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

1 Answer
George C.
Mar 30, 2018

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]}

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