Question

How do you sketch the graph `f(x)=2x^4-26x^2+72`?

Answer 1

See explanation...

Explanation:

`f(x) = 2x^4-26x^2+72`

`color(white)(f(x)) = 2((x^2)^2-13(x^2)+36)`

`color(white)(f(x)) = 2(x^2-4)(x^2-9)`

`color(white)(f(x)) = 2(x-2)(x+2)(x-3)(x+3)`

So the graph of this function intercepts the `x` axis at `x=+-2` and `x=+-3`.

It intercepts the `y` axis at `(0, 72)`, where it has a local maximum.

Note that `f(x)` is an even function, symmetric about the `y` axis.

`f'(x) = 8x^3-52x = 2((2x)^2-26)x`

So this quartic has local minima at:

`x = +-sqrt(26)/2 ~~ +-5.1/2 = +-2.55`

We find:

`f(+-sqrt(26)/2) = 2(13/2)^2-26(13/2)+72 = 169/2-169+72 = -25/2`

So this quartic function is a classic "W" shaped curve, symmetric about the `y` axis, passing through:

`(-3, 0), (-sqrt(26)/2, -25/2), (-2, 0), (0, 72), (2, 0), (sqrt(26)/2, -25/2), (3, 0)`

graph{2x^4-26x^2+72 [-10, 10, -20, 80]}