Question

Let `f` be the function given by `f(x) = 2x^4-4x^2+1`. What is an equation of the line tangent to the graph at `(-2,17)`?

Let `f` be the function given by `f(x) = 2x^4-4x^2+1`. What is an equation of the line tangent to the graph at `(-2,17)`?

Answer 1

`y = -48x - 79`

Explanation:

The line tangent to the graph `y=f(x)` at a point `(x_0, f(x_0))` is the line with slope `f'(x_0)` and passing through `(x_0, f(x_0))`.

In this case, we are given `(x_0, f(x_0)) = (-2, 17)`. Thus, we only need to calculate `f'(x_0)` as the slope, and then plug that into the point-slope equation of a line.

Calculating the derivative of `f(x)`, we get

`f'(x) = 8x^3-8x`

`=> f'(-2) = 8(-2)^3-8(-2) = -64+16 = -48`

So, the tangent line has a slope of `-48` and passes through `(-2, 17)`. Thus, it's equation is

`y - 17 = -48(x - (-2))`

`=> y = -48x - 79`