Question

Find the equation of the tangent line to the curve y=x^3 at the point (2,8)?

Answer 1

`y=12x-16`

Explanation:

`•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = 2"`

`dy/dx=3x^2`

`"at x = 2 "dy/dx=3(2)^2=12`

`rArry-8=12(x-2)larrcolor(blue)"point-slope form"`

`rArry=12x-16larrcolor(blue)"slope-intercept form"`