Question

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

Answer 1

`y-36 = 42(x-3)` or `y=42x-90`

Explanation:

We know the `x` value of the original function, `f(x)` find the `y` value by plugging in 3:

`f(3) = 2(3)^3-2(3)^2 = 36` so our point is: `(3,36)`

Now find the derivative of `f(x)` using the power rule:

`f'(x) = 6x^2 - 4x`

The derivative gives us the slope at any given point, by plugging in 3 into `f'(x)` we get the slope of `f(x)` at `x=3`

`f'(3) = 6(3)^2 - 4(3) = 42` so `m=42, (3, 36)`

We now have enough information to create a line in point slope form: `y-y_1 = m(x-x_1)`

`y-36 = 42(x-3)`
If you want, you can solve for y:
`y=42x-90`