Question

What is the equation of the line tangent to `f(x)=2x ^3-6x` at `x=1`?

Answer 1

`y=-4`

Explanation:

Let `y=2x^3-6x`

To find the gradient of the tangent line `m` at `x=1` take the 1st derivative:

`dy/dx=6x^2-6=m`

If `x=1rArr`

`m=6-6=0`

`:.y=2-6=-4`

The equation of the tangent line is of the form:

`y=mx+c`

`:.-4=0xxx+c`

`:.c=-4`

So the equation is:

`y=0xxx-4`

`y=-4`

The situation looks like this:

graph{(y-2x^3+6x)(y+4)=0 [-10, 10, -5, 5]}