Question: As illustrated, there are two islands, C and D, in the sea. A fishing boat at point A found island D to the northeast, at a distance of $30\sqrt{2}$ nautical miles. The boat then traveled from west to east for a period of time to reach point B, where it measured that island C was exactly to the north of point B, at a distance of $75$ nautical miles, and also measured that point B was $30\sqrt{5}$ nautical miles away from island D. What is the distance between islands C and D in nautical miles?

Solution:
\textbf{Solution:} Construct line DF such that DF $\perp$ BC at F. In the right triangle $\triangle BED$, where DE = 30 and BD = $30\sqrt{5}$, \\
thus, BE = $\sqrt{BD^{2}-DE^{2}}=60$, \\
since quadrilateral BFDE is a rectangle, \\
thus, DF = EB = 60 and BF = DE = 30, \\
therefore, CF = BC−BF = 45, \\
in the right triangle $\triangle CDF$, CD = $\sqrt{DF^{2}+CF^{2}}=\boxed{75}$, \\
therefore, the distance between islands C and D is \boxed{75} nautical miles.


Question: As shown in the figure, point $P$ is inside the square $ABCD$, and lines $AP$, $BP$, and $DP$ are connected, with $\angle 1 = \angle 2$. If $AB = m$ and $\tan \angle 2 = \frac{1}{2}$, what is the area of $\triangle ABP$? (Express in terms of $m$).

Solution: \textbf{Solution:} Therefore, the area of $\triangle ABP$ is $\boxed{\frac{m^2}{5}}$.


Question: As shown in the figure, point $P$ is a point inside the square $ABCD$, with lines $AP$, $BP$, $DP$ connected, and $\angle 1 = \angle 2$. If $AB = 2$, and the tangent of one of the acute angles in $\triangle PAB$ is 2, then what is the area of $\triangle PAD$?

Solution: \textbf{Solution:} Therefore, the area of $\triangle PAD$ is $\boxed{\frac{2}{5}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\cos B=\frac{4}{5}$, $AB=10$, and point $D$ is on side $BC$ with $AC=DC$. What is the length of $BD$?

Solution:
\textbf{Solution:} Given that $\angle C=90^\circ$ and $\cos B=\frac{4}{5}$,\\
it follows that $\frac{BC}{AB}=\frac{4}{5}$, hence $\frac{BC}{10}=\frac{4}{5}$,\\
solving this gives $BC=8$,\\
thus, $AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{10^{2}-8^{2}}=6$,\\
which means $CD=AC=6$,\\
therefore, $BD=BC-CD=8-6=\boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\cos B=\frac{4}{5}$, $AB=10$, and point D is a point on side $BC$, with $AC=DC$. Find the value of $\cot \angle BAD$?

Solution:
\textbf{Solution:} As shown in the figure, draw $DE \perp AB$ at E,\\
in $\triangle BED$, $BD=2$, $\cos B=\frac{4}{5}$,\\
$\therefore \frac{BE}{BD}=\frac{4}{5}$, which implies $\frac{BE}{2}=\frac{4}{5}$,\\
solving this gives $BE=\frac{8}{5}$,\\
$\therefore AE=10-\frac{8}{5}=\frac{42}{5}$, and $DE=\sqrt{BD^2-BE^2}=\sqrt{2^2-\left(\frac{8}{5}\right)^2}=\frac{6}{5}$,\\
in $\triangle ADE$, $\cot \angle BAD=\frac{AE}{DE}=\frac{\frac{42}{5}}{\frac{6}{5}}=\boxed{7}$.


Question: As shown in the figure, the parabola passes through points A$(-2, 0)$ and C$(0, -3)$, and its axis of symmetry is the line $x=\frac{1}{2}$. Find the equation of the parabola.

Solution:
\textbf{Solution:} Let the equation of the parabola be $y=ax^2+bx+c$,\\
since the axis of symmetry is the line $x=\frac{1}{2}$,\\
thus $-\frac{b}{2a}=\frac{1}{2}$,\\
thus $b=-a$,\\
thus $y=ax^2-ax+c$,\\
Substituting points A ($-2$, $0$) and C ($0$, $-3$),\\
thus $\begin{cases}c=-3 \\ 4a+2a+c=0\end{cases}$,\\
Solving yields $\begin{cases}c=-3 \\ a=\frac{1}{2}\end{cases}$,\\
thus $y=\frac{1}{2}x^2-\frac{1}{2}x-3$;\\
Hence, the equation of the parabola is $\boxed{y=\frac{1}{2}x^2-\frac{1}{2}x-3}$.


Question: As shown in the figure, in $\triangle ABC$, $AD\perp BC$ at point $D$, $E$ is a point on $AD$, and $AE: ED = 7: 5$. $CE$ is extended to intersect side $AB$ at point $F$, $AC = 13$, $BC = 8$, $\cos\angle ACB = \frac{5}{13}$. Find the value of $\tan\angle DCE$?

Solution: \textbf{Solution:} Since $AD \perp BC$, it follows that $\angle ADC=90^\circ$. In $\triangle ADC$, given that $AC=13$ and $\cos\angle ACB=\frac{5}{13}$, it can be deduced that $CD=5$ and therefore $AD=\sqrt{AC^{2}-CD^{2}}=\sqrt{13^{2}-5^{2}}=12$. Given that the ratio $AE:ED=7:5$, it follows that $ED=12\times \frac{5}{7+5}=5$. Consequently, $\tan\angle DCE=\frac{ED}{CD}=\boxed{1}$.


Question: As shown in the figure, in $\triangle ABC$, $AD \perp BC$ at point $D$, $E$ is a point on $AD$, and $AE:ED = 7:5$. Connect $CE$ and extend it to intersect side $AB$ at point $F$, $AC=13$, $BC=8$, $\cos\angle ACB = \frac{5}{13}$. What is the value of $\frac{AF}{BF}$?

Solution:
\textbf{Solution:} As shown in the figure, draw line $DG\parallel CF$ passing through point $D$ and intersecting $AB$ at point $G$, \\
since $BC=8, CD=5$, \\
thus $BD=BC-CD=3$. \\
Since $DG\parallel CF$, it follows that $\frac{BD}{CD}=\frac{BG}{FG}=\frac{3}{5}$, $\frac{AF}{FG}=\frac{AE}{DE}=\frac{7}{5}$, \\
therefore $AF=\frac{7}{5}FG$. Let $BG=3x$, then $FG=5x, AF=7x$. \\
Hence $BF=FG+BG=8x$, therefore $\frac{AF}{BF}=\boxed{\frac{7}{8}}$.


Question: Given the graph of the parabola $y = x^2 - 2x - 3$ as shown in the figure. What are the coordinates of the points where the parabola intersects the x-axis and y-axis?

Solution: \textbf{Solution:} Let $x=0$, then $y=-3$. Therefore, the intersection point of the parabola with the $y$-axis is \boxed{(0,-3)}. Let $y=0$, then $x^2-2x-3=0$. Solving this yields: $x_1=-1$, $x_2=3$. Therefore, the intersection points of the parabola with the $x$-axis are \boxed{(-1,0)} and \boxed{(3,0)}.


Question: Given the parabola $y=x^{2}-2x-3$ as shown in the graph. Based on the graph, answer the following: For what values of $x$ is $y>0$? For what values of $x$ is $y<0$?

Solution:
\textbf{Solution:} From the graph and the coordinates of the intersection points of the parabola with the x-axis, we know that\\
when $x>3$ or $x<-1$, $y>0$;\\
when $-1<x<3$, $y<0$.\\
$\therefore \boxed{when\ x>3\ or\ x<-1,\ y>0};\ \boxed{when\ -1<x<3,\ y<0}$.


Question: As shown in the figure, two balls are thrown from point A on the slope and pass through the basketball hoop at point B, landing at point O at the bottom of the slope. A rectangular coordinate system is established with O as the origin, and the coordinates of B are $\left(\frac{1}{2}, \frac{7}{8}\right)$. The slope of the slope is $1\colon 2$, and the height of point A from the ground is 1.5 meters. The trajectory of the balls can be described by a quadratic function. What is the expression of the quadratic function?

Solution:
Solution: Draw $AD\perp x$-axis through point A, with D as the foot of the perpendicular.\\
According to the problem, $AD:OD=1:2$ and $AD=1.5$,\\
$\therefore OD=2AD=3$,\\
$\therefore A(3, 1.5)$.\\
$\because$ The graph of the quadratic function passes through the origin O,\\
$\therefore$ let the expression of the quadratic function be $y=ax^{2}+bx$.\\
Substituting the coordinates of points A and B, we get:\\
$\begin{cases}9a+3b=1.5\\ \frac{1}{4}a+\frac{1}{2}b=\frac{7}{8}\end{cases}$,\\
Solving this, we find $\begin{cases}a=-\frac{1}{2}\\ b=2\end{cases}$,\\
$\therefore$ The expression of the quadratic function is $y=-\frac{1}{2}x^{2}+2x=\boxed{-\frac{1}{2}x^{2}+2x}$.


Question: As shown in the figure, two balls are thrown from point A on the slope, passing exactly through the basket at point B, and landing at point O at the bottom of the slope. Taking O as the origin, a Cartesian coordinate system is established. The coordinates of B are $\left(\frac{1}{2}, \frac{7}{8}\right)$, the slope ratio of the slope is $1\colon 2$, the height of point A from the ground is 1.5 meters, and the trajectory of the ball can be described by a quadratic function. What are the coordinates of the highest point the ball reaches?

Solution:
\textbf{Solution:} Since $y=-\frac{1}{2}x^{2}+2x=-\frac{1}{2}(x-2)^{2}+2$,\\
therefore, when $x=2$, $y$ reaches its maximum value, which is $2$,\\
thus, the highest point the ball reaches has the coordinates $\boxed{(2, 2)}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects with the graph of the inverse proportion function $y=\frac{m}{x}$ at points $P(2,a)$ and $Q(-1, -4)$; determine the expressions for the linear function and the inverse proportion function.

Solution:
\textbf{Solution:} Substitute $Q(-1, -4)$ into $y=\frac{4}{x}$, we get $m=4$\\
Therefore, $y=\frac{4}{x}$\\
Substitute $P(2, a)$ into $y=\frac{4}{x}$, we obtain $a=2$\\
Therefore, $P(2, 2)$\\
Substituting $Q(-1, -4)$ and $P(2, 2)$ into $y=kx+b$ yields:\\
$\begin{cases}
-k+b=-4\\
2k+b=2
\end{cases}$\\
Solving these, we get: $\begin{cases}
k=2\\
b=-2
\end{cases}$\\
Hence, the equation of the linear function is $\boxed{y=2x-2}$, and 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 $y=kx+b$ intersects with the graph of the inverse proportion function $y=\frac{m}{x}$ at points $P(2, a)$ and $Q(-1, -4)$. Based on the graph, directly write the solution set for the inequality $\frac{m}{x}>kx+b$ with respect to $x$.

Solution: \textbf{Solution:} The solution set for $x<-1$ or $0<x<2$ is $\boxed{x \in (-\infty, -1) \cup (0, 2)}$.


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. What is the expression of the parabola?

Solution:
\textbf{Solution}: Substitute point $A(-1, 0)$ and point $B(3, 0)$ into $y=ax^2+bx-3$, \\
$\therefore \begin{cases}
a-b-3=0 \\
9a+3b-3=0
\end{cases}$, \\
Upon solving, we get $\begin{cases}
a=1 \\
b=-2
\end{cases}$, \\
$\therefore y=x^2-2x-3=\boxed{x^2-2x-3}$.


Question: As shown in the figure, the parabola $y=ax^{2}+bx-3\ (a\ne 0)$ intersects the x-axis at points $A(-1, 0)$ and $B(3, 0)$, and intersects the y-axis at point C. Find a point $Q$ on the axis of symmetry such that the perimeter of $\triangle ACQ$ is minimized. Determine the coordinates of point $Q$.

Solution:
\textbf{Solution:}

Let the line $BC$ intersect the axis of symmetry at point $Q$,\\
since $y=x^{2}-2x-3=(x-1)^{2}-4$,\\
thus, the axis of symmetry of the parabola is the line $x=1$,\\
since points A and B are symmetric about the axis $x=1$,\\
thus $AQ=BQ$,\\
thus, $AC+AQ+CQ=AC+CQ+BQ\ge AC+BC$,\\
when points C, B, and Q are collinear, the perimeter of $\triangle ACQ$ is minimized,\\
since $C(0, -3)$, $B(3, 0)$,\\
let the equation of line $BC$ be $y=kx+b$,\\
thus $\begin{cases}b=-3 \\ 3k+b=0\end{cases}$, solving gives $\begin{cases}k=1 \\ b=-3\end{cases}$,\\
thus, the equation of line $BC$ is $y=x-3$,\\
thus, the coordinates of $Q\left(1, -2\right)$ are $\boxed{(1, -2)}$.


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. Let P be a moving point on the parabola in the fourth quadrant. Find the maximum value of the area $S$ of the triangle $\triangle BPC$ and the coordinates of the point P at this time.

Solution:
\textbf{Solution:} Draw $PD \perp x$-axis at point D through point P. Assume the coordinates of point P as $(x, y)$, \\
then $S = S_{\text{trapezoid }OCPD} + S_{\triangle DPB} - S_{\triangle OCB}$\\
$= \frac{1}{2}x\left[3-(x^{2}-2x-3)\right]+\frac{1}{2} \times (3-x) \times [- (x^{2}-2x-3)]-\frac{1}{2} \times 3 \times 3$\\
$= -\frac{3}{2}(x-\frac{3}{2})^{2}+\frac{27}{8}$\\
$\therefore$ when $x=\frac{3}{2}$, $S_{\text{max}}=\boxed{\frac{27}{8}}$.\\
At this point $y=(\frac{3}{2})^{2}-2 \times \frac{3}{2} -3 = -\frac{15}{4}$\\
Therefore, the maximum value of the area $S$ of $\triangle BPC$ is $\boxed{\frac{27}{8}}$, with the coordinates of point P being $\boxed{\left(\frac{3}{2}, -\frac{15}{4}\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=x-5$ intersects the x-axis at point A and the y-axis at point B. The parabola $y=ax^2+bx+c$ passes through points A and B, and has another intersection point C with the x-axis, where $\tan\angle CBO=\frac{1}{5}$. What is the equation of the parabola?

Solution:
\textbf{Solution:} To solve this, in the line $y=x-5$, when $x=0$, $y=-5$, \\
when $y=0$, $x=5$, \\
$\therefore B\left(0, -5\right)$, $A\left(5, 0\right)$, \\
$\therefore OB=5$, \\
$\because \tan\angle CBO=\frac{CO}{OB}=\frac{CO}{5}=\frac{1}{5}$, \\
$\therefore CO=1$, \\
$\therefore C\left(-1, 0\right)$, \\
Let the equation of the parabola be $y=a\left(x+1\right)\left(x-5\right)$, \\
Substituting $B\left(0, -5\right)$ into the equation yields: $-5=-5a$, \\
Solving this gives: $a=1$, \\
$\therefore$ the equation of the parabola is $y=\left(x+1\right)\left(x-5\right)=x^{2}-4x-5=\boxed{x^{2}-4x-5}$.


Question: As shown in the figure, in a Cartesian coordinate system, the line $y=x-5$ intersects the x-axis at point A and the y-axis at point B. The parabola $y=ax^2+bx+c$ passes through points A and B, with another intersection point C on the x-axis, and $\tan\angle CBO=\frac{1}{5}$. If point P is the vertex of the parabola, what is the area of the quadrilateral $APBC$?

