Answer
a) The maximum height attained by the shot is $2\text{ft}$ and the distance at which it occurs is $\text{9}\text{.2ft}$
b) The ball travels $\underline{\text{5}\text{.06 feet}}$ before hitting the ground.
c) The graph is shown below.
Work Step by Step
(a)
We have to compare the given equation $f\left( x \right)=-0.8{{x}^{2}}+3.2x+6$ with the standard equation $f\left( x \right)=a{{x}^{2}}+bx+c.$ , get:
$a=-0.8,\text{ }b=3.2\text{, }c=6$.
Because $a<0,$ the maximum height will occur at $x=\frac{-b}{2a}$.
$\begin{align}
& x=\frac{-3.2}{2\left( -0.8 \right)} \\
& =\frac{-3.2}{-1.6} \\
& =2
\end{align}$
So, the maximum height of the ball occurs at 2 feet from the thrower’s point of impact.
The distance travelled of the shot is
$\begin{align}
& f\left( 3 \right)=-0.8{{\left( 2 \right)}^{2}}+3.2\left( 2 \right)+6 \\
& =9.2
\end{align}$
Thus, the maximum height of the ball is $2\text{ft}$ from the point of release and its distance is $\text{9}\text{.2ft}$.
(b)
Since, the distance travelled by the ball just before hitting the ground corresponds to the maximum distance travelled by it, which is determined by calculating the x-intercept.
To find the x-intercept, replace $f\left( x \right)\text{ by 0}$ and solve the quadratic equation obtained.
$0=-0.8{{x}^{2}}+3.2x+6$
And solve the equation using quadratic formula.
$\begin{align}
& x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
& =\frac{-3.2\pm \sqrt{{{\left( 3.2 \right)}^{2}}-4\left( -0.8 \right)\left( 6 \right)}}{2\left( -0.8 \right)} \\
& =\frac{-3.2+\pm \sqrt{24.96}}{-1.6} \\
& =\frac{-3.2\pm 4.9}{-1.6}
\end{align}$
Get, $x=5.06\text{ or }x=-1.06$
As $x\ge 0,$ the horizontal distance of the ball will be $x=5.06\text{ feet}$.
The distance traveled by the ball just before hitting the ground is $\underline{\text{5}\text{.06 feet}}$.
(c)
We have the given equation $f\left( x \right)=-0.8{{x}^{2}}+3.2x+6$.
Compare to the standard equation $f\left( x \right)=a{{x}^{2}}+bx+c.$ ,one will get:
$a=-0.8,\text{ }b=3.2\text{, and }c=6$.
Since $a<0$ , the parabola opens downwards, and its vertex is given by $\left( \frac{-b}{2a},f\left( \frac{-b}{2a} \right) \right)$.
Calculate the vertex:
$\begin{align}
& x=\frac{-b}{2a} \\
& x=\frac{-3.2}{-1.6} \\
& x=2
\end{align}$
And
$\begin{align}
& y=f\left( 2 \right) \\
& y=-0.8\left( 4 \right)+3.2\left( 2 \right)+6 \\
& y=9.2
\end{align}$
Thus, the vertex is at $\left( 2,9.2 \right)$.
Plot the graph.
