Calculus 10th Edition

The slope at the point $(2, 8)$ is $-8$.
Similar to exercise 35, we must first expand ( no chain rule yet ) and hence we get $y = 2(x^2-8x+16) = 2x^2-16x+32$. The overall derivative of y is equal to the sum of the derivatives of the smaller function using the power rule $((x^n)'=nx^{n-1})$. The derivative of $2x^2$ is $(2)(2)(x^{2-1})$ equal to $4x$. The derivative of $-16x$ is $(-16)(1)(x^{1-1})$ which in simplest form is equal to $-16$. The derivative of $32$ is equal to $0$. Adding everything together we get the $y' = 4x-16$. To find the slope, plug in the x-coordinate into the derivative. Substituting $x=2$ gives $y' = 4(2)-16 = -8$ so the slope at the point is equal to -8.