## Thomas' Calculus 13th Edition

a. $12$ b. $y=12x-16$
Given $y=x^3, P(2,8)$ a. slope=$\frac{(2+h)^3-2^3}{h}=\frac{3\times4h+6h^2+h^3}{h}=12+6h+h^2=12$ (let $h$ be zero) b. tangent line: $y-8=12(x-2)$, which is $y=12x-16$