Answer
a) The maximum height is $1.5\text{ ft}$ and the required point is $7.8\text{ ft}$.
b) The maximum distance would be the positive horizontal intercept of the function, that is $4.61\text{ ft}$.
c) The graph is shown below.
Work Step by Step
(a)
The maximum height of the shot can be determined by the y-coordinate of the vertex and the distance where it reaches its maximum from its point of release can be determined by the x-coordinate of the vertex.
Let us consider the quadratic function of the form
$f\left( x \right)=a{{x}^{2}}+bx+c$
It can be converted into its standard form as
$f\left( x \right)=a{{\left( x+\frac{b}{2a} \right)}^{2}}+c-\frac{{{b}^{2}}}{4a}$
The quadratic function in its standard form can be written as
$f\left( x \right)=a{{\left( x-h \right)}^{2}}+k,\,\,\,\,\,\,\,\,a\ne 0$
Here, $a,h,k$ are constants and $x$ is a variable.
The graph of $f\left( x \right)$ is a parabola which is symmetric about the line $x=h$.
The coordinates of the vertex of the parabola are $\left( h,k \right)$.
Now, compare this with the standard form to get $h=-\frac{b}{2a}$ and $k=c-\frac{{{b}^{2}}}{4a}$.
The coordinates of the vertex of the parabola can also be written as $\left( -\frac{b}{2a},f\left( -\frac{b}{2a} \right) \right)$.
Consider the given function $f\left( x \right)=-0.8{{x}^{2}}+2.4x+6$.
So, compare with the quadratic function $f\left( x \right)=a{{x}^{2}}+bx+c$ to get $a=-0.8,b=2.4$.
The $x\text{-}$ coordinate of vertex is
$\begin{align}
& h=-\frac{b}{2a} \\
& =-\left( \frac{2.4}{-1.6} \right) \\
& =1.5
\end{align}$
And calculate $y\text{-}$ coordinate by determining $f\left( 1.5 \right)$. That is,
$\begin{align}
& f\left( 1.5 \right)=-0.8{{\left( 1.5 \right)}^{2}}+2.4\left( 1.5 \right)+6 \\
& =7.8
\end{align}$
The coordinates of the vertex are $\left( 1.5,7.8 \right)$.
Hence, the maximum height of the shot would be 7.8ft and it will at a distance of 1.5ft from its point of release.
(b)
The shot’s maximum horizontal distance can be found by the horizontal intercepts of the function. Thus, the maximum distance would be the positive horizontal intercept of the function.
Find the $x\text{-}$ intercepts by solve $f\left( x \right)=0$. That is, $-0.8{{x}^{2}}+2.4x+6=0$.
Find the roots of the equation use discriminant formula. The roots of a quadratic equation of the type $a{{x}^{2}}+bx+c=0$ are given by
$\alpha ,\beta =\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
And, the roots of the given equation are as follows:
$\begin{align}
& \alpha ,\beta =\frac{-2.4\pm \sqrt{5.76-4\left( -0.8 \right)\left( 6 \right)}}{-1.6} \\
& =\frac{-2.4\pm 4.99}{-1.6}
\end{align}$
Thus, the $x\text{-}$ intercepts would be $\left( -1.61,0 \right),\left( 4.61,0 \right)$.
The maximum distance would be the positive horizontal intercept of the function, which is $4.61\text{ ft}$.
Therefore, the maximum distance would be the positive horizontal intercept of the function, which is $4.61\text{ ft}$.
(c)
The coordinates of the vertex are $\left( 1.5,7.8 \right)$. The $x\text{-}$ intercepts would be $\left( -1.61,0 \right),\left( 4.61,0 \right)$.
Find $y\text{-}$ intercepts by solving $f\left( 0 \right)$. That is,
$\begin{align}
& f\left( 0 \right)=-0.08{{\left( 0 \right)}^{2}}+2.4\left( 0 \right)+6 \\
& =6
\end{align}$
So, $y\text{-}$ intercept is $\left( 0,6 \right)$.
The graph of the function is shown above.
