## Precalculus (10th Edition)

a) $-3$; b) $x^2-4x-2$; c) $x^2+4x+1$; d) $-x^2+4x-1$; e) $x^2-3$; f) $2x+h-4$
We are given the function: $f(x)=x^2-4x+1$ a) Compute $f(2)$: $f(2)=2^2-4(2)+1=-3$ b) Compute $f(x)+f(2)$: $f(x)+f(2)=x^2-4x+1-3=x^2-4x-2$ c) Compute $f(-x)$: $f(-x)=(-x)^2-4(-x)+1=x^2+4x+1$ d) Compute $-f(x)$: $-f(x)=-(x^2-4x+1)=-x^2+4x-1$ e) Compute $f(x+2)$: $f(x+2)=(x+2)^2-4(x+2)+1$ $=x^2+4x+4-4x-8+1$ $=x^2-3$ f) Compute for $h\not=0$: $\dfrac{f(x+h)-f(x)}{h}=\dfrac{(x+h)^2-4(x+h)+1-(x^2-4x+1)}{h}$ $=\dfrac{x^2+2hx+h^2-4x-4h+1-x^2+4x-1}{h}$ $=\dfrac{2hx+h^2-4h}{h}$ $=\dfrac{h(2x+h-4)}{h}$ $=2x+h-4$