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

Mar 30, 2018

See explanation...

#### Explanation:

Given:

$f \left(x\right) = 2 {x}^{4} - {x}^{2} + 5$

Complete the square as follows:

$f \left(x\right) = \frac{1}{8} \left(16 {x}^{4} - 8 {x}^{2} + 1\right) + \frac{39}{8}$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{1}{8} {\left(4 {x}^{2} - 1\right)}^{2} + \frac{39}{8}$

$\textcolor{w h i t e}{f \left(x\right)} = \frac{1}{8} {\left(2 x - 1\right)}^{2} {\left(2 x + 1\right)}^{2} + \frac{39}{8}$

So $f \left(x\right)$ has minimum value $\frac{39}{8}$ which it attains at $x = \pm \frac{1}{2}$

Also note that $f \left(0\right) = 5$

Since all of the terms of $f \left(x\right)$ 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 $\left(\pm \frac{1}{2} , \frac{39}{8}\right)$ and $\left(0 , 5\right)$.

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

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