$. If point $P$ is on the $y$-axis, and the area of $\triangle ABP$ is $6$, find the coordinates of point $P$.

Solution:
\textbf{Solution:}

Given that the graph of the inverse proportion function $y=-\frac{4}{x}$ passes through point B, and the x-coordinate of point B is 4,

$\therefore y=-1$, i.e., B(4, -1),

Let the coordinates of point P be (0, $a$),

Since the graph of the linear function $y=-\frac{1}{2}x+1$ intersects the y-axis at point C,

$\therefore$ C(0, 1),

$\therefore$ PC=$|1-a|$,

Since the area of $\triangle ABP$ is 6,

$\therefore$ $\frac{1}{2}\times |1-a|\times 6=6$,

Solving gives $a=-1$ or $a=3$,

$\therefore$ the coordinates of point P are (0, $-1$) or (0, $\boxed{3}$).


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

Solution:
\textbf{Solution:} Given that the parabola $y=ax^{2}+bx+3$ passes through points $A(1,0)$ and $B(-3,0)$,\\
we have $\begin{cases} a+b+3=0 \\ 9a-3b+3=0 \end{cases}$,\\
solving these equations yields $\begin{cases} a=-1 \\ b=-2 \end{cases}$,\\
therefore, the equation of the parabola is: \boxed{y=-x^{2}-2x+3}.


Question: As shown in the figure, the parabola $y=ax^2+bx+3$ ($a\neq 0$) intersects the x-axis at points $A(1,0)$ and $B(-3,0)$, and intersects the y-axis at point $C$. The line segment $BC$ is drawn and intersects the axis of symmetry of the parabola at point $E$, with the vertex being point $D$. If $M$ is a moving point on the parabola located above the line segment $BC$, what is the maximum area of $\triangle BCM$?

Solution:
\textbf{Solution:} Draw $MF\perp x$ axis from point M, intersecting BC at point G and the x-axis at point F, \\
when $x=0$, $y=3$, thus the coordinates of C are $(0,3)$, \\
it is easy to derive the equation of BC as $y=x+3$, \\
let $G(x, x+3)$, $M(x, -x^{2}-2x+3)$, \\
then $MG= -x^{2}-2x+3-(x+3)=-x^{2}-3x$, \\
Therefore, ${S}_{\triangle BCM}={S}_{\triangle BMG}+{S}_{\triangle CMG}=\frac{1}{2}MG\cdot BO=\frac{1}{2}(-x^{2}-3x)\times 3=-\frac{3}{2}x^{2}-\frac{9}{2}x$, \\
since $a=-\frac{3}{2}<0$, \\
Therefore, when $x=-\frac{b}{2a}=-\frac{3}{2}$, ${S}_{\triangle BCM}$ reaches its maximum value, which is $\boxed{\frac{27}{8}}$.


Question: As shown in the diagram, the parabola $y=ax^2+bx+3$ ($a\neq 0$) intersects the x-axis at points $A(1,0)$ and $B(-3,0)$, and intersects the y-axis at point $C$. Line $BC$ is drawn and it intersects the axis of symmetry of the parabola at point $E$, with the vertex being point $D$. Point $P$ is a moving point on the left side of the parabola's axis of symmetry, and point $Q$ lies on the ray $ED$. If the triangle formed by points $P$, $Q$, and $E$ is similar to $\triangle BOC$, please write down the coordinates of point $P$ directly.

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


Question: As shown in the figure, within the Cartesian coordinate system, the vertices of rectangle OABC are $A\left(0, 3\right)$, and $C\left(-1, 0\right)$. Now, rectangle OABC is rotated clockwise by $90^\circ$ around the origin O to obtain rectangle $O{A}^{\prime}{B}^{\prime}{C}^{\prime}$. The line $B{B}^{\prime}$ intersects the x-axis at point M and the y-axis at point N. The graph of the parabola $y=ax^{2}+bx+c$ passes through points C, M, and N. Please directly write down the coordinates of points B and ${B}^{\prime}$.

Solution: \textbf{Solution:} Point $B$ and point ${B}^{\prime}$ have coordinates $\boxed{B(-1, 3)}$ and $\boxed{{B}^{\prime}(3, 1)}$, respectively.


Question: As shown in the figure, in the Cartesian coordinate system, the vertices of rectangle $OABC$ are $A(0, 3)$ and $C(-1, 0)$. Now, rectangle $OABC$ is rotated $90^\circ$ clockwise around the origin $O$ to obtain rectangle $O{A}^{\prime }{B}^{\prime }{C}^{\prime }$. The line $B{B}^{\prime }$ intersects the x-axis at point $M$ and the y-axis at point $N$. The graph of the parabola $y=ax^{2}+bx+c$ passes through points $C$, $M$, and $N$. Find the equation of the parabola.

Solution:
\textbf{Solution:} Let the equation of line BB' be $y=kx+b$. Substituting the points B $(-1, 3)$, B' $(3, 1)$ into it, we obtain
\[
\begin{cases}
-k+b=3\\
3k+b=1
\end{cases},
\]
from which it follows that
\[
\begin{cases}
k=-\frac{1}{2}\\
b=\frac{5}{2}
\end{cases}
\]
$\therefore$ The equation of line BB' is $y=-\frac{1}{2}x+\frac{5}{2}$; setting $x=0$, we find $y=\frac{5}{2}$, and setting $y=0$, we find $x=5$, $\therefore$ the intersection points of line BB' with the x-axis and y-axis are M $(5, 0)$ and N $(0, \frac{5}{2})$, respectively. $\because$ C $(-1, 0)$, $\therefore$ let the equation of the parabola be $y=a(x-5)(x+1)$, $\because$ the parabola passes through point N, $\therefore$ $\frac{5}{2}=a\times (-5)\times 1$, $\therefore$ a $=-\frac{1}{2}$, $\therefore$ the equation of the parabola is $y=-\frac{1}{2}(x-5)(x+1)=-\frac{1}{2}x^2+2x+\frac{5}{2}$; hence, the equation of the parabola is $\boxed{y=-\frac{1}{2}x^2+2x+\frac{5}{2}}$.