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

\[\because y=x^2-4x-5=(x-2)^2-9,\]

\[\therefore\] the vertex \(P(2, -9)\),

Draw \(AP\) and \(BP\), and through point B draw \(BD\parallel x\) axis intersecting \(AP\) at \(D\),

Let the equation of line \(AP\) be \(y=kx+b(k\ne 0)\),

Substituting \(A(5, 0)\) and \(P(2, -9)\) yields: \[\left\{\begin{array}{l} 5k+b=0 \\ 2k+b=-9\end{array}\right.,\]

Solving this, we get: \[\left\{\begin{array}{l} k=3 \\ b=-15\end{array}\right.,\]

\[\therefore\] the equation of line \(AP\) is \(y=3x-15\),

Substituting \(y=-5\) into \(y=3x-15\) yields: \(-5=3x-15\),

Solving this, we get: \(x=\frac{10}{3}\),

\[\therefore D\left(\frac{10}{3}, -5\right),\]

\[\because AC=5-(-1)=6\], \(BD=\frac{10}{3}\),

\[\therefore {S}_{\text{quadrilateral}APBC}={S}_{\text{trapezoid}ADBC}+{S}_{\triangle BDP}\]

\[=\frac{1}{2}\times \left(6+\frac{10}{3}\right)\times 5+\frac{1}{2}\times \frac{10}{3}\times \left[-5-\left(-9\right)\right]\]

\[=\frac{1}{2}\times \frac{28}{3}\times 5+\frac{1}{2}\times \frac{10}{3}\times 4\]

\[=\frac{70}{3}+\frac{20}{3}\]

\[=\boxed{30}.\]


Question: As shown in the figure, in the Cartesian coordinate system, the line $y=x-5$ intersects the x-axis at point A and the y-axis at point B. The parabola $y=ax^2+bx+c$ passes through points A and B, and its other intersection with the x-axis is point C, where $\tan\angle CBO=\frac{1}{5}$. Is there a point Q on the parabola such that $\triangle ABQ$ is a right triangle with $AB$ as one of its legs? If it exists, find the coordinates of point Q; if not, explain why.

Solution:
\textbf{Solution}: Existence proved.\\
Let point Q$(m, m^{2}-4m-5)$.\\
(1) When point A is the vertex of the right angle, draw $AQ_{1}\perp AB$, intersecting the y-axis at point M and the parabola at point Q$_{1}$, and draw $Q_{1}N\perp y$-axis with the foot of the perpendicular being N.\\
Since $B(0, -5)$, $A(5, 0)$,\\
therefore $OA=OB=5$, $\angle BAO=45^\circ$,\\
Since $AQ_{1}\perp AB$,\\
thus $\angle MAO=45^\circ$,\\
thus $\angle AMO=45^\circ$,\\
thus $OA=OM$,\\
Since $Q_{1}N\perp y$-axis, $\angle Q_{1}MN=\angle AMO=45^\circ$,\\
thus $\angle NQ_{1}M=45^\circ$,\\
thus $Q_{1}N=MN$.\\
That is $-m=m^{2}-4m-5-5$,\\
Solving it we get $m_{1}=-2$, $m_{2}=5$ (discarded).\\
Thus, $Q_{1}(-2, 7)$.\\
(2) When point B is the vertex of the right angle, draw $BQ_{2}\perp AB$, intersecting the parabola at point $Q_{2}$, and draw $Q_{2}G\perp y$-axis with the foot of the perpendicular being G.\\
Since $OA=OB=5$,\\
thus $\angle ABO=45^\circ$,\\
Since $BQ_{2}\perp AB$,\\
thus $\angle GBQ_{2}=45^\circ$,\\
Since $Q_{2}G\perp y$-axis,\\
thus $\angle BQ_{2}G=45^\circ$,\\
thus $GQ_{2}=BG$,\\
That is $m=-5-m^{2}+4m+5$,\\
Solving it we get $m_{1}=3$, $m_{2}=0$ (discarded),\\
Therefore, the coordinates of point $Q_{2}(3, -8)$.\\
Therefore, the coordinates of point Q are $\boxed{(-2, 7)}$ or $\boxed{(3, -8)}$.


Question: The parabola $y=ax^2+bx-4\ (a\neq 0)$ intersects the x-axis at points $A(-2, 0)$ and $B(4, 0)$, and intersects the y-axis at point $C$. Connect $BC$. Point $P$ is a moving point on the parabola below the line segment $BC$ (not coinciding with points $B$ or $C$). A line parallel to the y-axis passing through point $P$ intersects $BC$ at $M$ and the x-axis at $N$. Let the x-coordinate of point $P$ be $t$. What is the analytic expression of this parabola?

Solution:
\textbf{Solution:} Substitute $A(-2, 0)$ and $B(4, 0)$ into $y=ax^2+bx-4$ to get\\
\[
\therefore \left\{\begin{array}{l}
4a-2b-4=0\\
16a+4b-4=0
\end{array}\right. \text{ which is } \left\{\begin{array}{l}
2a-b=2\\
4a+b=1
\end{array}\right. \therefore \left\{\begin{array}{l}
a=\frac{1}{2}\\
b=-1
\end{array}\right. \therefore \text{The equation of the parabola is: } \boxed{y=\frac{1}{2}x^2-x-4}
\]


Question: As shown in the figure, the parabola $y=ax^2+bx-4$ ($a\neq 0$) intersects the $x$-axis at points $A(-2, 0)$ and $B(4, 0)$, and intersects the $y$-axis at point $C$. Line segment $BC$ is drawn. Point $P$ is a moving point on the parabola below line segment $BC$ (not coinciding with points $B$, $C$). A line parallel to the $y$-axis is drawn through point $P$, intersecting line segment $BC$ at $M$ and the $x$-axis at $N$. Let the $x$-coordinate of point $P$ be $t$. Represent the length of segment $PM$ with an algebraic expression in terms of $t$, and find the maximum value of $PM$ and the coordinates of point $M$ at this time.

Solution:
\textbf{Solution:} Let $x=0$, we get $y=-4$,\\
$\therefore C(0, -4)$,\\
Assume the equation of line BC is $y=kx+b$,\\
then $\begin{cases}b=-4\\ 4k+b=0\end{cases}$,\\
$\therefore \begin{cases}k=1\\ b=-4\end{cases}$,\\
$\therefore$ the equation of line BC is: $y=x-4$,\\
Let the x-coordinate of P be t, and $PM\parallel y$ axis,\\
$\therefore P\left(t, \frac{1}{2}t^{2}-t-4\right)$, $M\left(t, t-4\right)$,\\
$\therefore PM=t-4-\left(\frac{1}{2}t^{2}-t-4\right)=-\frac{1}{2}t^{2}+2t$,\\
$\therefore$ when $t=2$, the maximum value of $PM=\boxed{2}$, at this time $M\left(2, -2\right)$.


Question: As shown in the diagram, in rectangle $ABCD$, $BC>CD$, and $BC$ and $CD$ are the two roots of the quadratic equation $x^2-14x+48=0$ respectively. Connect $BD$. A moving point $P$ starts from $B$ and moves towards $D$ along $BD$ at a speed of 1 unit per second. At the same time, another moving point $Q$ starts from $D$ and moves along the ray $DA$ at the same speed. When point $P$ reaches point $D$, point $Q$ stops moving. Let the duration of the motion be $t$ seconds. Construct $\text{Rt}\triangle PQM$ with $PQ$ as the hypotenuse, such that point $M$ falls on segment $BD$. What is the length of segment $BD$?

Solution:
\textbf{Solution:} Given $x^{2}-14x+48=0$,\\
$\therefore$ $(x-6)(x-8)=0$. Solving this gives $x_{1}=6$, $x_{2}=8$. \\
$\therefore$ $BC=8$, $CD=6$,\\
$\because$ Rectangle ABCD,\\
$\therefore$ $\angle C=90^\circ$,\\
$\therefore$ $BD=\sqrt{BC^{2}+CD^{2}}=\sqrt{8^{2}+6^{2}}=\boxed{10}$.


Question: As shown in the diagram, in rectangle $ABCD$, $BC>CD$, and $BC$ and $CD$ are the two roots of the quadratic equation ${x}^{2}-14x+48=0$. Connect $BD$. A moving point $P$ starts from $B$ and moves towards $D$ along $BD$ at a speed of $1$ unit per second. At the same time, another moving point $Q$ starts from $D$ and moves along the ray $DA$ at the same speed. When point $P$ reaches point $D$, point $Q$ stops moving. Let the movement time be $t$ seconds. Construct $\triangle PQM$ with $PQ$ as the hypotenuse, such that point $M$ lies on the segment $BD$. What is the maximum area of $\triangle PDQ$?

Solution:
\textbf{Solution:} Draw $PE \perp AD$ at point E, \\
$\therefore \angle DEP=\angle DAB=90^\circ$,\\
$\therefore PE \parallel AB$,\\
$\therefore \triangle DEP \sim \triangle DAB$,\\
$\therefore \frac{EP}{AB}=\frac{PD}{BD}$, given $AB=CD=6$, $DQ=BP=t$, then $PD=10-t$,\\
$\therefore \frac{EP}{6}=\frac{10-t}{10}$, solving this: $PE=6-\frac{3}{5}t$,\\
$\therefore S_{\triangle PQD}=\frac{1}{2}DQ \cdot PE=\frac{1}{2}t\left(6-\frac{3}{5}t\right)=-\frac{3}{10}(t-5)^2+\frac{15}{2}$,\\
$\because$ the parabola opens downwards,\\
when $t=5$, the maximum area is $\boxed{7.5}$.


Question: As shown in the diagram, in rectangle $ABCD$, $BC>CD$ with $BC$ and $CD$ being the two roots of the quadratic equation $x^2-14x+48=0$. Connect $BD$. A moving point $P$ starts from $B$ and moves towards $D$ along $BD$ at a speed of 1 unit per second, while another moving point $Q$ starts from $D$ and moves along ray $DA$ at the same speed. When point $P$ reaches point $D$, point $Q$ stops moving. Let the motion time be $t$ seconds. Construct triangle $\triangle PQM$ with $PQ$ as the hypotenuse such that point $M$ is on the line segment $BD$. When triangle $\triangle PQM$ is similar to $\triangle BCD$, what is the value of $t$?

Solution:
\textbf{Solution:} From the given information, $\because \angle QMD = \angle QMP = \angle C = 90^\circ$, $AD \parallel BC$, $\therefore \angle DBC = \angle QDM$, $\therefore \triangle DQM \sim \triangle BDC$, $\therefore \frac{QM}{DC} = \frac{QD}{BD} = \frac{DM}{CB}$, that is, $\frac{QM}{6} = \frac{t}{10} = \frac{DM}{8}$. Solving these equations gives: $QM = \frac{3}{5}t$, $DM = \frac{4}{5}t$.

$\because \triangle PQM \sim \triangle BCD$, $\therefore \frac{QM}{DC} = \frac{PM}{BC}$ or $\frac{QM}{BC} = \frac{PM}{DC}$, $\therefore QM = \frac{3}{4}PM$ or $QM = \frac{4}{3}PM$.

When point M is on segment PD, $PM = 10 - t - \frac{4}{5}t = 10 - \frac{9}{5}t$, $\therefore QM = \frac{3}{4}PM$ that is $\frac{3}{5}t = \frac{3}{4}\left(10 - \frac{9}{5}t\right)$. Solving this equation yields: $t = \boxed{\frac{50}{13}}$.

When $QM = \frac{4}{3}PM$, $\frac{3}{5}t = \frac{4}{3}\left(10 - \frac{9}{5}t\right)$. Solving this equation yields: $t = \boxed{\frac{40}{9}}$.

When point M is on PB, $PM = t + \frac{4}{5}t - 10 = \frac{9}{5}t - 10$, $QM = \frac{4}{3}PM$ that is $\frac{3}{5}t = \frac{4}{3}\left(\frac{9}{5}t - 10\right)$. Solving this equation yields: $t = \boxed{\frac{200}{27}}$.

When $QM = \frac{3}{4}PM$, $\frac{3}{5}t = \frac{3}{4}\left(\frac{9}{5}t - 10\right)$. Solving this equation yields: $t = \boxed{10}$.

Therefore, the possible values for $t$ are \boxed{\frac{50}{13}}, \boxed{\frac{40}{9}}, \boxed{\frac{200}{27}}, or \boxed{10}.


Question: As shown in the diagram, the parabola $y=x^2+bx+c$ intersects the $x$-axis at points $A$ and $B(3,0)$, and intersects the $y$-axis at point $C(0,3)$. Find the analytical expression of the parabola.

Solution:
\textbf{Solution:} Substitute points B$(3,0)$, C$(0,3)$ into the parabola $y=x^{2}+bx+c$, we get:
\[
\left\{
\begin{array}{l}
0=9+3b+c\\
3=c
\end{array}
\right.
\]
Solving this, we obtain:
\[
\left\{
\begin{array}{l}
b=-4\\
c=3
\end{array}
\right.
\]
$\therefore$ The equation of the parabola is \boxed{y=x^{2}-4x+3}.


Question: As shown in the figure, the parabola $y=x^2+bx+c$ intersects the x-axis at points $A$ and $B(3,0)$, and the y-axis at point $C(0,3)$. If point $M$ is a moving point on the parabola below the x-axis, and through point $M$ a line $MN$ is drawn parallel to the y-axis intersecting the line $BC$ at point $N$, what is the maximum value of the line segment $MN$?

Solution:
\textbf{Solution:} Let the coordinates of point M be $(m, m^2-4m+3)$. Assume the equation of line BC is $y=-x+3$, \\
Substituting point B $(3,0)$ into $y=-x+3$, \\
We get: $0=3k+3$, solving for $k$ we find $k=-1$, \\
$\therefore$ The equation of line BC is $y=-x+3$.\\
$\because MN\parallel y$-axis, \\
$\therefore$ The coordinates of point N are $(m, -m+3)$.\\
$\because$ The equation of the parabola is $y=x^2-4x+3=(x-2)^2-1$, \\
$\therefore$ The axis of symmetry for the parabola is $x=2$,\\
$\therefore$ The point $(1,0)$ lies on the graph of the parabola,\\
$\therefore$ $1<m<3$.\\
$\because$ The length of segment $MN=-m+3-(m^2-4m+3)=-m^2+3m=-\left(m-\frac{3}{2}\right)^2+\frac{9}{4}$,\\
$\therefore$ When $m=\frac{3}{2}$, segment $MN$ reaches its maximum value, which is $\boxed{\frac{9}{4}}$.


Question: As shown in the figure, the parabola $y=x^2+bx+c$ intersects the x-axis at points $A$ and $B(3,0)$, and intersects the y-axis at point $C(0,3)$. Given these conditions, when $MN$ reaches its maximum value, is there a point $P$ on the axis of symmetry $l$ of the parabola such that $\triangle PBN$ is an isosceles triangle? If such a point exists, please directly write down the coordinates of all such points $P$; if it does not exist, please explain why.

Solution: \textbf{Solution:} The coordinates of point $P$ are $\boxed{\left(2, \frac{1}{2}\right)}$, $\boxed{\left(2, -\frac{\sqrt{14}}{2}\right)}$, $\boxed{\left(2, \frac{\sqrt{14}}{2}\right)}$, $\boxed{\left(2, \frac{3-\sqrt{17}}{2}\right)}$, or $\boxed{\left(2, \frac{3+\sqrt{17}}{2}\right)}$.


