# Can you walk me through this problem? [closed]

The problem is:

If $f(x) = -4x + 5$, find $\dfrac{f(a + h) – f(a)}{h}$

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## closed as off-topic by glace, Kirill, 900 sit-ups a day, Davide Giraudo, HakimJun 25 at 20:57

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – glace, Kirill, 900 sit-ups a day, Davide Giraudo, Hakim
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What have you tried? For example, what is $f(a+h)$ for the function $f(x)$ you gave us? –  Tom Jun 25 at 19:27

If someone says $f(\color{red}{\mathrm{something}})$, they are telling you to take the formula of $f(x)$, and everywhere that appears $x$, you put $\color{red}{\mathrm{something}}$ instead. For example, if $f(x) = x^2 + 1$, $f(\color{red}{a+h})$ would be $(\color{red}{a+h})^2 + 1$. Can you do the same with the function you have there? (:
\eqalign{\tt f(\color{blue}x)=-4\color{blue}x+5 \ \ &\rlap{\rlap{=}=}{\ >}\tt \ \ f(\color{blue}{a+h})=-4\color{blue}{(a+h)}+5=-4a-4h+5 \\ &\rlap{\rlap{=}=}{\ >} \ \ \tt f(\color{blue}a)=-4\color{blue}a+5. } Therefore: \eqalign{\tt \dfrac{f(a+h)-f(a)}{h} &= \tt \dfrac{-4a-4h+5-(-4a+5)}{h}\\ &=\tt\require{cancel}\dfrac{\cancel{-4a}-4h+\cancel5+\cancel{4a}\cancel{-5}}{h}\\ \tt &=\tt \dfrac{-4h}{h} =-4.\tag{\tt h\neq0}\\ }