find the difference quotient and simplify. f(x)=-3x^2 - 6x + 2 the difference quotient of f(x) is

Explanation:

Step1: Recall the difference - quotient formula

The difference - quotient formula is $\frac{f(x + h)-f(x)}{h}$, where $h\neq0$. First, find $f(x + h)$. Given $f(x)=-3x^{2}-6x + 2$, then $f(x + h)=-3(x + h)^{2}-6(x + h)+2$. Expand $(x + h)^{2}=x^{2}+2xh+h^{2}$. So $f(x + h)=-3(x^{2}+2xh + h^{2})-6(x + h)+2=-3x^{2}-6xh-3h^{2}-6x-6h + 2$.

Step2: Substitute $f(x + h)$ and $f(x)$ into the difference - quotient formula

$\frac{f(x + h)-f(x)}{h}=\frac{(-3x^{2}-6xh-3h^{2}-6x-6h + 2)-(-3x^{2}-6x + 2)}{h}$. Remove the parentheses: $\frac{-3x^{2}-6xh-3h^{2}-6x-6h + 2 + 3x^{2}+6x - 2}{h}$.

Step3: Simplify the numerator

Combine like - terms in the numerator: $(-3x^{2}+3x^{2})+(-6x + 6x)+(-6xh-3h^{2}-6h)+(2 - 2)=-6xh-3h^{2}-6h$. So the expression becomes $\frac{-6xh-3h^{2}-6h}{h}$.

Step4: Factor out $h$ from the numerator and simplify

Factor out $h$ from the numerator: $\frac{h(-6x-3h - 6)}{h}$. Cancel out the common factor $h$ (since $h\neq0$), we get $-6x-3h - 6$.

Answer:

$-6x-3h - 6$