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This quadratic equation will now be solved by graphing. Before doing so, it should be rewritten in standard form. Now the function related to the equation can be written. Again, the solutions to the equation are the $x\text{-}$intercepts of the parabola. Identify the characteristics of the function to be able to draw it.

$g(t)=\text{-}16t^2+112t-196$
Direction	Vertex	Axis of Symmetry	$y\text{-}$intercept
$$\begin{array}{c}a<0\\ \Downarrow \\ \text{downward}\end{array}$$	$(3.5,0)$	$x=3.5$	$(0,\text{-}196)$

The reflection of the $y\text{-}$intercept across the axis of symmetry is at $(7,\text{-}196),$ which is also on the parabola. Now, the graph can be drawn.

This time, there is only one $x\text{-}$intercept.

The point of intersection is $(3.5,0).$ Therefore, the equation $\text{-}16t^2+112t-196=0$ has one solution, $t=3.5.$ This solution means that the rocket is at a height of $196$ meters after $3.5$ seconds.

The area of the square is the square of $0.5x,$ and the area of the triangle is half the product of $2$ and $0.5x+2.$

Area of Square	Area of Triangle
$(0.5x{)}^2$	$\frac{1}{2}(2)(0.5x+2)$
$0.25x^2$	$0.5x+2$

Since the geometric shapes have the same area, equating these expressions will produce a quadratic equation. To write it in standard form, the Properties of Equality will be used. This quadratic equation will now be solved by graphing. To do so, the quadratic function related to the equation will be written. Next, the graph of the related function will be drawn. Identify its characteristics.

$y=0.25x^2-0.5x-2$
Direction	Vertex	Axis of Symmetry	$y\text{-}$intercept
$$\begin{array}{c}a>0\\ \Downarrow \\ \text{upward}\end{array}$$	$(1,\text{-}2.25)$	$x=1$	$(0,\text{-}2)$

The reflection of the $y\text{-}$intercept across the axis of symmetry $(2,\text{-}2)$ is also on the parabola. With this information, the graph can be drawn.

From the graph, the $x\text{-}$intercepts of the function can be identified.

The parabola intersects the $x\text{-}$axis twice. The points of intersection are $(\text{-}2,0)$ and $(4,0).$ Therefore, the quadratic equation has two solutions, $x_1=\text{-}2$ and $x_2=4.$ Note that $x$ cannot be negative because $0.5x$ represents the side length of the square. Therefore, although $x=\text{-}2$ is a solution to the equation, only the solution $x=4$ makes sense in this context.

Area of the Square	Area of the Triangle
$f(x)=(0.5x{)}^2$	$g(x)=\frac{1}{2}\cdot 2\cdot (0.5x+2)$
$f(x)=0.25x^2$	$g(x)=0.5x+2$

To draw the graphs of these functions, a table of values will be made.

$x$	$f(x)=0.25x^2$	$g(x)=0.5x+2$
$\text{-}3$	$0.25(\text{-}3{)}^2=2.25$	$0.5(\text{-}3)+2=0.5$
$0$	$0.25(0{)}^2=0$	$0.5(0)+2=2$
$3$	$0.25(3{)}^2=2.25$	$0.5(3)+2=3.5$

Now, plot the points $(\text{-}3,2.25),$ $(0,0),$ and $(3,2.25)$ to graph $f(x).$ Similarly, plot the points $(\text{-}3,0.5),$ $(0,2),$ and $(3,3.5)$ to graph $g(x).$ The graphs of the functions intersect at $(\text{-}2,1)$ and $(4,4).$ These points mean that the geometric shapes have the same area when $x=\text{-}2$ or $x=4.$ However, since $x=\text{-}2$ makes the side length of the square negative, it should be discarded. Therefore, when $x=4,$ both figures have the same area, which is $4$ square units.

This resulting quadratic equation needs to be written in standard form. The solutions to this quadratic equation will give the horizontal distance to the collision point. To be able to graph the function, some of its characteristics should be determined.

$y=\text{-}0.7x^2+0.8x+8$
Direction	Vertex	Axis of Symmetry	$y\text{-}$intercept
$$\begin{array}{c}a<0\\ \Downarrow \\ \text{downward}\end{array}$$	$\left(\frac{4}{7},\frac{288}{35}\right)$	$x=\frac{4}{7}$	$(0,8)$

The reflection of the $y\text{-}$intercept across the axis of symmetry $\left(\frac{8}{7},8\right)$ is also on the parabola. Since these points are very close to each other, another point can be plotted by randomly choosing an $x\text{-}$value and calculating its corresponding $y\text{-}$value. Try $x=\text{-}3.$ Therefore, the point $(\text{-}3,\text{-}0.7)$ lies on the curve. The reflection of this point across the axis of symmetry is $\left(\frac{29}{7},\text{-}0.7\right),$ which also lies on the parabola. With this information, the graph can be drawn.

The $x\text{-}$intercepts of the graph can now be identified.

The parabola intersects the $x\text{-}$axis twice. From the graph only one of the intercepts is identified easily, which is $(4,0).$ The other intercept is between $\text{-}3$ and $\text{-}2.$ However, this intercept can be ignored because $x$ represents horizontal distance and distance cannot be negative. Therefore, in this context, only $x=4$ makes sense. This solution means that the horizontal distance from where Tiffaniqua and Ramsha stand to the point of collision is $4$ meters. To find how far Ramsha is to the collision point, the Distance Formula will be used. To do so, the $y\text{-}$coordinate of the collision point needs to be found. To accomplish this, substitute $4$ into the either of the quadratic functions, and evaluate. Therefore, the arrows collide at the point $(4,8).$ Now the distance from Ramsha to the point where the arrows collide can be found. Ramsha is $4\sqrt{5}$ meters away from the collision point of the arrows.