Find $f(a)$, $f(a + h)$, and the difference quotient, given $f(x)=7-8x+2x^2$

Question

Question is here:

I need help with part c.

I tried plugging everything in and simplifying to a point where my final answer was $(-8h+2ah+2h^2)/h$

My work:

$(7-8a-8h+2(a+h)^2-7+8a-2a^2)/h$ combine like terms

$(-8h+2(a+h)^2 -2a)/h$ foil/expand

$(-8h+2a^2+2ah+2h^2-2a)/h$ combine like terms

$-8h+2ah+2h^2/h$

Answer 1

There is a small algebra mistake, notice $2(a+h)^2=2a^2+\color{red}4ah+2h^2$. From here your last line turns into $\frac{-8h+4ah+2h^2}{h}$, which simplifies to $-8+4a+2h$.

Answer 2

You have also made an error in going from here:$$(7−8a−8h+2(a+h)^2−7+8a−\color{red}{2a^2})/h$$to here:$$(−8h+2(a+h)^2−\color{red}{2a})/h$$And, as @Darrell noted, you have not then expanded $2(a+h)^2$ correctly.

Answer 3

HINT: $$2(a+h)^2=2(a^2+h^2+2ah)=2a^2+2h^2+4ah$$ thus we get $\frac{-8h+4ah+2h^2}{h}=-8+4a+2h$

Answer 4

In your foil process you have $$\frac{-8h +2(a+h)^2-2a^2}{h}=\frac{-8h+2a^2+2ah+2h^2-2a^2}{h}$$.

This is actually supposed to be $$\frac{-8h +2(a+h)^2-2a}{h}=\frac{-8h+2(a^2+2ah+h^2)-2a^2}{h}$$ $$=\frac{-8h+2a^2+4ah+2h^2-2a^2}{h}$$ $$=\frac{h(4a+2h-8)}{h}$$ $$=4a+2h-8$$