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: 70


The equation of the parabola is $$y=f(x)=3x^2-2x+7$$

Work Step by Step

$$y=f(x)=ax^2+bx+c$$ - The derivative of $f(x)$ would be $$f'(x)=2ax+b$$ 1) The slope of the tangent line of $f(x)$ at $A(x,y)$ would equal $f'(x)$ According to the question: - $f(x)$ has slope 4 at $x=1$. That means $f'(1)=4$. So, $$2a\times1+b=4$$ $$2a+b=4\hspace{1cm}(1)$$ - $f(x)$ has slope -8 at $x=-1$. That means $f'(-1)=-8$. So, $$2a\times(-1)+b=-8$$ $$-2a+b=-8$$ $$2a-b=8\hspace{1cm}(2)$$ Summing (1) and (2), we have $$4a=12$$ $$a=3$$ Therefore, $b=2a-8=2\times3-8=-2$ So, $y=f(x)=3x^2-2x+c$ 2) The parabola passes through the point $(2,15)$. Therefore, $$3\times2^2-2\times2+c=15$$ $$8+c=15$$ $$c=7$$ Therefore, the equation of the parabola is $$y=f(x)=3x^2-2x+7$$
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