Question

For `f(x)=x^3` what is the equation of the tangent line at `x=-2`?

Answer 1

`y=12x+16`

Explanation:

`"We require the slope of the tangent and a point on it"`

`"slope of tangent "=f'(x)" at "x=-2`

`f'(x)=3x^2`

`f'(-2)=3(-2)^2=12`

`f(-2)=(-2)^3=-8rArr(-2,-8)larr" point"`

`y+8=12(x+2)`

`y=12x+16larrcolor(red)"equation of tangent"`

Answer 2

`12x-y+16=0`

Explanation:

Setting `x=-2` in the given function we get y-coordinate of point as follows

`y=f(-2)`

`=(-2)^3`

`=-8`

The coordinates of point are `(-2, -8)`

Now, differentiating given function w.r.t. `x`, we get slope of tangent `dy/dx` as follows

`dy/dx=f'(x)`

`=d/dx(x^3)`

`=3x^2`

hence the slope `m` of tangent at `x=-2`, is given as

`m=f'(-2)`

`=3(-2)^2`

`=12`

Now, the equation of tangent at the point `(x_1, y_1)\equiv(-2, -8)` & slope `m=12` is given by following formula

`y-y_1=m(x-x_1)`

y-(-8)=12(x-(-2))#

`y+8=12x+24`

`12x-y+16=0`