Question: As shown in the figure, in the Cartesian coordinate system, rectangle $OABC$ has vertices at $A\left(0, 3\right)$ and $C\left(-1, 0\right)$. Now, rotate rectangle $OABC$ clockwise by $90^\circ$ around the origin $O$ to obtain rectangle $O{A}^{\prime}{B}^{\prime}{C}^{\prime}$. The line $B{B}^{\prime}$ intersects the $x$-axis at point $M$ and the $y$-axis at point $N$. The parabola $y=ax^{2}+bx+c$ passes through points $C$, $M$, and $N$. Let point $P$ be a moving point on the parabola and above the line $B{B}^{\prime}$. Determine the coordinates of point $P$ and the maximum value of the area of $\triangle PMN$ when the area of $\triangle PMN$ is maximized.

Solution:
\textbf{Solution:} As shown in the figure, draw $PQ \perp x$-axis through point P, intersecting $BB'$ at point Q. Let $P\left(m, -\frac{1}{2}m^2+2m+\frac{5}{2}\right)$, then $Q\left(m, -\frac{1}{2}m+\frac{5}{2}\right)$,\\
$\therefore PQ=-\frac{1}{2}m^2+2m+\frac{5}{2}-\left(-\frac{1}{2}m+\frac{5}{2}\right)=-\frac{1}{2}m^2+\frac{5}{2}m=-\frac{1}{2}\left(m^2-5m+\frac{25}{4}\right)+\frac{25}{8}$;\\
When $m=\frac{5}{2}$, $PQ$ attains its maximum value of $\frac{25}{8}$, at this moment $-\frac{1}{2}\left(\frac{5}{2}-5\right)\left(\frac{5}{2}+1\right)=\frac{35}{8}$,\\
$\therefore P\left(\frac{5}{2}, \frac{35}{8}\right)$\\
$\therefore$ The maximum area of $\triangle PMN$ is: $S_{\triangle PMN}=\frac{1}{2}\times OM\times PQ=\frac{1}{2}\times 5\times \frac{25}{8}=\boxed{\frac{125}{16}}$.


Question: As shown in the figure, the line $y=\frac{1}{2}x$ intersects the graph of the inverse proportion function $y=\frac{k}{x}\ (x>0)$ at point $A(4, m)$, and point $B(n, 2n)$ is another point on the graph of the inverse proportion function. The line AB intersects the x-axis at point C. Find the analytical expression of the inverse proportion function and the coordinates of point B?

Solution:
\textbf{Solution:} Given that the line $y=\frac{1}{2}x$ intersects the graph of the inverse proportionality function $y=\frac{k}{x}\left(x>0\right)$ at point $A\left(4, m\right)$,\\
then $m=\frac{1}{2}\times 4=2$,\\
thus the coordinates of point A are $\left(4, 2\right)$,\\
thereby $k=2\times 4=8$,\\
hence the equation of the inverse proportionality function is: $y=\frac{8}{x}$,\\
Given that point $B\left(n, 2n\right)$ lies on the graph of the inverse proportionality function,\\
thus $2n=\frac{8}{n}$,\\
solving this gives: $n_{1}=2$, $n_{2}=-2$ (discard),\\
upon verification, these are the roots of the equation\\
therefore, the coordinates of point B are $\boxed{\left(2, 4\right)}$;


Question: As shown in the figure, the line $y=\frac{1}{2}x$ intersects the graph of the inverse proportion function $y=\frac{k}{x}\left(x>0\right)$ at point $A\left(4, m\right)$. Point $B\left(n, 2n\right)$ is another point on the graph of the inverse proportion function, and the line AB intersects the x-axis at point C. Find the area of $\triangle AOB$.

Solution:
\textbf{Solution:} Connect OB,\\
Assume the equation of line AB is: $y=-x+6$, substituting points $A(4, 2)$ and $B(2, 4)$ into it, we get:\\
$\left\{\begin{array}{l}
2=4a+b \\
4=2a+b
\end{array}\right.$, solving this yields: $\left\{\begin{array}{l}
a=-1 \\
b=6
\end{array}\right.$,\\
$\therefore$ The equation of line AB is: $y=-x+6$,\\
Then $C(6, 0)$,\\
Therefore, the area of $\triangle AOB ={S}_{\triangle BOC}-{S}_{\triangle AOC}=\frac{1}{2}\times 6\times 4-\frac{1}{2}\times 6\times 2=\boxed{6}$.


Question: As shown in the figure, the graph of the linear function $y = -x + b$ intersects with the graph of the inverse proportion function $y = \frac{k}{x}$ ($x < 0$) at point $A(-6, m)$, and intersects with the x-axis at point $B(-4, 0)$. What are the expressions for the linear function and the inverse proportion function?

Solution:
\textbf{Solution:} Given $B(-4, 0)$ lies on $y=-x+b$, we find that $b=-4$,\\
thus the expression of the linear function is $y=-x-4$.\\
Given $A(-6, m)$ lies on $y=-x-6$, we find that $m=-2$,\\
therefore $A(-6, -2)$.\\
Substituting $A(-6, -2)$ into $y=\frac{k}{x}$ yields $k=-12$,\\
therefore the expression of the inverse proportionality function is $\boxed{y=-\frac{12}{x}}$.


