Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 2 - Nonlinear Functions - 2.2 Quadratic Functions; Translation and Reflection - 2.2 Exercises - Page 64: 20

Answer

$$ f(x)=\frac{1}{2}x^{2}+6x+24 $$ There are no $x$-intercepts. The $y$-intercept is 24. Vertex parabola is $\left(-6 , 6\right)$. The axis is $ x=-6 $ , the vertical line through the vertex. Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(-6 , 6\right)$. gives the graph in Figure

Work Step by Step

$$ f(x)=\frac{1}{2}x^{2}+6x+24 $$ The $x$-intercepts can be found by letting $f(x)=0$ to get $$ f(x)=\frac{1}{2}x^{2}+6x+24=x^2+12x+48=0 $$ This does not appear to factor, so we’ll try the quadratic formula. $$ \begin{split} \:x_{1,\:2} &=\frac{-12\pm \sqrt{12^2-4\cdot \:1\cdot \:48}}{2\cdot \:1} \\ &=\frac{-12\pm \sqrt{-48}}{2} \end{split} $$ Since the radical is negative, then there are no $x$-intercepts. To find the $y$-intercept , set $x=0 $ $$ f(x)=\frac{1}{2}(0)^{2}+6(0)+24=24. $$ So the $y$-intercept is 24. The $x$-coordinate of the vertex is : $$ x=\frac{-b}{2a}=-\frac{-6}{2.\frac{1}{2}}=\frac{-6}{1}=-6 $$ Substituting this into the equation gives $$ f(x)=\frac{1}{2}(-6)^{2}+6.(-6)+24=18-36+24=6 $$ The vertex is $\left(-6 , 6\right)$. The axis is $x=-6 $ , the vertical line through the vertex. Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(-6 , 6\right)$. gives the graph in Figure
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