Question

How do you find the equation of the line tangent to `f(x)= x^3`, at (2,8)?

Answer 1

y = 12x - 16

Explanation:

One form of the equation of a straight line is y = mx + c , where m, represents the gradient and c , the y-intercept. To find m, require to obtain f'(x) and the value of f'(2).

`f'(x) = 3x^2 " and " f'(2) = 3(2)^2 = 12 = " m of tangent "`

equation is then : y = 12x + c

use (2 , 8 ) in equation to find c.

hence : 8 = 12(2) + c → c = 8 - 24 = - 16

`rArr " equation of tangent : " y = 12x - 16`