Question: In the figure, the graph of the linear function $y=-x+b$ intersects with the graph of the hyperbolic function $y=\frac{k}{x}$ ($x<0$) at point $A(-6,m)$, and intersects with the x-axis at point $B(-4,0)$. If the line $y=4$ intersects line $AB$ at point $C$, and intersects the hyperbola at point $D$, based on the graph, directly write out the solution set for the inequality $-x+b<\frac{k}{x}<4$.

Solution: \textbf{Solution:} Solving directly, the solution set of the inequality $-x+b<\frac{k}{x}<4$ is $\boxed{-6<x<-3}$.


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

Solution:
\textbf{Solution:} Substitute B ($-3$, $0$) and C ($0$, $3$) into $y=-x^2+bx+c$, yielding
\[
\left\{
\begin{array}{l}
-9-3b+c=0 \\
c=3
\end{array}
\right.
\]
Therefore,
\[
\left\{
\begin{array}{l}
b=-2 \\
c=3
\end{array}
\right.
\]
Therefore, the equation of the parabola is: \boxed{y=-x^2-2x+3}.


Question: As shown in the figure, it is known that the parabola $y = -x^2 + bx + c$ passes through the points A, B ($-3, 0$), and C ($0, 3$). Find a point M on the axis of symmetry of the parabola such that the sum of the distances from point M to points O and C is minimized. Determine the coordinates of point M at this time.

Solution:
\textbf{Solution:} \\
Since the axis of symmetry of the parabola is the line $x=-1$, construct point $C(0,3)$'s symmetric point $D(-2,3)$ about the line $x=-1$.\\
Find a point $M$ on the axis of symmetry of the parabola such that the sum of distances from point $M$ to points $O$ and $C$ is minimized, that is $OM+CM=OM+MD\leq OD$,\\
The sum is shortest when points $D$, $M$, and $O$ are collinear,\\
Let the equation of line $OD$ be: $y=kx$, passing through point $D$,\\
Therefore, $3=-2k, k=-\frac{3}{2}$,\\
Therefore, the equation of line $OD$ is: $y=-\frac{3}{2}x$.\\
When $x=-1$, $y=\frac{3}{2}$,\\
Therefore, the coordinates of $M$ are $\boxed{(-1, \frac{3}{2})}$.


Question: As shown in the figure, it is known that the parabola $y=-x^2+bx+c$ passes through points A, B ($-3$, $0$), and C ($0$, $3$). Let point P be a moving point on the axis of symmetry of the parabola. Directly write the coordinates of point P that make $\triangle BPC$ a right-angled triangle.

Solution: \textbf{Solution:} The coordinates of point P that make $\triangle BPC$ a right-angled triangle are $\boxed{P_1(-1, \frac{3+\sqrt{17}}{2})}$, $\boxed{P_2\left(-1, \frac{3-\sqrt{17}}{2}\right)}$, $\boxed{P_3\left(-1, 4\right)}$, and $\boxed{P_4\left(-1, -2\right)}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\frac{4}{x}$ (where $x>0$) at points $A(m, 4)$ and $B(2, n)$, and intersects with the coordinate axes at points $M$ and $N$, respectively. What is the analytical expression of the linear function?

Solution:
\textbf{Solution:} Given that A $(1, 4)$ and B $(2, 2)$ are on the graph of the inverse proportion function $y=\frac{4}{x}$,

\[
\therefore
\begin{cases}
4=\frac{4}{m} \\
n=\frac{4}{2}
\end{cases}
\text{, which implies }
\begin{cases}
m=1 \\
n=2
\end{cases},
\]
thus A $(1, 4)$ and B $(2, 2)$.

Substituting A $(1, 4)$ and B $(2, 2)$ into $y=kx+b$, we get:
\[
\begin{cases}
4=k+b \\
2=2k+b
\end{cases}
\]
which solves to:
\[
\begin{cases}
k=-2 \\
b=6
\end{cases},
\]
therefore, the equation of the linear function is \boxed{y=-2x+6}.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\frac{4}{x}$ ($x>0$) at points $A(m, 4)$ and $B(2, n)$. It also intersects with the coordinate axes at points $M$ and $N$. Based on the graph, write directly the solution set of the inequality $kx+b-\frac{4}{x}<0$.

Solution:
\textbf{Solution:} According to the graph, we directly obtain the solution set of the inequality $kx+b-\frac{4}{x}<0$ as $0<x<1$ or $x>2$.\\
Hence, the answer is $x\in(0,1)\cup(x>2)$.\\
Therefore, the solution set is $x\in\boxed{(0,1)\cup(2,\infty)}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\frac{4}{x}$ ($x>0$) at two points $A(m, 4)$ and $B(2, n)$. It also intersects with the coordinate axes at points $M$ and $N$ respectively. What is the area of $\triangle AOB$?

Solution:
\textbf{Solution:} For $y=-2x+6$, let $y=0$, then $-2x+6=0$,\\
solving this yields $x=3$,\\
$\therefore$ the coordinates of point N are $(3,0)$,\\
$\therefore$ ${S}_{\triangle AOB}={S}_{\triangle AON}-{S}_{\triangle BON}=\frac{1}{2}ON\cdot y_A-\frac{1}{2}ON\cdot y_B=\frac{1}{2}\times 3\times (4-2)=\boxed{3}$.


Question: As shown in the figure, the four vertices of rectangle DEFG are respectively on the sides of equilateral triangle ABC. Given that the side length of $\triangle ABC$ is 4, and denoting the area of rectangle DEFG as $S$, and the line segment BE as $x$. Find the expression of $S$ as a function of $x$.

Solution:
\textbf{Solution:}

In the equilateral triangle $ABC$, the length of line segment $BE$ is denoted as $x$,

$\therefore \angle B = \angle C = 60^\circ$.

$\therefore DE = BE \cdot \tan 60^\circ = \sqrt{3}x$.