Question: As shown in the figure, it is known that the parabola passes through points $O(0, 0)$ and $A(5, 5)$, and its axis of symmetry is $x=2$. Find the analytic equation of the parabola.

Solution:
Solution: Suppose the equation of the parabola is $y=ax^2+bx+c$, then:
\[
\begin{cases}
c=0 \\
5=25a+5b+c \\
-\frac{b}{2a}=2
\end{cases}
\]
Solving for:
\[
\begin{cases}
a=1 \\
b=-4 \\
c=0
\end{cases}
\]
$\therefore y=x^2-4x=\boxed{x^2-4x}$


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the parabola $y=x^{2}+x-2$ intersects the x-axis at points A and B, and intersects the y-axis at point C. Find the coordinates of the vertex D.

Solution:
\textbf{Solution:} Let $y=x^2+x-2=\left(x+\frac{1}{2}\right)^2-\frac{9}{4}$,\\
$\therefore$ the coordinates of the vertex D are $\boxed{\left(-\frac{1}{2}, -\frac{9}{4}\right)}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the parabola $y=x^2+x-2$ intersects the x-axis at points A and B, and intersects the y-axis at point C. Find the area of $\triangle ABC$.

Solution:
\textbf{Solution:} Let $y=0$, which gives ${x}^{2}+x-2=0$. Solving this, ${x}_{1}=-2$, ${x}_{2}=1$,\\
$\therefore$ points $A(-2, 0)$, $B(1, 0)$,\\
$\therefore$ $AB=3$,\\
Let $x=0$, which means $y=-2$,\\
$\therefore$ point $C(0, -2)$,\\
$\therefore$ $OC=2$,\\
$\therefore$ the area of $\triangle ABC=\frac{1}{2}\times AB\times OC=\frac{1}{2}\times 3\times 2=\boxed{3}$.


Question: As shown in the figure, the parabola intersects the x-axis at points $A$ and $B(3, 0)$, and intersects the y-axis at point $C(0, -3)$. The vertex of the parabola is $D$. What is the analytical equation of the parabola?

Solution:
Solution: (1) Substituting point B$(3,0)$ and C$(0,-3)$ into $y=ax^{2}-2x+c$, we get
\[
\left\{
\begin{array}{l}
9a-6+c=0\\
c=-3
\end{array}
\right.
\]
$\therefore$
\[
\left\{
\begin{array}{l}
a=1\\
c=-3
\end{array}
\right.
\]
$\therefore$ The equation of the parabola is $\boxed{y=x^{2}-2x-3}$.


Question: As shown in the figure, the parabola intersects the $x$-axis at points $A$ and $B(3,0)$, and intersects the $y$-axis at point $C(0,-3)$, with the vertex of the parabola being $D$. It is known that point $M$ is a moving point on the $x$-axis, and a perpendicular line is drawn from point $M$ to the parabola at point $G$. Is there such a point $M$ that makes the triangle with vertices $A$, $M$, and $G$ similar to $\triangle BCD$? If such a point exists, find the coordinates of point $M$; if it does not exist, please explain why.

Solution: \textbf{Solution:} Let $y=0$, then $x^{2}-2x-3=0$. Solving this, we find $x_{1}=-1$ and $x_{2}=3$. Therefore, $A(-1,0)$. Since $y=x^{2}-2x-3=(x-1)^{2}-4$, the vertex $D$ of the parabola has coordinates $(1,-4)$. Also, given $C(0,-3)$, $B(3,0)$, and $D(1,-4)$, we have $BD^{2}=2^{2}+4^{2}=20$, $CD^{2}=1^{2}+1^{2}=2$, $BC^{2}=3^{2}+3^{2}=18$. Therefore, $BD^{2}=CD^{2}+BC^{2}$, implying that $\triangle BDC$ is a right-angled triangle with $\angle BCD=90^\circ$. Let the coordinate of point $M$ be $(m,0)$, then the coordinate of point $G$ is $(m,m^{2}-2m-3)$. As shown in the figure, a perpendicular line to the $x$-axis through point $M$ intersects the parabola at point $G$. Therefore, $\angle AMG=\angle BCD=90^\circ$. Thus, for the triangle with vertices $A$, $M$, $G$ to be similar to $\triangle BCD$, it must satisfy the condition: $\frac{AM}{MG}=\frac{BC}{CD}$ or $\frac{AM}{MG}=\frac{CD}{BC}$. Since $AM=|-1-m|$ and $MG=|m^{2}-2m-3|$, with $CD=\sqrt{2}$ and $BC=3\sqrt{2}$, we have $\frac{BC}{CD}=3$ and $\frac{CD}{BC}=\frac{1}{3}$. Therefore, we get $\frac{|-1-m|}{|m^{2}-2m-3|}=3$ (1) and $\frac{|-1-m|}{|m^{2}-2m-3|}=\frac{1}{3}$ (2). Solving (1), we find $m=\boxed{\frac{8}{3}}$, or $m=\boxed{\frac{10}{3}}$, or $m=-1$ (discard). Solving (2) yields $m=0$, or $m=6$, or $m=-1$ (discard). Therefore, \boxed{M(0,0)}, \boxed{M(6,0)}, \boxed{M\left(\frac{8}{3},0\right)}, or \boxed{M\left(\frac{10}{3},0\right)}.


Question: As shown in the figure, a parabola intersects the $x$-axis at points $A$ and $B(3, 0)$, and intersects the $y$-axis at point $C(0, -3)$. The vertex of the parabola is $D$. Consider a point $P$ on the parabola below the line $BC$, a perpendicular line $PQ$ is drawn to $BC$ at point $Q$, and line $CP$ is connected. When one of the angles in $\triangle CPQ$ equals $\angle ACO$, find the coordinates of point $P$.

Solution:
\textbf{Solution:} As shown in the diagram, let there be a point P on the parabola below the line BC. Draw PQ perpendicular to BC at point Q, connect CP, and draw PH $\perp$ y-axis at point H, let P $(x,x^{2}-2x-3)$, \\
$\therefore$PH $=x$, CH $=|x^{2}-2x-3-(-3)|=|x^{2}-2x|$, \\
$\therefore \angle PQC = \angle CHP = 90^\circ$, \\
$\therefore$ Points P, Q, H, C are concyclic, \\
$\therefore \angle PCQ = \angle CPH$, \\
When $\angle PCQ = \angle ACO$, \\
$\therefore \angle HPC = \angle ACO$, \\
$\therefore \tan \angle HPC = \tan \angle ACO$, \\
$\therefore \frac{HC}{PH} = \frac{AO}{CO} = \frac{1}{3}$, \\
$\therefore$PH $= 3$HC, \\
$\therefore x = 3|x^{2}-2x|$, \\
Solving, we get $x_{1}=\frac{5}{3}, x_{2}=0$ (discard), or $x_{3}=\frac{7}{3}, x_{4}=0$ (discard), \\
$\therefore$ The coordinates of point P are $\boxed{\left(\frac{5}{3},-\frac{32}{9}\right)}$ or $\boxed{\left(\frac{7}{3},-\frac{20}{9}\right)}$.


Question: In the Cartesian coordinate system, the graph of the quadratic function $y = x^2 + bx + c$ passes through points $A(3, 0)$ and $B(2, -3)$. What is the expression of this quadratic function?

Solution:
Solution: By substituting the coordinates of points $A(3, 0)$ and $B(2, -3)$ into $y=x^{2}+bx+c$, we get:
\[
\begin{cases}
9+3b+c=0\\
4+2b+c=-3
\end{cases}
\]
Solving this, we find:
\[
\begin{cases}
b=-2\\
c=-3
\end{cases}
\]
Therefore, the expression for the quadratic function is \boxed{y=x^{2}-2x-3};


Question: In the Cartesian coordinate system, the graph of the quadratic function $y=x^2+bx+c$ passes through points $A(3, 0)$ and $B(2, -3)$. Determine the coordinates of the vertex of the quadratic function.

Solution:
\textbf{Solution:} Since $y=x^{2}-2x-3=x^{2}-2x+1-4=(x-1)^{2}-4$, \\
the vertex coordinates of the parabola are $\boxed{(1, -4)}$;


Question: In the Cartesian coordinate system, the graph of the quadratic function $y=x^2+bx+c$ passes through points $A(3, 0)$ and $B(2, -3)$. By translating the graph of this quadratic function to the right by a certain number of units, the resulting graph can be made to pass through the origin.

Solution:
\textbf{Solution:} When $y=0$, we have $x^2-2x-3=0$,\\
Solving this yields: $x_1=-1$, $x_2=3$,\\
$\therefore$ the coordinates of the intersection points of the parabola $y=x^2-2x-3$ with the x-axis are $C(-1,0)$ and $A(3,0)$,\\
$\therefore$ by shifting the graph of the quadratic function one unit to the right, the resultant graph can pass through the origin.\\
Thus, the answer is: by shifting the graph of the quadratic function $\boxed{1}$ unit to the right, the shifted graph can pass through the origin.


Question: As shown in the diagram, parabola $P_{1}\colon y=x^{2}+2x+m$ is translated to get parabola $P_{2}\colon y=x^{2}-5x+n$. The two parabolas intersect the $y$ axis at points $C$ and $D$ respectively. The intersection point $E$ of parabolas $P_{1}$ and $P_{2}$ has an abscissa of $1$. A line parallel to the $x$ axis is drawn through point $E$, intersecting parabolas $P_{1}$ and $P_{2}$ at points $A$ and $B$, respectively. Find the axis of symmetry of parabola $P_{1}$ and the abscissa of point $A$.

Solution:
\textbf{Solution:} To solve, the parabola $P_{1}\colon y=x^{2}+2x+m$ has its axis of symmetry as the line $x=-1$,\\
$\because AB\parallel x$ axis,\\
and $\because$ Points A and E are symmetric about the axis $x=-1$, and the x-coordinate of point E is 1,\\
$\therefore$ the x-coordinate of point A is $\boxed{-3}$;\\
Thus, the axis of symmetry of the parabola $P_{1}$ is the line $\boxed{x=-1}$ and the x-coordinate of point A is $-3$.


Question: As shown in the figure, the parabola $P_{1}\colon y=x^{2}+2x+m$ is translated to obtain parabola $P_{2}\colon y=x^{2}-5x+n$. The two parabolas intersect the $y$-axis at points $C$ and $D$, respectively. The intersection point $E$ of parabolas $P_{1}$ and $P_{2}$ has an abscissa of $1$. A line parallel to the $x$-axis passing through point $E$ intersects parabolas $P_{1}$ and $P_{2}$ at points $A$ and $B$, respectively. What is the length of segments $AB$ and $CD$?

Solution:
\textbf{Solution:} Given the information, the parabola $P_2\colon y=x^2-5x+n$ has an axis of symmetry represented by the line $x=\frac{5}{2}$,\\
$\because AB\parallel x$ axis,\\
$\therefore$ point B is symmetric to point E with respect to the axis of symmetry $x=\frac{5}{2}$,\\
$\therefore$ the x-coordinate of point B is 4,\\
$\therefore AB=4-(-3)=\boxed{7}$;\\
$\because$ point E is the intersection point of parabola $P_1$ and parabola $P_2$,\\
$1^2+2\times 1+m=1^2-5\times 1+n$,\\
$\therefore m=n-7$,\\
$\therefore n-m=7$,\\
Let $x=0$,\\
then $C(0, m), D(0, n)$,\\
$\therefore CD=n-m=\boxed{7}$;\\
Thus, the lengths of segments AB and CD are both 7.


Question: As shown in the figure, the parabola $y=-x^2+bx+c$ intersects the x-axis at points $A(-1,0)$ and $B(5,0)$, and intersects the y-axis at point $C$. The line $y=-x+n$ passes through point $B$. Point $P$ is a moving point on the parabola above the line $BC$. A perpendicular line from point $P$ to the x-axis meets point $F$, and intersects the line $BC$ at point $E$. Let the x-coordinate of point $P$ be $m$. What is the equation of the parabola?

Solution:
\textbf{Solution:} \\
Since the parabola $y=-x^2+bx+c$ intersects the x-axis at points A ($-1$, $0$) and B ($5$, $0$), \\
therefore, we can set the equation of the function as $y=-\left(x+1\right)\left(x-5\right)$ \\
thus, $y=-x^2+4x+5$ \\
Hence, the equation of the parabola is $\boxed{y=-x^2+4x+5}$.


Question: As shown in the figure, the parabola $y=-x^2+bx+c$ intersects the x-axis at points $A(-1,0)$ and $B(5,0)$, and intersects the y-axis at point $C$. The line $y=-x+n$ passes through point $B$. Point $P$ is a moving point on the parabola above line $BC$. A perpendicular line is drawn from point $P$, denoted as $PF\perp x$-axis, meeting point $F$, and intersects line $BC$ at point $E$. Assume the x-coordinate of point $P$ is $m$. If $PF=3EF$, what is the value of $m$?

Solution:
\textbf{Solution:} Let point C be $(0,5)$, then the slope of BC is $k$, with $y=kx+5$.\\
$\therefore 5k+5=0$, solving for $k$ gives $k=-1$.\\
$\therefore$ The equation for BC is $y=-x+5$.\\
Since $PF\perp x$-axis and passes through point F, intersecting line BC at point E,\\
$\therefore$ Let point P be $(m, -m^2+4m+5)$, and point E be $\left(m, -m+5\right)$.\\
$PF=-m^2+4m+5$, $EF=-m+5$.\\
We obtain $-m^2+4m+5=3(-m+5)$, solving for $m$ yields $m=\boxed{2}$, $m=5$ (discard).


Question: As shown in the figure, the parabola $y=-x^2+bx+c$ intersects the x-axis at points $A(-1,0)$ and $B(5,0)$, and intersects the y-axis at point $C$. The line $y=-x+n$ passes through point $B$. Point $P$ is a moving point on the parabola above line $BC$. Through point $P$, draw $PF\perp x$ axis intersecting line $BC$ at point $E$. Suppose the x-coordinate of point $P$ is $m$. If point $E'$ is the symmetric point of $E$ with respect to line $PC$, is there any point $P$ such that point $E'$ lies on the y-axis? If yes, find the coordinates of point $P$; if no, explain why.

Solution:
\textbf{Solution:} Construct ME $\perp$ y-axis at point M,\\
since points E$^{\prime}$ and E are symmetric about PC,\\
therefore $\angle$E$^{\prime}$CP$=\angle$PCE, CE=CE$^{\prime}$, PE=PE$^{\prime}$,\\
since PE$\parallel$y-axis,\\
therefore $\angle$E$^{\prime}$CP$=\angle$CPE$=\angle$PCE,\\
therefore CE=PE=EC,\\
therefore quadrilateral E$^{\prime}$PEC is a rhombus,\\
let point P $(m, -m^{2}+4m+5)$, E$(m, -m+5)$,\\
therefore PE$=-m^{2}+4m+5+m-5=-m^{2}+5m$;\\
since OB=OC=5,\\
therefore $\triangle$OBC is an isosceles right triangle,\\
therefore $\angle$MCE$=45^\circ$,\\
therefore $\triangle$MCE is an isosceles right triangle,\\
therefore CM=ME=m,\\
therefore CE$= \sqrt{2}m$,\\
therefore solving $-m^{2}+5m=\sqrt{2}m$: \\
$m = 5-\sqrt{2}$, m=0 (discard),\\
therefore the y-coordinate of point P is $-\left(5-\sqrt{2}\right)^{2}+4\left(5-\sqrt{2}\right)+5=6\sqrt{2}-2$,\\
therefore the coordinates of point P are $\boxed{\left(5-\sqrt{2}, 6\sqrt{2}-2\right)}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+c$ passes through the points $A(-1,0)$, $B(3,0)$, and $C(0,3)$. The vertex is denoted by $D$, and the axis of symmetry is the line $DE$. Determine the equation of the parabola.

