D.
Plugging 2 into original fxn gives you y-coordinate. (2,6)
Taking the derivative of y=(x^3)-x gives you f'(x)=3x^2-1. Plug in x=2 to find slope. use point slope form to get equation! :)
which of the following is an equation of the line tangent to the graph of y = (x^3)-x at the point where x =2:
a) y - 6 = 4(x-2)
b) y - 6 = 5(x-2)
c) y - 6 = 6(x-2)
d) y - 6 = 11(x-2)
e) y - 6 = 12(x-2)
thanks!
1 answer
1 answer