Since the four vertices of rectangle $DEFG$ are located on the sides of equilateral triangle $ABC$,

$\therefore DE = GF$, and $\angle BED = \angle CFG = 90^\circ$,

$\therefore \triangle BED \cong \triangle CFG \left(AAS\right)$.

$\therefore BE = CF = x$.

Since the side length of $\triangle ABC$ is 4,

$\therefore EF = 4 - 2x$.

$\therefore S = DE \cdot EF = \sqrt{3}x(4 - 2x) = -2\sqrt{3}x^2 + 4\sqrt{3}x \left(0 < x < 2\right)$.

Thus, the function expression for $S$ with respect to $x$ is $S = \boxed{-2\sqrt{3}x^2 + 4\sqrt{3}x}$.


Question: As shown in the figure, the four vertices of rectangle DEFG each lie on the sides of equilateral triangle ABC. It is known that the side length of $\triangle ABC$ is 4. Denote the area of rectangle DEFG as $S$, and the length of segment BE as $x$. When $S = \sqrt{3}$, find the value of $x$.

Solution:
\textbf{Solution:} When $S=\sqrt{3}$, we get $-2\sqrt{3}x^{2}+4\sqrt{3}x=\sqrt{3}$. \\
Solving gives $x=1\pm \frac{\sqrt{2}}{2}$. \\
Since $0<x<2$, \\
it follows that \boxed{x=1\pm \frac{\sqrt{2}}{2}}.


Question: As shown in the figure, the graph of the inverse proportion function $y=\frac{k}{x} (k\neq 0)$ intersects with the graph of the direct proportion function $y=2x$ at points $A(1, a)$ and $B$. Point $C$ is located in the fourth quadrant, with $CA\parallel y$-axis and $AB\perp BC$. What is the equation of the inverse proportion function and the coordinates of point $B$?

Solution:
\textbf{Solution:} Since point A$(1, a)$ is on $y=2x$,\\
it follows that $a=2$,\\
hence $A(1, 2)$.\\
Substituting A$(1, 2)$ into $y=\frac{k}{x}$ yields $k=2$,\\
therefore, the equation of the inverse proportion function is $y=\frac{2}{x}$.\\
Since points A and B are centrally symmetric with respect to the origin,\\
it follows that \boxed{B(-1, -2)}.


Question: As shown in the figure, the graph of the inverse proportion function $y=\frac{k}{x}\ (k\ne 0)$ intersects with the graph of the direct proportion function $y=2x$ at points $A(1, a)$ and $B$. Point $C$ is located in the fourth quadrant, with $CA\parallel y$-axis, and $AB\perp BC$. What is the area of $\triangle ABC$?

Solution:
\textbf{Solution:} As shown in the figure, draw $BH\perp AC$ at $H$, and let $AC$ intersect the x-axis at point $D$, \\
$\because AB\perp BC$, \\
$\therefore \angle ABC=90^\circ$, $\angle BHC=90^\circ$, \\
$\therefore \angle C=\angle ABH$, \\
$\because BH\parallel x$-axis, \\
$\therefore \angle AOD=\angle ABH$, \\
$\therefore \angle AOD=\angle C$, \\
$\therefore \tan C=\tan\angle AOD=\frac{AD}{OD}=2$, \\
$\because A(1,2), B(-1,-2)$, \\
$\therefore AH=4$, $BH=2$, $OD=1$, $AD=2$, \\
$\therefore AB=\sqrt{AH^2+BH^2}=\sqrt{4^2+2^2}=2\sqrt{5}$, the area of $\triangle AOD=\frac{1}{2}OD\cdot AD=1$, \\
$\because \angle AOD=\angle C$, $\angle ADO=\angle ABC=90^\circ$, \\
$\therefore \triangle ADO\sim \triangle ABC$, \\
$\therefore$ the ratio $\frac{S_{\triangle ADO}}{S_{\triangle ABC}}=\left(\frac{AD}{AB}\right)^2$, hence $\frac{1}{S_{\triangle ABC}}=\left(\frac{2}{2\sqrt{5}}\right)^2$, \\
Solving gives $S_{\triangle ABC}=\boxed{5}$.


Question: As shown in the diagram, $\triangle AOB$ is an isosceles right triangle with the hypotenuse $OB$ on the x-axis. The graph of the linear function $y=3x-4$ passes through point A and intersects the y-axis at point C. The graph of the inverse proportion function $y=\frac{k}{x} (x>0)$ also passes through point A. What is the analytical expression of the inverse proportion function?

Solution:
\textbf{Solution}: Construct $AM\perp y$-axis at M and $AN\perp x$-axis at N through point A, as shown in the figure,\\
$\therefore$ Quadrilateral $AMON$ is a rectangle,\\
$\because$ $\triangle AOB$ is an isosceles right triangle,\\
$\therefore$ $\angle AON=45^\circ$,\\
$\therefore$ $ON=NA$,\\
$\therefore$ Quadrilateral $AMON$ is a square,\\
$\therefore$ $AM=AN$,\\
Let $A\left(a, a\right)$,\\
$\because$ Point $A$ lies on the line $y=3x-4$,\\
$\therefore$ $a=3a-4$,\\
Solving gives $a=2$,\\
$\therefore$ $A\left(2, 2\right)$,\\
$\because$ The graph of the inverse proportion function $y=\frac{k}{x}\left(x>0\right)$ passes through point A,\\
$\therefore$ $2=\frac{k}{2}$,\\
$\therefore$ $k=4$,\\
$\therefore$ The analytical expression of the inverse proportion function is $\boxed{y=\frac{4}{x}}$;


Question: As shown in the figure, $\triangle AOB$ is an isosceles right triangle, with the hypotenuse $OB$ on the x-axis. The graph of the linear function $y=3x-4$ passes through point A and intersects the y-axis at point C. The graph of the inverse proportion function $y=\frac{k}{x}$ (where $x>0$) also passes through point A. Draw $OD \perp AC$ at point $D$ from point O. Find the value of $CD^2-AD^2$.