Solution:
\textbf{Solution:} Substitute $A(-1,0)$, $B(3,0)$, $C(0,3)$ into $y=ax^2+bx+c$,\\
$\therefore \begin{cases}
a-b+c=0\\
9a+3b+c=0\\
c=3
\end{cases}$,\\
Solving this, we get $\begin{cases}
a=-1\\
b=2\\
c=3
\end{cases}$,\\
$\therefore y=-x^2+2x+3$; hence, the equation of the parabola is $\boxed{y=-x^2+2x+3}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+c$ passes through points $A(-1,0)$, $B(3,0)$, and $C(0,3)$. Point $D$ is the vertex, and the axis of symmetry is the line $DE$. If point $M$ is a point on the parabola to the top right of $BD$, and the x-coordinate of $M$ is denoted as $m$, and the area of $\triangle MBD$ is $S$. Does $S$ have a maximum value? If yes, please find the maximum value and the coordinates of $M$. If no, please explain why.

Solution:
\textbf{Solution:}
Given that $y=-x^2+2x+3=-\left(x-1\right)^2+4$,\\
thus, the coordinates of point D are $(1,4)$, and the axis of symmetry is the line $x=1$,\\
let the equation of line BD be $y=kx+f$,\\
thus, $\left\{\begin{array}{l}
3k+f=0 \\
k+f=4
\end{array}\right.$,\\
solving this yields $\left\{\begin{array}{l}
k=-2 \\
f=6
\end{array}\right.$,\\
thus, $y=-2x+6$,\\
drawing line $MG\parallel y$ axis to intersect BD at point G from M,\\
since $M\left(m,-m^2+2m+3\right)$,\\
thus, $G\left(m,-2m+6\right)$,\\
thus, $MG=-m^2+4m-3$,\\
thus, $S=\frac{1}{2}\times(3-1)\times(-m^2+4m-3)=-m^2+4m-3=-\left(m-2\right)^2+1$,\\
thus, when $m=2$, $S$ reaches its maximum value $\boxed{1}$, at this time $M(2,3)$;


Question: As shown in the figure, the parabola $y=ax^2+bx+c$ passes through the points $A(-1,0)$, $B(3,0)$, and $C(0,3)$, where $D$ is the vertex and the axis of symmetry is the line $DE$. Find a point $P$ on the axis of symmetry of the parabola such that $\triangle PAC$ is an isosceles triangle. Please write down the coordinates of point $P$ directly.

Solution: \textbf{Solution:} The coordinates of point $P$ are $\boxed{(1,1)}$, $\boxed{\left(1, \sqrt{6}\right)}$, $\boxed{\left(1, -\sqrt{6}\right)}$, or $\boxed{(1,0)}$.


Question: As shown in the figure, in the Cartesian coordinate system, it is known that the parabola $y=x^{2}+bx+c$ intersects the line AB at points A and B, where A ($-3$, $-4$) and B ($0$, $-1$), and the axis of symmetry of the parabola intersects the x-axis at point Q. What is the function expression of this parabola?

Solution:
\textbf{Solution:} Since the parabola passes through point A ($-3$, $-4$) and point B ($0$, $-1$), \\
it follows that $\begin{cases} 9-3b+c=-4 \\ c=-1 \end{cases}$, hence $b=4$ and $c=-1$, \\
thus, the equation of the parabola is $\boxed{y=x^{2}+4x-1}$.


Question: As shown in the figure, in the Cartesian coordinate system, it is known that the parabola $y=x^{2}+bx+c$ intersects with line $AB$ at points $A$ and $B$, where $A(-3,-4)$ and $B(0,-1)$. The axis of symmetry of the parabola intersects with the x-axis at point $Q$. Point $P$ is any point on the parabola below line $AB$, and lines $PA$, $PB$, $QA$, $QB$ are connected. Find the maximum area of quadrilateral $PAQB$ and the coordinates of point $P$ at this time?

Solution:
\textbf{Solution:} Let the equation of line AB be $y=kx+b$. Substituting A$(-3, -4)$ and B$(0, -1)$ into it, we get
\[
\left\{
\begin{array}{l}
-3k+b=-4 \\
b=-1
\end{array}
\right.
\]
$\therefore k=1$, $b=-1$, thus $y=x-1$,

Draw a perpendicular line from point P to the x-axis, intersecting the line AB at point F.

Let point P be $(a,a^2+4a-1)$, then the coordinates of point F are $(a,a-1)$,

$S_{\triangle PAB}=\frac{1}{2} |PF|\cdot|x_B-x_A| =\frac{3}{2}(a-1-a^2-4a+1)$

$=\frac{3}{2}(-a^2-3a)=-\frac{3}{2}(a+\frac{3}{2})^2+\frac{27}{8}$,

When $a=-\frac{3}{2}$, $S_{\triangle PAB}$ reaches its maximum value of $\frac{27}{8}$. Since the area of $\triangle QAB$ is $\frac{9}{2}$,

$\therefore$ the maximum area of quadrilateral PAQB is $\boxed{\frac{63}{8}}$, at which point the coordinates of P are $\boxed{\left(-\frac{3}{2},-\frac{19}{4}\right)}$;


Question: As shown in the figure, in the Cartesian coordinate system, it is known that the parabola $y=x^{2}+bx+c$ intersects with line AB at points A and B, where A($-3$, $-4$) and B($0$, $-1$), and the axis of symmetry of the parabola intersects with the x-axis at point Q. The parabola is translated 2 units to the right to obtain the parabola $y=a_{1}x^{2}+b_{1}x+c_{1}$ ($a_{1}\neq 0$). The translated parabola intersects with the original parabola at point C. Point D is a point on the axis of symmetry of the original parabola. In the Cartesian coordinate system, is there a point E such that the quadrilateral with vertices B, C, D, E is a rhombus? If it exists, please find the coordinates of point E; if it does not exist, please explain the reason.

Solution:
\textbf{Solution:} The vertex of the parabola $y=x^{2}+4x-1$ is $(-2,-5)$, and its axis of symmetry is the line $x=-2$,\\
The parabola after translation is $y=x^{2}-5$, by
\[
\begin{cases}
y=x^{2}+4x-1\\
y=x^{2}-5
\end{cases}
\]
we get
\[
\begin{cases}
x=-1\\
y=-4
\end{cases}
\]
$\therefore$ The coordinates of point C are $(-1,-4)$, let the coordinates of point D be $(-2,t)$,\\
The quadrilateral formed by vertices B, C, D, E is a rhombus, that is, $\triangle BDC$ is an isosceles triangle,\\
(1) When BC is the diagonal, by the midpoint formula we get $E(1,-5-t)$, $\because BD=CD$,\\
$\therefore 1^{2}+(-4-t)^{2}=2^{2}+(-1-t)^{2}$, solving for $t=-2$, $\therefore E(1,-3)$;\\
(2) When BD is the diagonal, by the midpoint formula we get $E(-1,t+3)$, $\because BC=CD$,\\
$\therefore (-1)^{2}+(-3)^{2}=(-1)^{2}+(t+4)^{2}$,\\
solving for $t_{1}=-1$, $t_{2}=-7$ (discard), $\therefore E(-1,2)$;\\
(3) When BE is the diagonal, by the midpoint formula we get $E(-3,-3+t)$,\\
$\because BC=BD$, $\therefore (-1)^{2}+(-3)^{2}=(-2)^{2}+(t+1)^{2}$,\\
solving for $t_{1}=-1+ \sqrt{6}$, $t_{2}=-1- \sqrt{6}$,\\
$\therefore E(-3,-4+ \sqrt{6})$ or $E(-3,-4- \sqrt{6})$,\\
In summary, the coordinates of point E are $\boxed{(1,-3)}$ or $\boxed{(-1,2)}$ or $\boxed{(-3,-4+ \sqrt{6})}$ or $\boxed{(-3,-4- \sqrt{6})}$.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=-x+m$ intersects with the graph of the inverse proportionality function $y=\frac{k}{x}$ (where $x>0$) at points $A$ and $B$. Given $A(1,2)$, what are the equations of the linear and the inverse proportionality functions?

Solution: \textbf{Solution:} Substitute point A$(1,2)$ into $y=-x+m$, to obtain $-1+m=2$, $\therefore m=3$, $\therefore$ the linear function can be expressed as $y=-x+3$; Substitute point A$(1,2)$ into $y=\frac{k}{x}$, $\therefore k=1\times 2=2$, $\therefore$ the inverse proportional function can be expressed as $\boxed{y=-x+3}$ and $\boxed{y=\frac{2}{x}}$;


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=-x+m$ intersects with the graph of the inverse proportion function $y=\frac{k}{x}$ ($x>0$) at points $A$ and $B$. It is known that $A(1,2)$. Connect $AO$ and $BO$, what is the area of $\triangle AOB$?

Solution:
\textbf{Solution:} Solve the system of equations
\[
\left\{\begin{array}{l}
y=-x+3\\
y=\frac{2}{x}
\end{array}\right.
\]
to get
\[
\left\{\begin{array}{l}
x=1\\
y=2
\end{array}\right.
\]
or
\[
\left\{\begin{array}{l}
x=2\\
y=1
\end{array}\right.
\]
$\therefore$ For point B $(2, 1)$, let the intersection of the line $y=-x+3$ with the $y$-axis be point C, $\therefore$ C $(0, 3)$. Therefore, the area of $\triangle AOB$ equals the area of $\triangle COB$ minus the area of $\triangle COA$, which is $\frac{1}{2}\times3\times2-\frac{1}{2}\times3\times1=\boxed{1.5}$.


Question: Given the partial graph of the parabola $y = -x^2 + bx - c$ as shown in the figure. Find the values of $b$ and $c$.

Solution:
Solution: Substituting $(1, 0)$ into $y = -x^2 + bx - c$ yields
\[
\left\{
\begin{array}{l}
-1 + b + c = 0 \\
-c = 3
\end{array}
\right.
\]
Solving yields $b = \boxed{-2}$, $c = \boxed{-3}$.


Question: As shown in the figure, it is known that part of the graph of the parabola $y = -x^2 + bx - c$. Find the axis of symmetry and the maximum value of $y$ for the parabola, respectively.

Solution:
\textbf{Solution:} Given the function $y = -x^{2} - 2x + 3$,\\
it can be rewritten as $y = -(x + 1)^{2} + 4$,\\
therefore, the axis of symmetry of the parabola is the line $\boxed{x = -1}$,\\
thus, the maximum value of $y$ is $\boxed{4}$.


Question: As shown in the figure, it is known that the parabola $y=-x^2+2x+4$ intersects the y-axis at point $C$, and its vertex is point $D$. Find the coordinates of points $C$ and $D$.

Solution:
\textbf{Solution:} The equation of the parabola is $y=-x^2+2x+4=-(x-1)^2+5$,\\
$\therefore$ the coordinates of point $D$ are \boxed{(1,5)}.\\
Let $x=0$, then $y=4$,\\
$\therefore$ the coordinates of point $C$ are \boxed{(0,4)}.


Question: As shown in the figure, it is known that the parabola $y=-x^2+2x+4$ intersects the $y$-axis at point $C$ and has the vertex at $D$. Connect $CD$. Point $Q$ is on the line $CD$ in the first quadrant. Draw $QM\perp x$-axis, intersecting the $x$-axis at point $M$. If the abscissa of point $Q$ is $x$, and the area of $\triangle QMO$ is $S$, what is the function expression of $S$ in terms of $x$?

Solution:
\textbf{Solution:} Let the equation of line $CD$ be $y=ax+b$. Substituting points $D(1, 5)$ and $C(0, 4)$ into this, we get
\[
\left\{
\begin{array}{l}
a+b=5 \\
b=4
\end{array}
\right.
\]
Solving this system yields
\[
\left\{
\begin{array}{l}
a=1 \\
b=4
\end{array}
\right.
\]
Therefore, the equation of line $CD$ is $y=x+4$.\\
Since the x-coordinate of point $Q$ is $x$,\\
Therefore, $Q(x, x+4)$, $M(x, 0)$,\\
Since point $Q$ is in the first quadrant,\\
Therefore, $QM=x+4$, $MO=x$,\\
Therefore, $S=\frac{1}{2}QM\cdot MO=\frac{1}{2}x(x+4)=\boxed{\frac{1}{2}x^2+2x}$.


Question: As shown in Figure 1, the parabola $y = ax^2 + bx + 3$ intersects the x-axis at points $A(3,0)$ and $B(-1,0)$, and intersects the y-axis at point $C$. Point $P$ is a moving point on the parabola in the first quadrant, and point $F$ is a moving point on the y-axis. Connect $PA$, $PF$, $AF$. What is the function expression corresponding to this parabola?

Solution:
\textbf{Solution:} \\
Since the parabola $y=ax^{2}+bx+3$ intersects the $x$-axis at points $A(3,0)$ and $B(-1,0)$, \\
we have $\begin{cases}9a+3b+3=0 \\ a-b+3=0 \end{cases}$ \\
Solving these, we get $\begin{cases}a=-1 \\ b=2 \end{cases}$ \\
Therefore, the corresponding function expression for the parabola is $y=-x^{2}+2x+3=\boxed{-x^{2}+2x+3}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+3$ intersects the $x$-axis at points $A(3,0)$ and $B(-1,0)$, and intersects the $y$-axis at point $C$. Point $P$ is a moving point on the parabola in the first quadrant, and point $F$ is a moving point on the $y$-axis. Lines $PA$, $PF$, and $AF$ are connected. When the coordinates of point $F$ are $(0,-4)$, what is the maximum value of the area of $\triangle AFP$?

Solution:
\textbf{Solution:} As shown in Figure 1, we draw PQ through point P parallel to the y-axis to meet line AF at point Q, \\
let the equation of line AF be $y = kx + d$, \\
since A(3, 0), F(0, -4), \\
thus $\begin{cases} 3k+d=0 \\ d=-4 \end{cases}$, solving gives: $\begin{cases} k=\frac{4}{3} \\ d=-4 \end{cases}$, \\
therefore, the equation of line AF is $y=\frac{4}{3}x-4$, \\
let $P(t, -t^{2}+2t+3)$ (-1<t<3), hence Q(t, $\frac{4}{3}t-4$), \\
thus $PQ = -t^{2}+2t+3-(\frac{4}{3}t-4) = -t^{2}+\frac{2}{3}t+7$, \\
therefore, $S_{\triangle AFP}$ = $\frac{1}{2}$PQ·OA = $\frac{1}{2}(-t^{2}+\frac{2}{3}t+7)×3$ = -$\frac{3}{2}$(t-$\frac{1}{3})^{2}+\frac{32}{3}$, \\
since $-\frac{3}{2}<0$, -1<t<3, \\
therefore, when t=$\frac{1}{3}$, the maximum area of $\triangle AFP$ is $\boxed{\frac{32}{3}}$;


Question: As shown in Figure 1, the parabola $y=ax^2+bx+3$ intersects the $x$-axis at points $A(3,0)$ and $B(-1,0)$, and intersects the $y$-axis at point $C$. Point $P$ is a moving point on the parabola in the first quadrant, and point $F$ is a moving point on the $y$-axis. Connect $PA$, $PF$, and $AF$. As shown in Figure 2, is there a point $F$ such that $\triangle AFP$ is an isosceles right triangle with $AP$ as one of its legs? If such a point F exists, find the coordinates of all such points $F$; if not, please explain why.

