Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises - Page 182: 84

Answer

The lines intersect at $(0,-0.25)$.

Work Step by Step

The only perpendicular lines that can be tangent to a curve symmetrical over the y-axis are lines with slopes of $1$ and $-1$. Knowing this, we can find the equations of these lines by first equating 1 or -1 to the derivative of $x^2$ which is $2x$. This results in $\frac{1}{2}$ or $-\frac{1}{2}$, respectively. Plugging these numbers into the equation $x^2$ outputs $\frac{1}{4}$ or $-\frac{1}{4}$, respectively. Now using point-slope form, we can derive the equation $y-\frac{1}{4}=1(x-\frac{1}{2})$ or $y+\frac{1}{4}=-1(x+\frac{1}{2})$ which simplified will result in $y=x-0.25$ and $y=-x-0.25$, respectively. These lines intersect at the point $(0,-0.25)$.
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