Solution:
\textbf{Solution:} Given $A\left(2, 2\right)$,\\
therefore $AO^{2}=2^{2}+2^{2}=8$,\\
substitute $x=0$ into $y=3x-4$ to obtain $y=-4$,\\
therefore $C\left(0, -4\right)$,\\
therefore $OC=4$,\\
in $\triangle COD$, we have $CD^{2}=OC^{2}-OD^{2}$ (1),\\
in $\triangle AOD$, we have $AD^{2}=OA^{2}-OD^{2}$ (2),\\
Subtracting (2) from (1) yields: $CD^{2}-AD^{2}=OC^{2}-OA^{2}=16-8=\boxed{8}$.


Question: As shown in the figure, the side $OC$ of the rhombus $OABC$ is on the positive half of the $x$-axis, and the coordinates of point $B$ are $\left(8, 4\right)$. What is the length of the side of this rhombus?

Solution:
\textbf{Solution}: As shown in the figure, point B is drawn such that $BE\perp x$ axis at point E. Let the side length of the diamond be $x$, \\
$\because B(8,4)$, \\
$\therefore CE=8-x$, $BE=4$, \\
In $\triangle CBE$, $CB^{2}=CE^{2}+BE^{2}$, \\
thus, $x^{2}=(8-x)^{2}+4^{2}$, solving for $x$ gives $x=\boxed{5}$, \\
$\therefore$ the side length of the diamond is \boxed{5}.


Question: As shown in the figure, the sides of rhombus $OABC$ are such that side $OC$ is on the positive semi-axis of the $x$-axis, and the coordinates of point $B$ are $ \left(8, 4\right)$. If the graph of the inverse proportion function $y=\frac{k}{x}\left(x>0\right)$ passes through point $A$ and intersects side $BC$ at point $D$, find the coordinates of point $D$.

Solution:
\textbf{Solution:} Given that the side length of the diamond is 5, \\
it follows that A(3, 4), \\
therefore, $k=3\times4=12$, the inverse proportional function can be expressed as $y=\frac{12}{x}$.\\
Given points C(5, 0) and B(8, 4), \\
let the equation of line CB be $y=kx+b$, \\
then $\begin{cases}5k+b=0\\ 8k+b=4\end{cases}$, \\
solving this, we get $\begin{cases}k=\frac{4}{3}\\ b=-\frac{20}{3}\end{cases}$, \\
therefore, the equation of line CB is: $y=\frac{4}{3}x-\frac{20}{3}$,\\
From $\begin{cases}y=\frac{12}{x}\\ y=\frac{4}{3}x-\frac{20}{3}\end{cases}$\\
we solve to get $\begin{cases}x=\frac{5+\sqrt{61}}{2}\\ y=\frac{2\sqrt{61}-10}{3}\end{cases}$ or $\begin{cases}x=\frac{5-\sqrt{61}}{2}\\ y=-\frac{2\sqrt{61}+10}{3}\end{cases}$ (discarded for being irrelevant to the problem),\\
therefore, the coordinates of point D are $\boxed{\left(\frac{5+\sqrt{61}}{2}, \frac{2\sqrt{61}-10}{3}\right)}$.


Question: As illustrated, the parabola $y=ax^2+bx+c$ intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the y-axis at point C, also passing through point $D(2, -3)$. Points P and Q are moving points on the parabola $y=ax^2+bx+c$. What is the equation of the parabola?

Solution:
\textbf{Solution:} The expression for the function is: $y=a(x+1)(x-3)$. Substituting the coordinates of point D into the above equation and solving gives: $a=1$, \\
Hence, the equation of the parabola is: $y=x^2-2x-3$. \boxed{y=x^2-2x-3}


Question: Given the parabola \(y=ax^2+bx+c\) intersects the x-axis at points \(A(-1, 0)\) and \(B(3, 0)\), and intersects the y-axis at point C. It also passes through point \(D(2, -3)\). Points P and Q are moving points on the parabola \(y=ax^2+bx+c\). When point P is located below the line OD, what is the maximum area of \(\triangle POD\)?

Solution:
\textbf{Solution:} Let the line PD intersect the y-axis at point G, assume point $P\left(m, m^{2}-2m-3\right)$,\\
Substituting the coordinates of points P, D into the linear function equation: $y=mx+t$ and solving, we get the expression for the line PD as: $y=mx-3-2m$, thus $OG=3+2m$,\\
$S_{\triangle POD}=\frac{1}{2}\times OG(x_{D}-x_{P})=\frac{1}{2}(3+2m)(2-m)=-m^{2}+\frac{1}{2}m+3$,\\
Since $-1<0$, it follows that $S_{\triangle POD}$ has a maximum value, when $m=\frac{1}{4}$, its maximum value is $\boxed{\frac{49}{16}}$;


Question: As shown in the figure, the parabola $y=ax^2+bx+c$ intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the y-axis at point C, and passes through the point $D(2, -3)$. Points P and Q are moving points on the parabola $y=ax^2+bx+c$. The line OQ intersects the segment BC at point E. When $\triangle OBE$ is similar to $\triangle ABC$, what are the coordinates of point Q?