Solution:
\textbf{Solution:} Let $P(m, -m^{2}+2m+3)$ ($-1<m<3$), $F(0, n)$, \\
since $A(3,0)$, \\
therefore $OA=3$, $OF=|n|$, \\
(1) When $AP=AF$, $\angle PAF=90^\circ$, draw $PD\perp x$-axis at point $D$, \\
then $\angle ADP=90^\circ=\angle AOF$, \\
thus $\angle PAD+\angle APD=90^\circ$, \\
since $\angle PAD+\angle FAO=90^\circ$, \\
therefore $\angle APD=\angle FAO$, \\
in $\triangle APD$ and $\triangle FAO$, \\
$\left\{\begin{array}{l}
\angle ADP=\angle AOF \\
\angle APD=\angle FAO \\
AP=AF
\end{array}\right.$ \\
thus $\triangle APD \cong \triangle FAO$ (AAS), \\
thus $PD=OA$, $AD=OF$, \\
since $PD=-m^{2}+2m+3$, $AD=3-m$, \\
therefore $-m^{2}+2m+3=3$, \\
solving for $m$: $m=0$ or $2$, \\
when $m=0$, $P(0,3)$, $AD=3$, \\
thus $OF=3$, i.e., $|n|=3$, \\
since point $F$ is on the negative $y$-axis, \\
thus $n=-3$, \\
therefore $F(0,-3)$; \\
when $m=2$, $P(2,3)$, $AD=1$, \\
thus $OF=1$, i.e., $|n|=1$, \\
since point $F$ is on the negative $y$-axis, \\
thus $n=-1$, \\
therefore $F(0,-1)$; \\
(2) When $AP=PF$, $\angle APF=90^\circ$, draw $PD\perp x$-axis at point $D$, $PG\perp y$-axis at point $G$, \\
then $\angle PDA=\angle PDO=\angle PGF=90^\circ$, \\
since $\angle PDO=\angle PGF=\angle DOG=90^\circ$, \\
thus quadrilateral $PDOG$ is a rectangle, \\
thus $\angle FPG+\angle FPD=90^\circ$, \\
since $\angle APD+\angle FPD=\angle APF=90^\circ$, \\
therefore $\angle FPG=\angle APD$, \\
in $\triangle FPG$ and $\triangle APD$, \\
$\left\{\begin{array}{l}
\angle PGF=\angle PDA \\
\angle FPG=\angle APD \\
PF=AP
\end{array}\right.$, \\
thus $\triangle FPG \cong \triangle APD$ (AAS), \\
thus $PG=PD$, $FG=AD$, \\
since $PD=-m^{2}+2m+3$, $AD=3-m$, $PG=m$, \\
therefore $-m^{2}+2m+3=m$, \\
solving for $m$: $m=\frac{1-\sqrt{13}}{2}$ (discard) or $m=\frac{1+\sqrt{13}}{2}$, \\
when $m=\frac{1+\sqrt{13}}{2}$, $P\left(\frac{1+\sqrt{13}}{2}, \frac{1+\sqrt{13}}{2}\right)$, \\
thus $FG=AD=3-m=3-\frac{1+\sqrt{13}}{2}=\frac{5-\sqrt{13}}{2}$, \\
thus $F\left(0, \sqrt{13}-2\right)$; \\
In summary, the coordinates of point $F$ are $(0,-3)$ or $(0,-1)$ or $\boxed{(0,\sqrt{13}-2)}$.


Question: As shown in the figure, it is known that the parabola $y=ax^2+bx-1\ (a\ne 0)$ intersects the x-axis at points $A(-2, 0)$ and $B(2, 0)$, and intersects the y-axis at point D. The line through point $C(0, 1)$ intersects the parabola at points A and F. What is the equation of the parabola?

Solution:
\textbf{Solution:} Substituting the points $A\left(-2, 0\right)$, $B\left(2, 0\right)$ into the equation $y=ax^{2}+bx-1\left(a\ne 0\right)$, we get
\[
\begin{cases}
4a-2b-1=0 \\
4a+2b-1=0
\end{cases}
\]
Solving these, we find
\[
\begin{cases}
a=\frac{1}{4} \\
b=0
\end{cases}
\]
$\therefore$ The equation of the parabola is $\boxed{y=\frac{1}{4}x^{2}-1}$


Question: As shown in the figure, it is known that the parabola $y=ax^2+bx-1$ ($a\neq 0$) intersects the x-axis at points $A(-2, 0)$ and $B(2, 0)$, and intersects the y-axis at point D. The line through point $C(0, 1)$, line AC, intersects the parabola at points A and F. Point P is a movable point on the parabola below line AF. A line parallel to the y-axis passing through point P meets line AC at point Q, and a line parallel to the x-axis passing through point P meets the y-axis at point E. Determine the maximum value of $PQ+PE$ and the corresponding coordinates of point P.

Solution:
Solution: Let the equation of line AC be $y=mx+n$, \\
$\left\{\begin{array}{l}
-2m+n=0\\
n=1
\end{array}\right.$, solving this system gives $\left\{\begin{array}{l}
m=\frac{1}{2}\\
n=1
\end{array}\right.$, \\
$\therefore$ the equation of line AC is $y=\frac{1}{2}x+1$, \\
Let P$(x, \frac{1}{4}x^{2}-1)$, then Q$(x, \frac{1}{2}x+1)$, E$(0, \frac{1}{4}x^{2}-1)$, \\
When point P is to the right of the y-axis, \\
$\therefore PQ+PE=(\frac{1}{2}x+1)-(\frac{1}{4}x^{2}-1)+x=-\frac{1}{4}(x-3)^{2}+\frac{17}{4}$, \\
When $x=3$, $PQ+PE$ reaches its maximum value $\boxed{\frac{17}{4}}$, \\
At this time, the coordinates of point P are $(3, \frac{5}{4})$; \\
When point P is to the left of the y-axis, \\
$\therefore PQ+PE=(\frac{1}{2}x+1)-(\frac{1}{4}x^{2}-1)+(0-x)=-\frac{1}{4}(x+1)^{2}+\frac{9}{4}$, \\
When $x=-1$, $PQ+PE$ reaches its maximum value $\frac{9}{4}$, \\
At this time, the coordinates of point P are $(-1, -\frac{3}{4})$; \\
In summary, $PQ+PE$ reaches its maximum value $\boxed{\frac{17}{4}}$, the coordinates of point P are $(3, \frac{5}{4})$;


Question: As shown in the figure, it is known that the parabola $y=ax^2+bx-1\ (a\neq 0)$ intersects the x-axis at points $A(-2, 0)$ and $B(2, 0)$, and intersects the y-axis at point D. A line passing through point $C(0, 1)$ intersects the parabola at points A and F. Under these conditions, shift the parabola $y=ax^2+bx-1\ (a\neq 0)$ to the right by 2 units and then downward by 1 unit to obtain a new parabola $y_1$. Let point M be a point on the axis of symmetry of $y_1$, and point N be another point on $y_1$. If the quadrilateral with vertices D, P, M, N is a parallelogram, write down the coordinates of all possible points M, and select one of these points M to explain the process of finding its coordinates.

Solution:
\textbf{Solution:}
Since $y=\frac{1}{4}x^{2}-1$ intersects the y-axis at point D, \\
$\therefore$ D(0, $-1$),\\
Since the parabola $y=ax^{2}+bx-1\ (a\ne 0)$ is first translated 2 units to the right and then 1 unit downward to get the new parabola $y_{1}=\frac{1}{4}(x-2)^{2}-2$, \\
$\therefore$ the x-coordinate of point M is 2, \\
(1) When DP is the diagonal, the x-coordinate of point N is 1, \\
$\therefore$ the y-coordinate of point N is $-\frac{7}{4}$, \\
$\therefore$ translating point N 1 unit to the left and $\frac{3}{4}$ units upward yields point D; then applying the same translation to point P yields point M, i.e., M(2, 2); \\
(2) When DM is the diagonal, the x-coordinate of point N is $-1$, \\
$\therefore$ the y-coordinate of point N is $\frac{1}{4}$, \\
$\therefore$ translating point D 1 unit to the left and $\frac{5}{4}$ units upward yields point N; then applying the same translation to point P yields point M, i.e., M(2, $\frac{5}{2}$); \\
(3) When DN is the diagonal, the x-coordinate of point N is 5, \\
$\therefore$ the y-coordinate of point N is $\frac{1}{4}$, \\
$\therefore$ translating point P 2 units to the right and 1 unit downward yields point N; then applying the same translation to point D yields point M, i.e., M(2, $-2$); \\
Overall, the coordinates of M are (2, 2), (2, $\frac{5}{2}$), or (2, $-2$). Choosing M(2, 2) for the solution, we find the coordinates of M to be M(2, $\boxed{2}$).


Question: The partial graph of a quadratic function is shown in the figure, with the axis of symmetry being the line \(x=-1\). Find the analytic expression of this quadratic function.

Solution:
\textbf{Solution:} Let the equation of the parabola be $y=a(x+1)^2+k$, \\
Substituting $(-3,0)$ and $(0,3)$ into $y=a(x+1)^2+k$, we get $\begin{cases}0=4a+k\\ 3=a+k\end{cases}$, \\
Solving this, we obtain $\begin{cases}a=-1\\ k=4\end{cases}$, \\
$\therefore y=-\boxed{(x+1)^2+4}$.


Question: The partial graph of a quadratic function is shown in the figure, with its axis of symmetry being the line $x = -1$. What are the coordinates of the vertex of this graph?

Solution:
\textbf{Solution:} Since $y=-\left(x+1\right)^{2}+4$,\\
the vertex of the parabola is at $\boxed{(-1,4)}$.


Question: The partial graph of the quadratic function is shown in the figure, with the axis of symmetry being the line $x=-1$. By observing the graph, find the range of the independent variable $x$ when $y>0$.

Solution:
\textbf{Solution:} Since the parabola passes through the point ($-3$, $0$) and its axis of symmetry is the line $x = -1$, \\
it follows that the parabola also passes through the point ($1$, $0$), \\
therefore, when \boxed{-3 < x < 1}, we have $y > 0$.


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=x^{2}-2x+c$ intersects with the line $y=x+1$ at points $A$ and $C$, and the coordinates of point $A$ are $(-1,0)$. What are the coordinates of point $C$?

Solution:
Solution: Since point A $(-1,0)$ lies on the parabola $y=x^{2}-2x+c$, \\
$\therefore$ $0=-1^{2}-2(-1)+c$, \\
$\therefore$ $c=-3$, \\
$\therefore$ the equation of the parabola is $y=x^{2}-2x-3$, \\
By solving the system with the line $y=x+1$, we get $\left\{\begin{array}{l}y=x^{2}-2x-3 \\ y=x+1\end{array}\right.$, \\
which yields $\left\{\begin{array}{l}x=4\\ y=5\end{array}\right.$ or $\left\{\begin{array}{l}x=-1\\ y=0\end{array}\right.$, \\
$\therefore$ the coordinates of point C are $\boxed{(4,5)}$;


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=x^{2}-2x+c$ intersects with the line $y=x+1$ at points $A$ and $C$, and the coordinates of point $A$ are $(-1,0)$. If point $P$ is a moving point on the parabola below line $AC$, what is the maximum distance from point $P$ to line $AC$?

Solution:
\textbf{Solution}: Draw $PM\perp x$-axis from point $P$ to meet $AC$ at point $M$, as shown in the diagram: Let $P(m,m^{2}-2m-3)$ ($-1<m<5$), then $M(m,m+1)$,\\
$\therefore PM=m+1-(m^{2}-2m-3)=-m^{2}+3m+4$,\\
$\therefore S_{\triangle ACP}=\frac{1}{2}\times 5(-m^{2}+3m+4)=-\frac{5}{2}(m-\frac{3}{2})^{2}+\frac{125}{8}$,\\
$\therefore$when $m=\frac{3}{2}$, $S_{\triangle ACP}$ is maximized to $\frac{125}{8}$,\\
$\because$point $A(-1,0)$, point $C(4,5)$,\\
$\therefore AC=\sqrt{(4+1)^{2}+5^{2}}=5\sqrt{2}$,\\
Let the distance from point $P$ to line $AC$ be $h$,\\
$\therefore S_{\triangle ACP}=\frac{1}{2}\times 5\sqrt{2}\times h=\frac{125}{8}$,\\
$\therefore h=\frac{25\sqrt{2}}{8}$,\\
$\therefore$the maximum distance from point $P$ to line $AC$ is $\boxed{\frac{25\sqrt{2}}{8}}$;


Question: As shown in the figure, in the Cartesian coordinate system, the parabola $y=x^{2}-2x+c$ intersects the line $y=x+1$ at points $A$ and $C$, and the coordinates of point $A$ are $(-1, 0)$. If point $E$ is on the parabola, and point $F$ is on the axis of symmetry of the parabola, is there such a point $E$ that the quadrilateral formed by vertices $A$, $C$, $E$, $F$ is a parallelogram? If such a point exists, please write down the coordinates of point $E$ directly; if it does not exist, please explain the reason.

Solution:
\textbf{Solution:} Exist, the reason is as follows:\\
$\because y=x^{2}-2x-3=(x-1)^{2}-4$,\\
$\therefore$ the axis of symmetry of the parabola is the line $x=1$,\\
Let the coordinates of point $F$ be $(1,n)$, and the coordinates of point $E$ be $(x,x^{2}-2x-3)$,\\
Consider three cases:\\
(1) When $AC$ is the diagonal of the parallelogram,\\
$-1+4=1+x$,\\
Solving this gives $x=2$,\\
$\therefore$ the coordinates of point $E$ are $(2,-3)$;\\
(2) When $AF$ is the diagonal of the parallelogram,\\
$-1+1=x+4$,\\
Solving this gives $x=-4$,\\
$\therefore$ the coordinates of point $E$ are $(-4,21)$;\\
(3) When $AE$ is the diagonal of the parallelogram,\\
$-1+x=4+1$,\\
Solving this gives $x=6$,\\
$\therefore$ the coordinates of point $E$ are $(6,21)$;\\
In conclusion, the coordinates of point $E$ are \boxed{(2,-3)}, \boxed{(-4,21)}, or \boxed{(6,21)}.


Question: As shown in the figure, it is known that the parabola $y=x^2+bx+c$ intersects the x-axis at points $A$ and $B$, where the coordinates of point $A$ are $(-3,0)$, and it intersects the y-axis at point $C$. Point $D(-2,-3)$ is on the parabola. What is the equation of the parabola?

Solution:
\textbf{Solution:} Solve from the problem's statement, we obtain
\[
\left\{\begin{array}{l}
9-3b+c=0\\
4a-2b+c=-3
\end{array}\right.
\]
Solving yields:
\[
\left\{\begin{array}{l}
b=2\\
c=-3
\end{array}\right.
\]
$\therefore$ The equation of the parabola is $\boxed{y=x^{2}+2x-3}$.


