University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 203: 102


The $x-$intercept is $(-4/3,0)$ and the $y-$intercept is $(0,16)$.

Work Step by Step

$$y=x^3$$ 1) First, find the derivative of $y$: $$y'=3x^2$$ We have $y'(-2)=3\times(-2)^2=3\times4=12$ 2) At the point $(-2,-8)$, the tangent to the curve has the following equation: $$y=y'(-2)x+y_0$$ $$y=12x+y_0$$ To find $y_0$, we replace $x$ and $y$ with $-2$ and $-8$ respectively: $$-8=12\times(-2)+y_0$$ $$-8=-24+y_0$$ $$y_0=16$$ Therefore, the tangent to the curve at $(-2,-8)$ is $$y=12x+16$$ 3) Find the $x-$intercept: for $y=0$: $$12x+16=0$$ $$x=-\frac{16}{12}=-\frac{4}{3}$$ Find the $y-$intercept: for $x=0$: $$y=12\times0+16=16$$ So the $x-$intercept is $(-4/3,0)$ and the $y-$intercept is $(0,16)$.
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