Answer
$\frac{f(x+h) - f(x)}{h} = 4x + 2h - 1$
Work Step by Step
Since we know that $f(x) = 2x^{2} - x - 1$, to find $\frac{f(x + h) - f(x)}{h}$, we're simply missing the value of $f(x + h)$ to finally substitute and simplify:
$f(x + h) = 2[x + h]^{2} - [x + h] - 1$
$f(x + h) = 2[x^{2} + 2xh + h^{2}] - x - h - 1$
$f(x + h) = 2x^{2} + 4xh + 2h^{2} - x - h - 1$
And now, we can substitute the appropriate values:
$\frac{f(x+h) - f(x)}{h} = \frac{[2x^{2} + 4xh + 2h^{2} - x - h - 1] - [2x^{2} - x - 1]}{h}$
$\frac{f(x+h) - f(x)}{h} = \frac{2x^{2} + 4xh + 2h^{2} - x - h - 1 - 2x^{2} + x + 1}{h} = \frac{2x^{2} - 2x^{2} + 4xh + 2h^{2} - x + x - h - 1 + 1}{h}$
$\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^{2} - h}{h} = \frac{h(4x + 2h - 1)}{h} = 4x + 2h - 1$