Solution:
\textbf{Solution:} Given that $OB=OC=3$, it follows that $\angle OCB=\angle OBC=45^\circ$, \\
since $\angle ABC=\angle OBE$, the triangle $\triangle OBE$ and $\triangle ABC$ are similar. This leads to two cases: \\
(1) When $\angle ACB=\angle BOQ$, we have $AB=4$, $BC=3\sqrt{2}$, $AC=\sqrt{10}$, \\
draw $AH$ perpendicular to $BC$ at point H, \\
the area of $\triangle ABC$ is $\frac{1}{2}\times AH\times BC=\frac{1}{2}AB\times OC$, solving for $AH$ gives: $AH=2\sqrt{2}$, \\
thus $CH=\sqrt{2}$ \\
therefore $\tan\angle ACB=2$, \\
the equation of line OQ is: $y=-2+x$... (2), \\
solving (1) and (2) gives: $x=\pm \sqrt{3}$, \\
thus point $\boxed{Q(\sqrt{3}, -2\sqrt{3})}$ or $\boxed{(−\sqrt{3}, 2\sqrt{3})}$; \\
(2) When $\angle BAC=\angle BOQ$, \\
$\tan\angle BAC=\frac{OC}{OA}=3=\tan\angle BOQ$, \\
the equation of line OQ is: $y=-3+x$... (3), \\
solving (1) and (3) gives: $x=\frac{-1\pm \sqrt{13}}{2}$, \\
thus point $\boxed{Q\left(\frac{-1+\sqrt{13}}{2}, \frac{-1-\sqrt{13}}{2}\right)}$ or $\boxed{\left(\frac{-1-\sqrt{13}}{2}, \frac{3+3\sqrt{13}}{2}\right)}$; \\
In summary, the point $\boxed{Q(\sqrt{3}, -2\sqrt{3})}$ or $\boxed{(−\sqrt{3}, 2\sqrt{3})}$ or $\boxed{\left(\frac{-1+\sqrt{13}}{2}, \frac{-1-\sqrt{13}}{2}\right)}$ or $\boxed{\left(\frac{-1-\sqrt{13}}{2}, \frac{3+3\sqrt{13}}{2}\right)}$.


Question: As shown in the figure, the linear function $y_{1}=x+b$ and the inverse proportion function $y_{2}=\frac{k}{x}$ intersect at points $A\left(1, a\right)$ and $D\left(-4, -1\right)$, and intersect the $y$-axis and $x$-axis at points $B$ and $C$, respectively. What is the expression of the inverse proportion function?

Solution:
\textbf{Solution:} Since point $D(-4, -1)$ lies on the graph of the inverse proportion function $y=\frac{k}{x}$,\\
therefore $k=-4\times (-1)=4$,\\
thus, the expression for the inverse proportion function is $\boxed{y=\frac{4}{x}}$.


Question: As shown in the figure, the linear function $y_{1}=x+b$ intersects with the hyperbolic function $y_{2}=\frac{k}{x}$ at points $A(1, a)$ and $D(-4, -1)$, and intersects with the $y$-axis and $x$-axis at points $B$ and $C$, respectively. Draw $AE\perp y$-axis at point $E$, and connect $DE$. What is the area of $\triangle ADE$?

Solution:
\textbf{Solution:} Since point $A(1, 4)$ lies on the graph of the inverse proportion function $y=\frac{4}{x}$,\\
it follows that $a=4$,\\
hence the coordinates of point $A$ are $(1, 4)$,\\
since $AE \perp y$-axis,\\
it follows that $AE=1, OE=4$,\\
Draw $DM\perp AE$ intersecting the extension of $AE$ at point $M$, and intersecting the $x$-axis at point $N$.\\
Hence, $\angle NME=90^\circ$,\\
since $AE \perp y$-axis,\\
it follows that $\angle MEO=90^\circ$,\\
since $\angle EON=90^\circ$\\
it follows that $\angle NME=\angle MEO=\angle EON=90^\circ$,\\
hence, quadrilateral EONM is a rectangle,\\
therefore $MN=OE=4, ND=1$,\\
hence, $MD=5$,\\
therefore, ${S}_{\triangle ADE}=\frac{1}{2}\times AE\times DM=\frac{1}{2}\times 1\times 5=\boxed{\frac{5}{2}}$;


Question: As shown in the figure, the linear function $y_{1}=x+b$ and the inverse proportion function $y_{2}=\frac{k}{x}$ intersect at points $A(1, a)$ and $D(-4, -1)$, and intersect the $y$-axis and $x$-axis at points $B$ and $C$, respectively. Based on the graph, please directly write down the range of $x$ values when $y_{1}>y_{2}$.

Solution:
\textbf{Solution:} When $y_{1}>y_{2}$, the graph of the linear function $y_{1}=x+3$ is above the graph of the hyperbolic function $y_{2}=\frac{4}{x}$,\\
since the intersection points of the two functions are $A(1, 4)$ and $D(-4, 1)$,\\
to the right of point D, to the left of the y-axis, and to the right of point A meet the condition,\\
therefore, \boxed{-4<x<0} or \boxed{x>1}.


Question: As shown in the figure, it is known that the line $y=-\frac{2}{3}x+2$ intersects the x-axis at point A and the y-axis at point B. The parabola $y=-\frac{2}{3}x^2+bx+c$ passes through points A and B. What is the expression of this parabola?

Solution:
\textbf{Solution:} From $y=-\frac{2}{3}x+2$, we obtain:\\
When $x=0$, $y=2$; when $y=0$, $x=3$,\\
$\therefore \boxed{A(3,0)}, \boxed{B(0,2)}$,\\
Substituting the coordinates of $A$, $B$ into $y=-\frac{2}{3}x^{2}+bx+c$ yields:\\
$\left\{\begin{array}{l}
-\frac{2}{3}\times 9+3b+c=0 \\
c=2
\end{array}\right.$,\\
Solving this gives: $\left\{\begin{array}{l}
b=\frac{4}{3} \\
c=2
\end{array}\right.$,\\
$\therefore$ The equation of the parabola is: \boxed{y=-\frac{2}{3}x^{2}+\frac{4}{3}x+2}.


