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

Oct 12, 2016

See explanation...

#### Explanation:

$f \left(x\right) = 2 {x}^{4} - 26 {x}^{2} + 72$

$\textcolor{w h i t e}{f \left(x\right)} = 2 \left({\left({x}^{2}\right)}^{2} - 13 \left({x}^{2}\right) + 36\right)$

$\textcolor{w h i t e}{f \left(x\right)} = 2 \left({x}^{2} - 4\right) \left({x}^{2} - 9\right)$

$\textcolor{w h i t e}{f \left(x\right)} = 2 \left(x - 2\right) \left(x + 2\right) \left(x - 3\right) \left(x + 3\right)$

So the graph of this function intercepts the $x$ axis at $x = \pm 2$ and $x = \pm 3$.

It intercepts the $y$ axis at $\left(0 , 72\right)$, where it has a local maximum.

Note that $f \left(x\right)$ is an even function, symmetric about the $y$ axis.

$f ' \left(x\right) = 8 {x}^{3} - 52 x = 2 \left({\left(2 x\right)}^{2} - 26\right) x$

So this quartic has local minima at:

$x = \pm \frac{\sqrt{26}}{2} \approx \pm \frac{5.1}{2} = \pm 2.55$

We find:

$f \left(\pm \frac{\sqrt{26}}{2}\right) = 2 {\left(\frac{13}{2}\right)}^{2} - 26 \left(\frac{13}{2}\right) + 72 = \frac{169}{2} - 169 + 72 = - \frac{25}{2}$

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

$\left(- 3 , 0\right) , \left(- \frac{\sqrt{26}}{2} , - \frac{25}{2}\right) , \left(- 2 , 0\right) , \left(0 , 72\right) , \left(2 , 0\right) , \left(\frac{\sqrt{26}}{2} , - \frac{25}{2}\right) , \left(3 , 0\right)$

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