Question: As shown in the figure, it is known that the parabola $y=x^2+bx+c$ intersects the x-axis at points $A$ and $B$, where the coordinate of point $A$ is $(-3,0)$, and it intersects the y-axis at point $C$. Point $D(-2,-3)$ lies on the parabola. There is a moving point $P$ on the axis of symmetry of the parabola. What is the minimum perimeter of $\triangle PAD$?

Solution:
\textbf{Solution:} First, connect BD intersecting the axis of symmetry at point P, \\
$\because$ points A and B are symmetric about the axis of symmetry, \\
$\therefore$ AP=BP, \\
$\therefore$ AP+PD+AD=BP+PD+AP=BD+AD, \\
$\because$ the shortest distance between two points is a line segment, \\
$\therefore$ at this time, the perimeter of $\triangle PAD$ is the smallest when it equals BD+AD, \\
when y=0, solving $x^{2}+2x-3=0$ gives $x_{1}=-3$, $x_{2}=1$, \\
$\therefore$ point B (1,0), \\
$\because$ A (-3,0) and D (-2,-3), \\
$\therefore$ $BD=\sqrt{(-2-1)^{2}+3^{2}}=3\sqrt{2}$, \\
$AD=\sqrt{(-3+2)^{2}+3^{2}}=\sqrt{10}$ \\
$\therefore$ the perimeter of $\triangle PAD$ is $3\sqrt{2}+\sqrt{10}=\boxed{3\sqrt{2}+\sqrt{10}}$.


Question: As shown in the figure, it is known that the parabola $y = x^{2} + bx + c$ intersects the $x$-axis at points $A$ and $B$, where the coordinates of point $A$ are $(-3, 0)$. It intersects the $y$-axis at point $C$, and point $D(-2, -3)$ lies on the parabola. There is a moving point $M$ on the axis of symmetry of the parabola. When $\triangle MAD$ is an isosceles triangle, directly write the coordinates of point $M$.

Solution:
\textbf{Solution:} Let point M be ($-1$, $n$),

since $A (-3,0)$ and $D (-2,-3)$,

thus, $AM^{2} = (-1+3)^{2}+n^{2} = n^{2}+4$,

$AD^{2} = (-3+2)^{2}+9 = 10$,

$MD^{2} = (-1+2)^{2} + (n+3)^{2} = n^{2} + 6n + 10$,

When $AM=AD$, $n^{2}+4=10$, solving yields:

$n_{1}=\sqrt{6}$, $n_{2}=-\sqrt{6}$,

thus, the coordinates of point M are $(-1, \sqrt{6})$ or $(-1, -\sqrt{6})$;

When $AM=MD$, $n^{2}+4 = n^{2}+6n+10$, solving yields:

$n = -1$, thus, the coordinates of point M are $(-1, -1)$;

When $AD=DM$, $n^{2}+6n+10=10$, solving yields:

$n_{1}=0$, $n_{2}=-6$;

thus, the coordinates of point P are $(-1,0)$ and $(-1,-6)$,

Let the equation of line AD be $y=kx+b$,

thus, $\left\{\begin{array}{l}
-3k+b=0 \\
-2k+b=-3
\end{array}\right.$, solving yields:

$\left\{\begin{array}{l}
k=-3 \\
b=-9
\end{array}\right.$,

thus, the equation of line AD is $y=-3x-9$,

when $x=-1$, $y=3-9=-6$,

thus, point $(-1, -6)$ is on line AD,

thus, point $(-1, -6)$ does not meet the problem requirement, discard it,

thus, when $\triangle MAD$ is an isosceles triangle, the coordinates of point M are \boxed{(-1, \sqrt{6})} or \boxed{(-1, -\sqrt{6})} or \boxed{(-1, -1)} or \boxed{(-1, 0)}.


Question: As shown in the figure, it is known that the graph of the inverse proportional function $y_1 = \frac{k}{x}$ intersects with the graph of the linear function $y_2 = ax + b$ at points $A(1,4)$ and $B(m,-2)$. If a point $C$ is found on the $x$-axis such that the area of $\triangle ABC$ equals $18$, determine the coordinates of point $C$.

Solution:
\textbf{Solution:} Assume the line AB intersects the x-axis at point D, it is easy to find D (-1, 0).\\
Assume C(x, 0), $\because s_{\triangle ABC}=s_{\triangle ADC}+s_{\triangle BCD}$,\\
$\therefore \frac{1}{2}y_A|x+1|+\frac{1}{2}|y_B||x+1|=18$,\\
$\frac{1}{2}\times 4\times |x+1|+\frac{1}{2}\times 2\times |x+1|=18$\\
$3|x+1|=18$,\\
Solving, we get: $x=5$ or $x=-7$,\\
$\therefore \boxed{C(5, 0)}$ or $\boxed{C(-7, 0)}$.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$, respectively. It intersects with the graph of the inverse proportional function $y=\frac{m}{x}$ at points $C$ and $D$, with $DE\perp x$ axis at point $E$. Given that the coordinates of point $C$ are $(8, -2)$ and $DE=4$, find the expressions for the inverse proportional function and the linear function?

Solution:
Solution: Since point C (8, -2) lies on the graph of the inverse proportion function $y=\frac{m}{x}$,

it follows that $m=-16$,

thus, the equation of the inverse proportion function is $y=-\frac{16}{x}$.

Since point D lies on the graph of $y=-\frac{16}{x}$, and DE=4,

it follows that $y=4$,

thus, point D is (-4, 4).

Also, since C, D lie on the graph of $y=kx+b$,

it follows that $\left\{\begin{array}{l}8k+b=-2 \\ -4k+b=4\end{array}\right.$,

solving this gives $\left\{\begin{array}{l}k=-\frac{1}{2} \\ b=2\end{array}\right.$,

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 $y=kx+b$ intersects the $x$ axis and the $y$ axis at points $A$ and $B$ respectively, and intersects the graph of the hyperbolic function $y=\frac{m}{x}$ at points $C$ and $D$. $DE \perp x$ axis at point $E$. Given that the coordinates of point $C$ are $(8, -2)$, and $DE=4$. Find the area of $\triangle ABO$.

Solution:
\textbf{Solution:} Substituting $y=0$ into $y=-\frac{1}{2}x+2$, we find $x=4$,\\
$\therefore A(4,0)$, $\therefore OA=4$. Substituting $x=0$ into $y=-\frac{1}{2}x+2$, we find $y=2$,\\
$\therefore B(0,2)$, $\therefore OB=2$,\\
$\therefore$ The area of $\triangle ABO$ is $\frac{1}{2}\cdot OA\cdot OB=\frac{1}{2}\times 4\times 2=\boxed{4}$.


Question: As shown in the figure, in the Cartesian coordinate system, the graph of the linear function $y=kx+b$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and intersects the graph of the inverse proportion function $y=\frac{m}{x}$ at points $C$ and $D$. $DE\perp x$-axis at point $E$. Given that the coordinate of point $C$ is $(8, -2)$ and $DE=4$, answer directly based on the graph: For which values of $x$ is the value of the linear function greater than that of the inverse proportion function?

Solution:
\textbf{Solution:} Since a linear function and an inverse proportion function intersect at point D ($-4$, $4$) and point C ($8$, $-2$), \\
it follows that for \boxed{$x<-4$} or \boxed{$0<x<8$}, the value of the linear function is greater than the value of the inverse proportion function.


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

Solution:
\textbf{Solution:} Since the vertex $C$ of the parabola is on the positive half of the x-axis,\\
let the equation of the parabola be $y=a(x-h)^{2}$.\\
Substituting points $A(0, 4)$ and $B(5, 9)$ into $y=a(x-h)^{2}$, we have:\\
$\begin{cases}
a(0-h)^{2}=4\\
a(5-h)^{2}=9
\end{cases}$,\\
Solving this gives $h=-10$ (discard) or $h=2$,\\
thus $a=1$,\\
therefore, the equation of the parabola is: \boxed{y=(x-2)^{2}}.


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

Solution:
\textbf{Solution:} Substituting $y=0$ into $y=(x-2)^{2}$ yields:\\
$(x-2)^{2}=0$,\\
$\therefore x=2$,\\
$\therefore$ the coordinates of point $C$ are $\boxed{\left(2, 0\right)}$;


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

Solution:
\textbf{Solution:} Let the equation of $AB$ be: $y=kx+b$,\\
Substituting points $A\left(0, 4\right)$ and $B\left(5, 9\right)$ into $y=kx+b$ gives:\\
$\begin{cases}
b=4\\
5k+b=9
\end{cases}$,\\
Solving yields: $\begin{cases}
k=1\\
b=4
\end{cases}$,\\
$\therefore$ The equation of $AB$ is: $y=x+4$,\\
$\because$ Point $P$ is a point on segment $AB$, and point $M$ is a point on the parabola $y=(x−2)^{2}$, with $1\le x\le 4$, and $PM\parallel$ to the $y$ axis,\\
$\therefore$ When $x=1$, $P\left(1, 5\right)$, $M\left(1, 5\right)$,\\
$\therefore PM=5−1=4$,\\
When $x=4$, $P\left(4, 8\right)$, $M\left(4, 4\right)$,\\
$\therefore PM=8−4=4$,\\
When $x=2$, $P\left(2, 6\right)$, $M\left(2, 0\right)$,\\
$\therefore PM=6−0=6$,\\
Let $P\left(n, n+4\right)$, $M\left(n, n^{2}−4n+4\right)$,\\
$\therefore PM=n+4−\left(n^{2}−4n+4\right)$\\
$=−n^{2}+5n$\\
$=−\left(n−\frac{5}{2}\right)^{2}+\frac{25}{4}$,\\
$\therefore$ When $n=\frac{5}{2}$, the maximum value of $PM$ is: $\boxed{\frac{25}{4}}$,\\
$\therefore$ The maximum value of $PM$ is $\frac{25}{4}$, and the minimum value is $\boxed{4}$.


Question: As shown in Figure 1, the graph of the quadratic function $y=-\frac{1}{2}x^{2}+bx+c$ intersects the x-axis at points $A$ and $B$, with the coordinates of point $B$ being $(1,0)$, and intersects the y-axis at point $C(0,2)$. Connect $AC$ and $BC$. What is the expression of this quadratic function and the coordinates of point $A$?

Solution:
\textbf{Solution:} Solve by substituting B(1, 0) and C(0, 2) into $y=-\frac{1}{2}x^{2}+bx+c$,\\
$\therefore \left\{\begin{array}{l}
-\frac{1}{2}+b+c=0\\
c=2
\end{array}\right.$,\\
we get $\left\{\begin{array}{l}
b=-\frac{3}{2}\\
c=2
\end{array}\right.$,\\
$\therefore y=-\frac{1}{2}x^{2}-\frac{3}{2}x+2$,\\
Let $y=0$, then $-\frac{1}{2}x^{2}-\frac{3}{2}x+2=0$,\\
we find $x=-4$ or $x=1$,\\
$\therefore A(\boxed{-4},0)$;


Question: As shown in Figure 1, the graph of the quadratic function $y = -\frac{1}{2}x^{2} + bx + c$ intersects the x-axis at points $A$ and $B$, with point $B$ having coordinates $(1, 0)$, and intersects the y-axis at point $C(0, 2)$. Connect points $AC$ and $BC$. Point $P$ is a moving point on the parabola above line segment $AC$.

Solution:
\textbf{Solution:}

(1) Let the equation of line AC be $y=k_{1}x+b_{1}$,\\
$\therefore \left\{\begin{array}{l}
-4k_{1}-b_{1}=0\\
b_{1}=2
\end{array}\right.$,\\
we get $\left\{\begin{array}{l}
k_{1}=\frac{1}{2}\\
b_{1}=2
\end{array}\right.$,\\
$\therefore y=\frac{1}{2}x+2$,\\
Draw $PG\parallel y$-axis from P to intersect AC at point G,\\
Let $P(t, -\frac{1}{2}t^{2}-\frac{3}{2}t+2)$, thus $G(t, \frac{1}{2}t+2)$,\\
$\therefore PG=-\frac{1}{2}t^{2}-\frac{3}{2}t+2-\frac{1}{2}t-2=-\frac{1}{2}t^{2}-2t$,\\
$\therefore S_{\triangle APC}=\frac{1}{2}\times 4\times (-\frac{1}{2}t^{2}-2t)=-t^{2}-4t=3$,\\
Solving we get $t=-1$ or $t=-3$,\\
$\therefore P(-1,3)$ or $P(-3,2)$;\\
(2) Let the equation of line BC be $y=k_{2}x+b_{2}$,\\
$\therefore \left\{\begin{array}{l}
k_{2}+b_{2}=0\\
b_{2}=2
\end{array}\right.$,\\
we get $\left\{\begin{array}{l}
k_{2}=-2\\
b_{2}=2
\end{array}\right.$,\\
$\therefore y=-2x+2$,\\
$\because PD\parallel BC$,\\
Let $P(t, -\frac{1}{2}t^{2}-\frac{3}{2}t+2)$, thus the equation of line PD is $y=-2x-\frac{1}{2}t^{2}+\frac{1}{2}t+2$,\\
$\therefore D(-\frac{1}{4}t^{2}+\frac{1}{4}t+1,0)$,\\
$\therefore PD=\sqrt{\left(\frac{1}{4}t^{2}+\frac{3}{4}t-1\right)^{2}+\left(-\frac{1}{2}t^{2}-\frac{3}{2}t+2\right)^{2}}=-\frac{\sqrt{5}}{4}(t^{2}+3t-4)=-\frac{\sqrt{5}}{4}(t+\frac{3}{2})^{2}+\frac{25\sqrt{5}}{16}$,\\
$\therefore$ when $t=-\frac{3}{2}$, $PD$ achieves its maximum value $\boxed{\frac{25\sqrt{5}}{16}}$.


Question: As shown in the figure, the line $y = -x + 2$ passes through point $A(2, 0)$ on the $x$-axis and intersects the parabola $y = ax^2$ at points $B$ and $C$, where the coordinates of point $B$ are $(1, 1)$. What is the equation of the parabola?

Solution:
\textbf{Solution:} Since point B$(1,1)$ lies on the parabola $y=ax^{2}$,\\
it follows that $1=a$,\\
therefore, the equation of the parabola is $\boxed{y=x^{2}}$.


Question: As shown in the figure, the line $y = -x + 2$ passes through point $A(2, 0)$ on the $x$-axis and intersects the parabola $y = ax^2$ at points $B$ and $C$, with the coordinates of point $B$ being $(1, 1)$. Connect $OC$. What is the area of $\triangle AOC$?

Solution:
\textbf{Solution:} Given the problem, we know the analytical expression for line AB is $y=-x+2$.\\
By solving the system of equations formed by the two functions, $\left\{\begin{array}{l}y=x^{2}\\ y=-x+2\end{array}\right.$,\\
we get: $\left\{\begin{array}{l}x=1\\ y=1\end{array}\right.$ or $\left\{\begin{array}{l}x=-2\\ y=4\end{array}\right.$,\\
$\therefore$ the coordinates of point C are (-2, 4).\\
$\therefore$ the area of $\triangle AOC$ is $\frac{1}{2}\times 2\times 4=\boxed{4}$;


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $y=mx$ intersects with the hyperbola $y=\frac{n}{x}$ at points $A(-1, a)$ and $B$. $BC$ is perpendicular to the $x$-axis, with the foot of the perpendicular being $C$, and the area of $\triangle AOC$ is $1$. What are the values of $m$ and $n$?

