Question

Find an equation for the line thatis tangent to the curve y=5x^3-5x at the point (1,0)

Answer 1

Hello,

The gradient of a tangent can be found by differentiating the equation:
$\frac{d}{dx}(5x^3-5x) = 15x^2-5$

Subbing in the x value of 1, the gradient therefore is 15-5 = 10, giving us m in our general equation for a line y=mx+c, i.e. y = 10x + c

We know at x=1, y is 0. Subbing this in:
0 = 10 + c, and solving for c = -10

Therefore, the equation of the tangential line at (1,0) is $y=10x-10$.