Question

How do you find the equation of the line tangent to the graph of `y=x^2-2x` at the point (-2,8)?

Answer 1

`y=-6x-4`

Explanation:

Take the derivative of the given function

`(dy)/(dx)=2x-2`

Find `(dy)/(dx)` for given value of `x`

`(dy)/(dx)=2(-2)-2=-4-2=-6`

Now create the point slope form of a line and the solve for `y`

`y-8=-6(x-(-2))`

`y-8=-6(x+2)`

`y-8=-6x-12`

`y=-6x-4`

Answer 2

`y=-6x-4`

Explanation:

`y=x^2-2x`

`dy/dx=2x-2`

The general equation for the tangent is given by:

`y=mx+c`

`m=2x-2`

Since `x=-2` then:

`m=(2xx-2)-2=-6`

Since `y=8`:

`8=(-6xx-2)+C`

`C=-4`

So the equation of the tangent becomes:

`y=-6x-4`

Answer 3

You can also perform the linearization of the function.

`f_T(a) = f'(a)(x-a) + f(a)`
where `f_T` is the tangent line, `f` is the function at `x = a`, and `f'` is the derivative at `x = a`.

Thus:
`f'(-2) = [2x - 2]|_(x = -2) = -6`
`f(-2) = [x^2 - 2x]|_(x = -2) = 8`

Therefore, after taking a derivative and evaluating two functions at `x = -2`, without even using the value of `y`, you still get:

`color(blue)(f_T(-2)) = -6(x+2) + 8 color(blue)(= -6x - 4)`