Solution:
\textbf{Solution:} Since the line $y=mx$ intersects the hyperbola $y=\frac{n}{x}$ at points A($-1$, $a$) and B,\\
it follows that the x-coordinate of point B is $1$, that is, C($1$, $0$),\\
since the area of $\triangle AOC$ is $1$,\\
it follows that A($-1$, $2$),\\
Substituting A($-1$, $2$) into $y=mx$ and $y=\frac{n}{x}$, we obtain $m=\boxed{-2}$, $n=\boxed{-2}$;


Question: As shown in the graph, in the Cartesian coordinate system $xOy$, the line $y=mx$ intersects the hyperbola $y=\frac{n}{x}$ at points $A(-1, a)$ and $B$, with $BC\perp x$-axis and the foot of the perpendicular being $C$. The area of $\triangle AOC$ is $1$. Find the analytical expression of line $AC$.

Solution:
\textbf{Solution:} Let the analytical expression of line $AC$ be $y=kx+b$,\\
since $y=kx+b$ passes through points $A(-1,2)$ and $C(1,0)$,\\
we have $\begin{cases} -k+b=2 \\ k+b=0 \end{cases}$,\\
solving which gives $k=-1$, $b=1$,\\
therefore, the analytical expression of line $AC$ is $\boxed{y=-x+1}$.


Question: As shown in the figure, if the graph of the linear function $y=-x+5$ intersects with the graph of the reciprocal function $y=\frac{k}{x}$ ($k\ne 0$) at points $A(1, n)$ and $B$ in the first quadrant, what is the equation of the reciprocal function?

Solution:
\textbf{Solution:} Since $A(1, 4)$ lies on the line $y=-x+5$,\\
it follows that $n=-1+5=4$,\\
therefore $A(1, 4)$,\\
substituting $A(1, 4)$ into $y=\frac{k}{x}$ yields $k=4$,\\
therefore, the equation of the inverse proportion function is $\boxed{y=\frac{4}{x}}$.


Question: As shown in the figure, if the graph of the linear function $y=-x+5$ intersects with the graph of the inverse proportion function $y=\frac{k}{x}$ $(k\ne 0)$ at points $A(1, n)$ and $B$ in the first quadrant, what is the area of $\triangle ABO$?

Solution:
\textbf{Solution}: Given $\begin{cases}y=\frac{4}{x} \\ y=-x+5\end{cases}$, we find $\begin{cases}x=1 \\ y=4\end{cases}$ or $\begin{cases}x=4 \\ y=1\end{cases}$, \\
$\therefore B(4, 1)$, \\
The line $y=-x+5$ intersects the $y$-axis at $E(0, 5)$, \\
$\therefore S_{\triangle ABO}=S_{\triangle OBE}-S_{\triangle AOE}=\frac{1}{2}\times 5\times 4-\frac{1}{2}\times 5\times 1=\boxed{7.5}$.


Question: As shown in the figure, the concentration level of students during class can be represented by an attention index. The variation of the attention index $y$ of a class of students during a math class with time $x$ (in minutes) is shown in the figure. At the beginning of the class, the attention index is $30$, and at the 2nd minute, the attention index is $40$. Within the first $10$ minutes, the attention index $y$ is a linear function of time $x$. After $10$ minutes, the attention index $y$ becomes an inverse proportional function of $x$. When $0\leq x\leq 10$, what is the functional relationship between $y$ and $x$?

Solution:
\textbf{Solution:} Solve for when $0 \leq x \leq 10$, let $y=kx+b$.\\
Substituting the points $(0,30)$ and $(2,40)$ into the equation yields:\\
$\left\{\begin{array}{l}
b=30\\
2k+b=40
\end{array}\right.$\\
Solving this system, we find: $k=5$ and $b=30$,\\
Therefore, $y=5x+30=\boxed{5x+30}$.


Question: As shown in the figure, the level of concentration of students during a class can be represented by the attention index. The variation of the attention index $y$ of a class of students during a math class over the class time $x$ (in minutes) is depicted in the figure. At the beginning of the class, the attention index is $30$, and at the $2$nd minute, the attention index is $40$. Within the first $10$ minutes, the attention index $y$ is a linear function of time $x$. After $10$ minutes, the attention index $y$ becomes inversely proportional to $x$. When $10\leq x\leq 40$, what is the functional relationship between $y$ and $x$?

Solution:
\textbf{Solution:} When $10\leq x\leq 40$, let $y=\frac{m}{x}$. Substituting $(10,80)$ into it, we get: $m=800$.\\
Thus $y=\frac{800}{x}$.\\
Therefore, the function relationship of $y$ with respect to $x$ is $\boxed{y=\frac{800}{x}}$.


Question: As shown, the parabola $y=ax^2 + bx - 4a$ ($a\neq 0$) passes through points $A(-1, 0)$ and $C(0, 4)$, and intersects the $x$-axis at another point $B$. Lines $AC$ and $BC$ are drawn. What is the equation of the parabola?

Solution:
\textbf{Solution:} Given the parabola $y=ax^2+bx-4a$ passes through two points,\\
$\therefore \begin{cases} a-b-4a=0 \\ -4a=4 \end{cases}$,\\
$\therefore \begin{cases} a=-1 \\ b=3 \end{cases}$,\\
$\therefore$ the equation of the parabola is $\boxed{y=-x^2+3x+4}$.


Question: As shown in the figure, the parabola $y=ax^{2}+bx-4a$ (where $a\neq 0$) passes through points $A(-1, 0)$ and $C(0, 4)$, and intersects with the x-axis at another point $B$. Lines $AC$ and $BC$ are drawn. A line parallel to the x-axis, $y=-14$, intersects the parabola at points $D$ and $E$. Find the length of the line segment $DE$.

Solution:
\textbf{Solution:} When $y=-14$, $-x^2+3x+4=-14$,\\
rearrange to get: $x^2-3x-18=0$,\\
solving gives $x_1=-3$, $x_2=6$,\\
$\therefore DE=\left|x_1-x_2\right|=\left|-3-6\right|=\boxed{9}$;


Question: As shown in the figure, the parabola $y=ax^2+bx-4a$ (where $a \neq 0$) passes through points $A(-1,0)$ and $C(0,4)$, and intersects the x-axis at another point $B$. Lines $AC$ and $BC$ are drawn. Point $P$ is on the line segment $OB$ (not coinciding with points $B$ or $O$). A line through $P$ perpendicular to the x-axis intersects the parabola at point $M$, and lines $CM$ and $BM$ are drawn. Find the maximum area of $\triangle BCM$ and the coordinates of point $M$ at this condition.

Solution:
\textbf{Solution:} Let $y=0$, then we have $-x^2+3x+4=0$, which simplifies to $x^2-3x-4=0$.\\
Solving this, we get $x=-1$ or $x=4$,\\
$\therefore$ the coordinates of point B are $(4,0)$;\\
Let the equation of line BC be $y=kx+n$,\\
$\begin{cases}
b=4 \\
4k+b=0
\end{cases}$,\\
$\therefore$
$\begin{cases}
k=-1 \\
b=4
\end{cases}$,\\
$\therefore$ the equation of line BC is $y=-x+4$,\\
Let PM intersect segment BC at point E, with M$(m, -m^2+3m+4)$, then E$(m, -m+4)$\\
ME$=-m^2+4m$,\\
$\therefore$ the area of $\triangle BMC=S_{\triangle BME}+S_{\triangle CME}$\\
$=\frac{1}{2}\times 4(-m^2+4m)$\\
$=-2m^2+8m$\\
$=-2(m^2-4m+4)+8$\\
$=-2(m-2)^2+8$ $(0<m<4)$\\
Since $-2<0$,\\
$\therefore$ when $m=2$, $S_{\triangle BMC}$ reaches its maximum value of $\boxed{8}$,\\
$\therefore$ the coordinates of M are $\boxed{(2,6)}$.


Question: As illustrated in the figure, rectangle $OCBA$ is positioned within the Cartesian coordinate system. Point $C$ is located on the positive half of the $x$-axis, and point $A$ is on the positive half of the $y$-axis. It is known that point $B$ has the coordinates $(4,2)$. The graph of the inverse proportion function $y=\frac{k}{x}$ passes through the midpoint $D$ of $AB$ and intersects $BC$ at point $E$. Suppose the equation of line $DE$ is $y=mx+n$. Connect $OD$ and $OE$. What are the expression of the inverse proportion function $y=\frac{k}{x}$ and the coordinates of point $E$?

Solution:
\textbf{Solution:} Given $\because$ quadrilateral OABC is a rectangle, and point B is (4, 2),\\
$\therefore$ AB = 4, BC = 2,\\
$\because$ the midpoint of AB is D,\\
$\therefore$ D is (2, 2),\\
$\because$ the graph of the inverse proportion function $y=\frac{k}{x}$ passes through the midpoint D of AB,\\
$\therefore$ 2 = $\frac{k}{2}$,\\
$\therefore$ k = 4,\\
$\therefore$ the equation of the inverse proportion function is: $y=\frac{4}{x}$;\\
when $x=4$, $y=\frac{4}{4}=1$,\\
$\therefore$ the coordinates of point E are $\boxed{(4, 1)}$;


Question: As shown, rectangle $OCBA$ is positioned in the Cartesian coordinate system, where point $C$ is on the positive half of the $x$-axis, and point $A$ is on the positive half of the $y$-axis. Given that the coordinates of point $B$ are $(4,2)$, the graph of the inverse proportion function $y=\frac{k}{x}$ passes through the midpoint $D$ of $AB$ and intersects $BC$ at point $E$. Let the equation of line $DE$ be $y=mx+n$. Connecting $OD$ and $OE$, directly write out the solution set for the inequality $\frac{k}{x}>mx+n$.

Solution: \textbf{Solution:} The solution set is \boxed{0<x<2} or \boxed{x>4}.


Question: As shown in the figure, rectangle $OCBA$ is located in the Cartesian coordinate system. Point $C$ is on the positive x-axis, and point $A$ is on the positive y-axis. It is known that the coordinates of point $B$ are $(4,2)$. The graph of the inverse proportion function $y=\frac{k}{x}$ passes through the midpoint $D$ of $AB$ and intersects $BC$ at point $E$. Assume the equation of line $DE$ is $y=mx+n$. Connect points $OD$ and $OE$. Point $M$ is on the positive y-axis. If the area of $\triangle MBO$ is equal to the area of $\triangle ODE$, what are the coordinates of point $M$?

Solution:
\textbf{Solution}: Given, $D(2,2)$, $E(4,1)$,\\
$\therefore$ the area of $\triangle ODE$ is $2\times 4-\frac{1}{2}\times 2\times 2-\frac{1}{2}\times 2\times 1-\frac{1}{2}\times 4\times 1=3$,\\
Let $M(0,m)$, and the area of $\triangle MBO$ $=\frac{1}{2}|m|\times 4=3$,\\
$\therefore m=\pm\frac{3}{2}$,\\
$\therefore \boxed{M(0,\frac{3}{2})}$, and $(0,-\frac{3}{2})$ (discard).


Question: It is known that the rectangle $OCBA$ is located as shown in the diagram within the Cartesian coordinate system. Point $C$ is on the positive semi-axis of $x$, and point $A$ is on the positive semi-axis of $y$. It is known that point $B$ has coordinates $(4,2)$, and the graph of the inverse proportion function $y=\frac{k}{x}$ passes through the midpoint $D$ of $AB$ and intersects $BC$ at point $E$. Suppose the equation of line $DE$ is $y=mx+n$. Connect $OD$ and $OE$. Let point $P$ be a point on the $x$-axis, and point $Q$ be a point on the graph of the inverse proportion function $y=\frac{k}{x}$. Are there points $P$ and $Q$ such that the quadrilateral with vertices $P$, $Q$, $D$, and $E$ forms a parallelogram? If such points exist, provide the coordinates of point $Q$ directly; if not, explain why.

Solution: \textbf{Solution:} Yes, there exists. The coordinates of point $Q$ are $\boxed{(-4, -1)}$ or $\boxed{\left(\frac{4}{3}, 3\right)}$.


Question: As shown in the figure, the parabola passes through points \(A(4, 0)\), \(B(1, 0)\), and \(C(0, -2)\). Find the analytical expression of this parabola.

Solution:
\textbf{Solution:} Given that the parabola passes through point $C(0, -2)$,\\
we can assume the equation of the parabola to be $y=ax^2+bx-2$.\\
Substituting points $A(4, 0)$, $B(1, 0)$ into it,\\
we get $\begin{cases}16a+4b-2=0\\ a+b-2=0\end{cases}$,\\
from which we find $\begin{cases}a=-\frac{1}{2}\\ b=\frac{5}{2}\end{cases}$,\\
therefore, the equation of this parabola is $y=-\frac{1}{2}x^2+\frac{5}{2}x-2=\boxed{-\frac{1}{2}x^2+\frac{5}{2}x-2}$.


Question: As shown in the figure, the parabola passes through points $A\left(4, 0\right)$, $B\left(1, 0\right)$, and $C\left(0, -2\right)$. Let $P$ be a moving point on the parabola, and a perpendicular line is drawn from $P$ to the $x$-axis, with the foot of the perpendicular being $M$. Is there a point $P$ on the parabola such that the triangle with vertices $A$, $P$, and $M$ is similar to $\triangle OAC$? If such a point exists, find the coordinates of point $P$; if not, please explain why.

Solution:
\textbf{Solution:} A solution exists. Let the x-coordinate of point $P$ be $m$, then the y-coordinate of $P$ is $-\frac{1}{2}m^{2}+\frac{5}{2}m-2$. When $1<m<4$, $AM=4-m$, $PM=-\frac{1}{2}m^{2}+\frac{5}{2}m-2$.\\
Since $\angle COA=\angle PMA=90^\circ$,\\
Therefore, (1) when $\frac{AM}{PM}=\frac{AO}{OC}$, since $C$ is on the parabola, $OC=2$, $OA=4$, thus $\frac{AM}{PM}=\frac{AO}{OC}=2$, $\triangle APM \sim \triangle ACO$, that is, $4-m=2\left(-\frac{1}{2}m^{2}+\frac{5}{2}m-2\right)$.\\
Solving yields $m_{1}=2$, $m_{2}=4$ (discard), thus $P(2, 1)$.\\
(2) When $\frac{AM}{PM}=\frac{OC}{OA}=\frac{1}{2}$, $\triangle APM \sim \triangle CAO$, that is, $2(4-m)=-\frac{1}{2}m^{2}+\frac{5}{2}m-2$.\\
Solving yields $m_{1}=4$, $m_{2}=5$ (neither fits the conditions, discard)\\
Therefore, when $1<m<4$, $P(2, 1)$, when $m>4$, $AM=m-4$, $PM=\frac{1}{2}m^{2}-\frac{5}{2}m+2$,\\
(1) $\frac{PM}{AM}=\frac{OC}{OA}=\frac{1}{2}$ or (2) $\frac{PM}{AM}=\frac{OA}{OC}=2$,\\
Substituting $P\left(m, -\frac{1}{2}m^{2}+\frac{5}{2}m-2\right)$ yields: $2\left(\frac{1}{2}m^{2}-\frac{5}{2}m+2\right)=m-4$, $2\left(m-4\right)=\frac{1}{2}m^{2}-\frac{5}{2}m+2$,\\
The solutions are: the first equation yields $m=-2-2\sqrt{3}<4$ (discard) $m=-2+2\sqrt{3}<4$ (discard), the second equation yields $m=5$, $m=4$ (discard)\\
Solving for $m=5$, $-\frac{1}{2}m^{2}+\frac{5}{2}m-2=-2$, then $P(5, -2)$, when $m<1$, $AM=4-m$, $PM=\frac{1}{2}m^{2}-\frac{5}{2}m+2$.\\
(1) $\frac{PM}{AM}=\frac{OC}{OA}=\frac{1}{2}$ or $\frac{PM}{AM}=\frac{OA}{OC}=2$,\\
Yields: $2\left(\frac{1}{2}m^{2}-\frac{5}{2}m+2\right)=4-m$, $2\left(4-m\right)=\frac{1}{2}m^{2}-\frac{5}{2}m+2$,\\
The solutions are: the first equation yields $m=0$ (discard), $m=4$ (discard), the second equation yields $m=4$ (discard), $m=-3$,\\
For $m=-3$, $-\frac{1}{2}m^{2}+\frac{5}{2}m-2=-14$, then $P(-3, -14)$,\\
In summary, the points $P$ that meet the conditions are $\boxed{(2, 1)}$, $\boxed{(5, -2)}$, or $\boxed{(-3, -14)}$.


