Question

How can help me with this operation about Pre-Calculus.Simplify

Answer 1

If we want to compute $f(2+h)$, we have to substitute every occurrence of $x$ with $2+h$ in the definition of the function:

$f(x)=x^2-11x \implies f(2+h) = (2+h)^2-11(2+h) = 4+4h+h^2-22-11h = h^2-7h-18$

Similarly, computing $f(2)$ means to substitute the input with 2:

$f(x)=x^2-11x \implies f(2) = 2^2-11\cdot 2 = 4-22 = -18$

So, we have

$f(2+h) - f(2) = h^2-7h-18 - (-18) = h^2-7h$

And finally

$\dfrac{f(2+h)-f(2)}{h} = \dfrac{h^2-7h}{h} = h-7$