Question: As shown in the figure, the line $y=x+b$ intersects the x-axis at point $C(4,0)$ and the y-axis at point $B$, and intersects with the hyperbola $y=\frac{m}{x}$ (where $x<0$) at point $A(-1,n)$. The line segment $OA$ is drawn. Find the analytic expression of the line and the hyperbola?

Solution:
\textbf{Solution:} By substituting point $C$ into $y=x+b$, we get $b=-4$,\\
therefore $y=x-4$;\\
substituting point $A$ into $y=x-4$, we find $n=-5$,\\
therefore $A(-1, -5)$,\\
thus $m=-1\times (-5)=5$,\\
therefore $y=\frac{5}{x}$\\
therefore, the equations of the line and the hyperbola are \boxed{y=x-4} and \boxed{y=\frac{5}{x}}, respectively.


Question: As shown in the figure, the line $y=x+b$ intersects the $x$-axis at point $C(4,0)$ and the $y$-axis at point $B$, and intersects the hyperbola $y=\frac{m}{x}$ (where $x<0$) at point $A(-1,n)$. Connect $OA$. What is the area of $\triangle OAC$?

Solution:
\textbf{Solution:} Since point $B$ passes through the y-axis,\\
$ \therefore x=0$,\\
$ \therefore y=-4$,\\
$ \therefore {S}_{\triangle AOC}={S}_{\triangle AOB}+{S}_{\triangle BOC}$,\\
$ =\frac{1}{2}\times 4\times 1+\frac{1}{2}\times 4\times 4$,\\
$ =\boxed{10}$;


Question: As shown in the figure, the line $y=x+b$ intersects the $x$ axis at point $C(4,0)$ and the $y$ axis at point $B$, and it intersects the hyperbola $y=\frac{m}{x}$ (where $x<0$) at point $A(-1,n)$. Connect $OA$. What is the sine value of $\angle OAB$?

Solution:
\textbf{Solution:} Construct $OM \perp AC$ at point $M$ through point $O$,\\
when $x=0$, $y=-4$, that is, $B(0, -4)$.\\
Since $OC=OB=4$,\\
Therefore, $\triangle OCB$ is an isosceles right triangle,\\
Thus, $\angle OBC=\angle OCB=45^\circ$\\
Therefore, in $\triangle OMB$ $\sin 45^\circ=\frac{OM}{OB}$,\\
Thus, $OM=4\times \frac{\sqrt{2}}{2}=2\sqrt{2}$.\\
Therefore, in the right triangle $AOM$,\\
$AO=\sqrt{(-1-0)^{2}+(-5-0)^{2}}=\sqrt{26}$,\\
$\sin\angle OAB=\frac{OM}{OA}=\boxed{\frac{2\sqrt{2}}{\sqrt{26}}}=\boxed{\frac{2\sqrt{13}}{13}}$.


Question: In the Cartesian coordinate system $xOy$, points $A$ and $B$ lie on the line $y=\frac{1}{2}x$, as shown in the figure. The graph of the quadratic function $y=ax^2+bx-2$ also passes through points $A$ and $B$, and intersects the $y$-axis at point $C$. If $BC$ is parallel to the $x$-axis, and the $x$-coordinate of point $A$ is $2$, what is the analytical expression of this quadratic function?

Solution:
\textbf{Solution:} Given that the quadratic function $y=\frac{1}{4}x^{2}+bx-2$ intersects the $y$-axis at point $C$,

thus the coordinates of point $C$ are $(0, -2)$,

since $BC\parallel x$-axis,

thus the ordinate of point $B$ is $-2$,

since points $A$ and $B$ lie on the line $y=\frac{1}{2}x$ and the abscissa of point $A$ is $2$,

thus the coordinates of point $A$ are $(2, 1)$, and those of point $B$ are $(-4, -2)$,

since the graph of the quadratic function also passes through points $A(2, 1)$ and $B(-4, -2)$, we get

\[
\left\{
\begin{array}{l}
4a+2b-2=1\\
16a-4b-2=-2
\end{array}
\right.,
\]

solving this system of equations, we find $a=\frac{1}{4}$, $b=1$,

therefore, the explicit formula of the quadratic function is $\boxed{y=\frac{1}{4}x^{2}+x-2}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, points $A$ and $B$ lie on the line $y=\frac{1}{2}x$. The graph of the quadratic function $y=ax^2+bx-2$ also passes through points $A$ and $B$, and intersects the $y$-axis at point $C$. If $BC$ is parallel to the $x$-axis, and the $x$-coordinate of point $A$ is $2$. Suppose the axis of symmetry of the graph of this quadratic function intersects $BC$ at point $D$, and point $E$ is on the negative $x$-axis. If the triangle formed by points $E$, $O$, and $B$ is similar to $\triangle OBD$, and the ratio of similarity is not $1$, what are the coordinates of point $E$?

Solution:
\textbf{Solution:} From (1), we get that the axis of symmetry of the quadratic function $y=\frac{1}{4}x^{2}+x-2$ is the line $x=-2$, \\
$\therefore$ the coordinates of point $D$ are $(-2, -2)$, \\
$\therefore$ $OB=2\sqrt{5}$, $BD=2$, \\
$\because$ $BC\parallel x$-axis, \\
$\therefore$ $\angle OBD=\angle BOE$, \\
$\therefore$ The triangle formed by points $E$, $O$, and $B$ is similar to $\triangle OBD$, there are two possibilities: \\
If $\frac{BO}{OB}=\frac{BD}{OE}$, then $\triangle BOD\sim \triangle OBE$, \\
Obviously, these two similar triangles have a similarity ratio of $1$, \\
This contradicts the given information that the similarity ratio is not $1$; thus, this case should be discarded; \\
When $\frac{BO}{OE}=\frac{BD}{OB}$, then $\triangle BOD\sim \triangle OEB$, \\
$\therefore$ $\frac{2\sqrt{5}}{OE}=\frac{2}{2\sqrt{5}}$, \\
$\therefore$ $OE=10$, \\
Since point $E$ is on the negative half of the x-axis, \\
$\therefore$ the coordinates of point $E$ are $\boxed{(-10, 0)}$.


