Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.2 - Quadratic Functions - Exercise Set - Page 331: 60

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.
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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.
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