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: 75

Answer

The equation of the parabola is $$y=2x^2-x$$

Work Step by Step

$$y=f(x)=ax^2+bx$$ - Derivative of $f(x)$: $$f'(x)=2ax+b$$ 1) $f'(x)$ is the slope of the tangent line to the parabola at point $A(x, y)$. So, since the tangent line to the parabola at point $(1,1)$ has the equation: $$y=3x-2$$ we find that $f'(1)=3$, or in other words, $$2a\times1+b=3$$ $$2a+b=3\hspace{1cm}(1)$$ 2) There is a tangent line to the parabola at point $(1,1)$. That means point $(1,1)$ must lie in the parabola. Therefore, $$a\times1^2+b\times1=1$$ $$a+b=1\hspace{1cm}(2)$$ Subtract (2) from (1), we have $$(2a-a)+(b-b)=3-1$$ $$a=2$$ So, $$b=1-a=1-2=-1$$ Therefore, the equation of the parabola is $$y=2x^2-x$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.