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: 58

Answer

a) The maximum height attained by the shot is $\text{26}\text{.25 feet}$, and the distance at which it occurs is $\text{33}\text{.70 feet}$. b) The maximum horizontal distance is $\underline{\text{55}\text{.25 feet}}$ c) The shot was released from $\underline{\text{6}\text{.1 feet}}$.

Work Step by Step

(a) Compare the equation$f\left( x \right)=-0.04{{x}^{2}}+2.1x+6.1$ to the standard equation and get: $a=-0.04,\text{ }b=2.1,\ \text{and }c=6.1$. Since, $a<0,$ the maximum height will occur at $x=\frac{-b}{2a}$. $\begin{align} & x=\frac{-2.1}{2\left( -0.04 \right)} \\ & =\frac{-2.1}{-0.08} \\ & =26.25 \end{align}$ So, the maximum height of the shot occurs at 26.25 feet from the point of release. The distance traveled of the shot is: $\begin{align} & f\left( 26.25 \right)=-0.04{{\left( 26.25 \right)}^{2}}+2.1\left( 26.25 \right)+6.1 \\ & =33.70 \end{align}$ Therefore, the maximum height of the shot occurs at $\text{26}\text{.25 feet}$ from the point of release and the maximum height is $\text{33}\text{.65 feet}$. (b) The maximum horizontal displacement 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. $0=-0.04{{x}^{2}}+2.1x+6.1$ And solving the equation using the quadratic formula, we get, $\begin{align} & x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ & =\frac{-2.1\pm \sqrt{{{\left( 2.1 \right)}^{2}}-4\left( -0.04 \right)\left( 6.1 \right)}}{2\left( -0.04 \right)} \\ & =\frac{-2.1\pm \sqrt{5.386}}{-0.08} \\ & =\frac{-2.1\pm 2.3}{-0.08} \end{align}$ Now, the value thus obtained is $x=55.25\text{ or }x=-2.5$ As $x\ge 0,$ the horizontal distance of the shot will be $x=55.25\text{ feet}$. Hence, the maximum horizontal distance is $55.25\text{ feet}$. (c) The height from which the shot is released is given by its y-intercept, that is $f\left( 0 \right)$. So, $f\left( 0 \right)=-0.04{{\left( 0 \right)}^{2}}+2.1\left( 0 \right)+6.1$feet Thus, the height of the shot will be $6.1\text{ feet}$. Hence, the height at which the shot was released is $\text{6}\text{.1 feet}$.
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