Question

How do you sketch `2x^4-x^2+5`?

Answer 1

See explanation...

Explanation:

Given:

`f(x) = 2x^4-x^2+5`

Complete the square as follows:

`f(x) = 1/8(16x^4-8x^2+1)+39/8`

`color(white)(f(x)) = 1/8(4x^2-1)^2+39/8`

`color(white)(f(x)) = 1/8(2x-1)^2(2x+1)^2+39/8`

So `f(x)` has minimum value `39/8` which it attains at `x=+-1/2`

Also note that `f(0) = 5`

Since all of the terms of `f(x)` are of even degree, it is even and thus symmetric about the `y` axis.

So this quartic is a classic "W" shape, with turning points at `(+-1/2, 39/8)` and `(0, 5)`.

If we want any more guidance, we can just evaluate `f(x)` for other values of `x`, e.g. `f(1) = 2-1+5 = 6`, so the graph passes through `(+-1, 6)` ...

graph{2x^4-x^2+5 [-2.508, 2.492, 3.72, 6.22]}