Question: In the Cartesian coordinate system $xOy$, points $A$ and $B$ lie on the line $y=\frac{1}{2}x$, as shown in the figure. The graph of the quadratic function $y=ax^2+bx-2$ also passes through points $A$ and $B$, and intersects the $y$-axis at point $C$. If line $BC$ is parallel to the $x$-axis, and the $x$-coordinate of point $A$ is $2$. Let the vertex of the graph of this quadratic function be $M$, calculate the value of $\tan\angle AMC$?

Solution:
\textbf{Solution:} Draw $CH\perp AM$ at point $C$, with the foot of the perpendicular being $H$.\\
Let the equation of line $AM$ be $y=kx+b$,\\
then $\left\{\begin{array}{l}2k+b=1 \\ -2k+b=-3\end{array}\right.$,\\
which gives us $k=1$ and $b=-1$.\\
Therefore, the equation of line $AM$ is $y=x-1$.\\
Let the intersections of line $AM$ with the $x$-axis and $y$-axis be points $P$ and $Q$ respectively,\\
then the coordinates of point $P$ are $(1, 0)$, and the coordinates of point $Q$ are $(0, -1)$,\\
therefore, $\triangle OPQ$ is an isosceles right triangle, $\angle OQP=45^\circ$, $OQ=OP=1$,\\
since $\angle OQP=\angle HQC$,\\
it follows that $\angle HQC=45^\circ$,\\
given that the coordinates of point $C$ are $(0, -2)$,\\
it follows that $CQ=1$,\\
therefore, $HC=HQ=\frac{\sqrt{2}}{2}$,\\
and $MQ=\sqrt{(-2-0)^2+(-3-(-1))^2}=2\sqrt{2}$,\\
therefore, $MH=MQ-HQ=\frac{3\sqrt{2}}{2}$,\\
thus, $\tan\angle AMC=\frac{HC}{MH}=\boxed{\frac{1}{3}}$.


Question: As the diagram shows, a certain school in Xi'an city has carried out "fumigation disinfection" in all classrooms to further prevent the "novel coronavirus". It is known that during the burning and releasing process of the disinfectant, the concentration of the disinfectant in the classroom air, \(y\) (mg) per cubic meter, and the burning time \(x\) (min) have the following functional relationship: when \(x < 6\), \(y\) is directly proportional to \(x\); when \(x \geq 6\), \(y\) is inversely proportional to \(x\). Based on the information provided by the graph, find the functional relation between \(y\) and \(x\) when \(x \geq 6\).

Solution:
\textbf{Solution}: From the graph, it is known that when $x \geq 6$, $y$ is an inverse proportion function of $x$, let $y=\frac{k}{x}$ ($k \neq 0$).\\
$\because$ point $(15,4)$,\\
$\therefore k=15 \times 4=60$,\\
The functional relationship between $y$ and $x$ is $\boxed{y=\frac{60}{x}}\left(x\geq 6\right)$.


Question: To further prevent the "novel coronavirus", a school in Xi'an has disinfected all classrooms through the method of "fumigation". It is known that during the combustion release process of the medicine, the concentration of the medicine in the air of the classrooms per cubic meter, \(y\) (mg), is related to the burning time, \(x\) (min), as shown in the graph. Here, when \(x < 6\), \(y\) is directly proportional to \(x\), and when \(x \ge 6\), \(y\) is inversely proportional to \(x\). Based on the information provided by the graph, solve the following question: Find the coordinates of point A.

Solution:
\textbf{Solution:} Solve: When $x=6$, we have $10y=60$. Solving for $y$, we get $y=6$. \\
$\therefore$, the coordinates of point A are $(6, \boxed{6})$.


Question: As shown in the figure, the graph of the inverse proportion function $y=\frac{2m}{x}$ and the graph of the linear function $y=kx-1$ intersect at points $A(m, 2m)$ and $B$. Find the expression for the linear function.

Solution:
\textbf{Solution:} Since point $A(m,2m)$ lies on the graph of the inverse proportion function, it follows that $2m = \frac{2m}{m}$, so $m=1$, hence $A(1,2)$.\\
Furthermore, since $A(1,2)$ lies on the graph of the linear function $y=kx-1$, it follows that $2=k-1$, which gives $k=3$,\\
thus, the expression for the linear function is: $\boxed{y=3x-1}$.


Question: As shown in the figure, the hyperbola represented by the inverse proportion function $y=\frac{2m}{x}$ and the straight line represented by the linear function $y=kx-1$ intersect at points $A(m, 2m)$ and $B$. Find the coordinates of point $B$ and, based on the graph, directly write out the range of $x$ values that satisfy the inequality $\frac{2m}{x}<kx-1$.

Solution:
\textbf{Solution:} From
\[
\left\{
\begin{array}{l}
y=\frac{2}{x}\\
y=3x-1
\end{array}
\right.
\]
we get \boxed{B\left(-\frac{2}{3},-3\right)}.

Therefore, by the graphical representation, the range of $x$ that satisfies $\frac{2m}{x}<kx-1$ is \boxed{-\frac{2}{3}<x<0} or \boxed{x>1}.


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

Solution:
\textbf{Solution}: Substitute $(-3, 0)$ and $(0, -3)$ into $y=x^{2}+bx+c$ to obtain:\\
\[
\begin{cases}
9-3b+c=0\\
c=-3
\end{cases},
\]
Solving this gives:
\[
\begin{cases}
b=\boxed{2}\\
c=\boxed{-3}
\end{cases},
\]
Therefore, the values of $b$ and $c$ are $2$ and $-3$, respectively.