Find the slope and equation of the tangent line to the graph of the function at the given value of x.
f(x)=x^4-20x^2+64; x=-1

The slope is the differential of the function.

Recall, if y = x^n,  (dy/dx) =  nx^(n-1).

y=x^4-20x^2+64;  x = -1.  To differentiate this, we do it for each term.

(dy/dx) = (4)(x^(4 -1))  - (2)(20x^(2-1) + 0*64x^(0-1)        (Note 64 = 64x^0, x^0 =1)
           =  (4)x^(3) - 40x^(1)  + 0
            =    4x^3  -  40x^1. 

(dy/dx)  =    4x^3  -  40x.   Note at x = -1.


(dy/dx), x = -1,   =     4(-1)^3  -  40(-1)                         
                         =    -4 +  40  = 40 - 4 = 36.

Slope at x = -1 is 36.
Cheers.

The correct answer is .44