f'(x) = 4x^3-40x
f'(-1) = -4+40 = 36
f(-1) = 45
so the line is
y-45 = 36(x+1)
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
1 answer
f(x)=x^4-20x^2+64;x=-1
1 answer