Question: As shown in the figure, the parabola passes through points $A(4, 0)$, $B(1, 0)$, and $C(0, -2)$. There exists a point $D$ on the parabola above line $AC$ for which the area of $\triangle DCA$ is maximized. Determine the coordinates of point $D$.

Solution:
\textbf{Solution:} Let the $x$-coordinate of point $D$ be $t(0<t<4)$, then the $y$-coordinate of point $D$ is $\left|-\frac{1}{2}t^{2}+\frac{5}{2}t-2\right|$.\\
Draw a line parallel to the $y$-axis through point $D$ that intersects $AC$ at point $E$.\\
From the given information, the equation of line $AC$ is $y=\frac{1}{2}x-2$.\\
$\therefore$ The coordinates of point $E$ are $\left(t, \frac{1}{2}t-2\right)$.\\
$\therefore DE=-\frac{1}{2}t^{2}+\frac{5}{2}t-2-\left(\frac{1}{2}t-2\right)=-\frac{1}{2}t^{2}+2t$,\\
$\therefore S_{\triangle DAC}=S_{\triangle DCE}+S_{\triangle DEA}=\frac{1}{2}DE\cdot h+\frac{1}{2}DE\cdot (4-h)=\frac{1}{2}DE\cdot 4$,\\
$\therefore S_{\triangle DAC}=\frac{1}{2}\times \left(-\frac{1}{2}t^{2}+2t\right)\times 4=-t^{2}+4t=-(t-2)^{2}+4$,\\
$\therefore$ When $t=2$, the area of $\triangle DAC$ is maximized,\\
$\therefore D\left(2, 1\right)$. Hence, the coordinates of point $D$ are $\boxed{(2, 1)}$.


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

Solution:
\textbf{Solution:} Since the vertex $C$ of the parabola is on the positive x-axis, \\
let the equation of the parabola be $y=a(x-h)^{2}$. \\
Substituting points $A(0, 4)$ and $B(5, 9)$ into $y=a(x-h)^{2}$, we get: $\left\{\begin{array}{l}a(0-h)^{2}=4 \\ a(5-h)^{2}=9\end{array}\right.$, \\
Solving this yields: $h=-10$ (discard) or $h=2$, \\
thus $a=1$, \\
Therefore, the equation of the parabola is: $\boxed{y=(x-2)^{2}}$;


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

Solution:
\textbf{Solution:} Substitute $y=0$ into $y=(x-2)^{2}$ to obtain $(x-2)^{2}=0$,\\
$\therefore x=2$,\\
$\therefore$ the coordinates of point $C$ are $\boxed{(2, 0)}$;


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

Solution:
\textbf{Solution:} Let the equation of line $AB$ be: $y=kx+b$, \\
Substituting points $A\left(0, 4\right)$ and $B\left(5, 9\right)$ into $y=kx+b$ yields: $\left\{\begin{array}{l}b=4 \\ 5k+b=9\end{array}\right.$, \\
Solving this, we get: $\left\{\begin{array}{l}k=1 \\ b=4\end{array}\right.$, \\
Therefore, the equation of line $AB$ is: $y=x+4$, \\
Since point $P$ is on the line segment $AB$ and point $M$ is on the parabola $y=\left(x−2\right)^{2}$, with $1\le x\le 4$, and $PM\parallel y$ axis, \\
Therefore, when $x=1$, $P\left(1, 5\right)$, $M\left(1, 5\right)$, \\
Thus, $PM=5−1=4$, \\
When $x=4$, $P\left(4, 8\right)$, $M\left(4, 4\right)$, \\
Thus, $PM=8−4=4$, \\
When $x=2$, $P\left(2, 6\right)$, $M\left(2, 0\right)$, \\
Thus, $PM=6−0=6$, \\
Let $P\left(n, n+4\right)$, $M\left(n, n^{2}−4n+4\right)$, \\
Thus, $PM=n+4−\left(n^{2}−4n+4\right)=−n^{2}+5n=−\left(n−\frac{5}{2}\right)^{2}+\frac{25}{4}$, \\
Therefore, when $n=\frac{5}{2}$, the maximum value of $PM$ is: $\boxed{\frac{25}{4}}$, \\
Therefore, the maximum value of $PM$ is $\frac{25}{4}$, and the minimum value is $\boxed{4}$.


Question: As shown in the figure, in the Cartesian coordinate system, it is known that the parabola $y=x^2-2x-3$ has its vertex at point $A$, intersects the y-axis at point $C$, and the line segment $CB$ is parallel to the x-axis, intersecting the parabola at another point $B$. Find the coordinates of point $B$ and the equation of the line $AC$.

Solution:
Given the problem, we have $y=x^{2}-2x-3=(x-1)^{2}-4$, \\
$\therefore$ the vertex $A(1, -4)$, \\
Let $x=0$, then $y=-3$, \\
$\therefore C(0, -3)$, \\
$\because CB\parallel x$-axis, \\
$\therefore B(2, -3)$, \\
Let the equation of line $AC$ be $y=kx+b$, \\
$\left\{\begin{array}{l}
k+b=-4 \\
b=-3
\end{array}\right.$, \\
Solving yields $\left\{\begin{array}{l}
k=-1 \\
b=-3
\end{array}\right.$, \\
$\therefore y=-x-3$.

The coordinates of point $B$ are $B=\boxed{(2, -3)}$, the equation of line $AC$ is $\boxed{y=-x-3}$.


Question: As shown in the figure, in the Cartesian coordinate system, it is known that the parabola $y=x^2-2x-3$ has its vertex at point $A$, intersects the $y$-axis at point $C$, and the line segment $CB\parallel x$-axis intersects the parabola at another point $B$. When the independent variable $x$ of the quadratic function $y=x^2-2x-3$ satisfies $m\le x\le m+2$, the maximum value of this function is $p$, the minimum value is $q$, and $p-q=2$. What is the value of $m$?

Solution:
\textbf{Solution:} Given the problem, the axis of symmetry of the parabola $y=x^2-2x-3$ is the line $x=1$,

(1) When $m>1$, \\
at $x=m$, we have $q=m^2-2m-3$, \\
at $x=m+2$, we have $p=(m+2)^2-2(m+2)-3$, \\
$\therefore p-q=(m+2)^2-2(m+2)-3-m^2+2m+3=2$, \\
solving this gives $m=\frac{1}{2}$ (discard);

(2) When $m+2<1$, i.e., $m<-1$, \\
at $x=m$, we have $p=m^2-2m-3$, \\
at $x=m+2$, we have $q=(m+2)^2-2(m+2)-3$, \\
$\therefore p-q=m^2-2m-3-(m+2)^2+2(m+2)+3=2$, \\
solving this gives $m=-\frac{1}{2}$ (discard);

(3) When $m\le 1\le m+1$, i.e., $0\le m\le 1$, \\
at $x=1$, we have $q=-4$, \\
at $x=m+2$, we have $p=(m+2)^2-2(m+2)-3$, \\
$\therefore p-q=(m+2)^2-2(m+2)-3+4=2$, \\
solving this gives $m=\sqrt{2}-1$ or $m=-\sqrt{2}-1$ (discard);

(4) When $m+1<1\le m+2$, i.e., $-1\le m<0$, \\
at $x=1$, we have $q=-4$, \\
at $x=m$, we have $p=m^2-2m-3$, \\
$\therefore p-q=m^2-2m-3+4=2$, \\
solving this gives $m=\sqrt{2}+1$ (discard) or $m=-\sqrt{2}+1$,

To conclude: The values for $m$ are $m=\boxed{\sqrt{2}-1}$ or $m=\boxed{1-\sqrt{2}}$.


Question: As shown in the figure, in the Cartesian coordinate system, it is known that the parabola $y=x^2-2x-3$ has its vertex at point $A$, intersects the $y$-axis at point $C$, and the segment $CB\parallel x$-axis intersects the parabola at another point $B$. The parabola $y=x^2-2x-3$ is translated so that its vertex always moves along the line $AC$. When the translated parabola has only one common point with the ray $BA$, let the $x$-coordinate of the vertex of the parabola at this time be $n$. Please write down the range of values of $n$ directly.

Solution: \textbf{Solution:} The range of values for $n$ is $1<n\le 4$ or $n=\boxed{\frac{7}{8}}$.


Question: As shown in the figure, the parabola $y=-\frac{1}{2}x^2+bx+c$ intersects the $x$-axis at points $A$ and $B(4, 0)$, and intersects the $y$-axis at point $C$. The line $y=-\frac{1}{2}x+n$ passes through points $B$ and $C$. Point $D$ is a moving point on line $BC$. A perpendicular line $DQ$ is drawn from point $D$ to the $x$-axis, with the foot of the perpendicular being $Q$, intersecting the parabola at point $P$. Find the equation of the parabola and the coordinates of point $A$.

Solution:
\textbf{Solution:} Substitute $B(4, 0)$ into $y=-\frac{1}{2}x+2$,\\
$\therefore -2+2=0$,\\
we find $n=2$,\\
$\therefore y=-\frac{1}{2}x+2$,\\
let $x=0$, then $y=2$,\\
$\therefore C(0, 2)$,\\
Substituting $B(4, 0)$ and $C(0, 2)$ into $y=-\frac{1}{2}x^2+\frac{3}{2}x+2$,\\
$\therefore \left\{\begin{array}{l}c=2 \\ -8+4b+c=0\end{array}\right.$,\\
we find $\left\{\begin{array}{l}c=2 \\ b=\frac{3}{2}\end{array}\right.$,\\
$\therefore y=-\frac{1}{2}x^2+\frac{3}{2}x+2$,\\
let $y=0$, then $-\frac{1}{2}x^2+\frac{3}{2}x+2=0$,\\
we find $x=-1$ or $x=4$,\\
$\therefore \boxed{A\left(-1, 0\right)}$.


Question: As shown in the figure, the parabola $y=-\frac{1}{2}x^2+bx+c$ intersects the x-axis at points $A$ and $B(4, 0)$, and intersects the y-axis at point $C$. The line $y=-\frac{1}{2}x+n$ passes through points $B$ and $C$. Point $D$ is a moving point on line $BC$. A perpendicular line $DQ$ is drawn from point $D$ to the x-axis with $Q$ as the foot of the perpendicular, and it intersects the parabola at point $P$. When point $P$ is located above line $BC$ and the area of $\triangle PBC$ is the maximum, what are the coordinates of $P$?

Solution:
\textbf{Solution:} Let $P\left(t, -\frac{1}{2}t^{2}+\frac{3}{2}t+2\right)$. Then $D\left(t, -\frac{1}{2}+2\right)$, \\
$\therefore PD=-\frac{1}{2}t^{2}+\frac{3}{2}t+2+\frac{1}{2}t-2=-\frac{1}{2}t^{2}+2t$, \\
$\therefore S_{\triangle PBC}=\frac{1}{2}\times 4\times \left(-\frac{1}{2}t^{2}+2t\right)=-t^{2}+4t=-(t-2)^{2}+4$, \\
$\therefore$ when $t=2$, the area of $\triangle PBC$ is maximized, \\
At this time $P(2, 3)$, hence the coordinates of $P$ are $\boxed{(2, 3)}$.


Question: As shown in the figure, the parabola $y=-\frac{1}{2}x^2+bx+c$ intersects the $x$-axis at points $A$ and $B(4, 0)$, and intersects the $y$-axis at point $C$. The line $y=-\frac{1}{2}x+n$ passes through points $B$ and $C$. Point $D$ is a movable point on line $BC$. A perpendicular line $DQ$ from point $D$ to the $x$-axis is drawn, with $Q$ being the foot of the perpendicular and intersecting the parabola at point $P$. When point $D$ is moved 5 units to the right to arrive at point $E$, and segment $DE$ intersects with the parabola at only one point, please directly write down the range of values for the $x$-coordinate $m$ of point $D$.

Solution: \textbf{Solution:} The range of the x-coordinate $m$ of point $D$ is $m=-\frac{9}{4}$ or $0<m\leq 4$. Therefore, the answer is $m=\boxed{-\frac{9}{4}}$ or $\boxed{0<m\leq 4}$.


Question: As shown in the diagram, the parabola $y=ax^2+bx+6$ passes through points $A(-2, 0)$ and $B(4, 0)$, and intersects the $y$-axis at point $C$. Point $D$ is a moving point on the parabola with the $x$-coordinate $m(1<m<4)$. Connect $AC$, $BC$, $DB$, and $DC$. Find the equation of the parabola.

Solution:
\textbf{Solution:}
Given the equation of the intersection of the parabola as: $y=a(x+2)(x-4)=a(x^2-2x-8)=ax^2-2ax-8a$,\\
we have $-8a=6$. Solving this gives: $a=-\frac{3}{4}$,\\
thus the equation of the parabola is: $y=-\frac{3}{4}x^2+\frac{3}{2}x+6=\boxed{-\frac{3}{4}x^2+\frac{3}{2}x+6}$.


Question: As shown in the figure, the parabola $y=ax^2+bx+6$ passes through points $A(-2, 0)$, $B(4, 0)$ and intersects the $y$-axis at point $C$. Let point $D$ be a moving point on the parabola with the $x$-coordinate of $m(1<m<4)$. Connect $AC$, $BC$, $DB$, and $DC$. When the area of $\triangle BCD$ is $\frac{3}{4}$ of the area of $\triangle AOC$, what is the value of $m$?

Solution:
\textbf{Solution:} Given the expression for the parabola, point $C(0, 6)$ is known,\\
from the coordinates of points $B$ and $C$, the equation of line $BC$ can be obtained: $y=-\frac{3}{2}x+6$,\\
as shown in the figure, a line parallel to the $y$-axis through point $D$ intersects line $BC$ at point $H$,\\
let point $D$ be $(m, -\frac{3}{4}m^2+\frac{3}{2}m+6)$, then point $H$ is $(m, -\frac{3}{2}m+6)$,\\
thus the area of $\triangle BDC$ is $\frac{1}{2}HD\times OB=2\left(-\frac{3}{4}m^2+\frac{3}{2}m+6+\frac{3}{2}m-6\right)=2\left(-\frac{3}{4}m^2+3m\right)$,\\
$\therefore \frac{3}{4}{S}_{\triangle ACO}=\frac{3}{4}\times \frac{1}{2}\times 6\times 2=\frac{9}{2}$,\\
i.e., $2\left(-\frac{3}{4}m^2+3m\right)=\frac{9}{2}$,\\
Solving yields: $m=1$ or $3$ (discard $1$),\\
hence $m=\boxed{3}$;