University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.3 - Differentiation Rules - Exercises - Page 136: 59


$a=1$, $b=1$ and $c=0$.

Work Step by Step

$$y=f(x)=ax^2+bx+c$$ The derivative of $f(x)$ is $$f'(x)=2ax+b$$ Since the curve $f(x)$ is tangent to $y=x$ at the origin, $f'(0)$ is equal to the slope of $y=x$, which is $1$: $$f'(0)=1$$ $$2a\times0+b=1$$ $$b=1$$ Also, $f(x)$ being tangent to $y=x$ at the origin means that the graph of $f(x)$ passes through the point $(0,0)$: $$a\times0^2+b\times0+c=0$$ $$c=0$$ Finally, since $f(x)$ passes through $(1,2)$, we have: $$a\times1^2+b\times1+c=2$$ $$a+b+c=2$$ Substitute $b=1$ and $c=0$ here: $$a+1+0=2$$ $$a=1$$
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