## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 2 - Section 2.1 - Functions - Exercises - Page 157: 49

#### Answer

$8a+4h-5$

#### Work Step by Step

$f(x)=3-5x+4x^{2}$ Find $f(a)$ by substituting $x$ with $a$: $f(a)=3-5a+4a^{2}$ Find $f(a+h)$ by substituting $x$ with $a+h$: $f(a+h)=3-5(a+h)+4(a+h)^{2}=...$ $...=3-5a-5h+4(a^{2}+2ah+h^{2})=...$ $...=3-5a-5h+4a^{2}+8ah+4h^{2}$ We have $f(a)$ and $f(a+h)$, we can now substitute them in the expression $\dfrac{f(a+h)-f(a)}{h}$. $\dfrac{3-5a-5h+4a^{2}+8ah+4h^{2}-3+5a-4a^{2}}{h}=...$ Combine like terms in the numerator: $...=\dfrac{-5h+8ah+4h^{2}}{h}=...$ Take out common factor $h$ from the numerator and simplify: $...=\dfrac{h(-5+8a+4h)}{h}=8a+4h-5$

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