How do you graph f(x)=x^4-3x^2+2x?

Mar 29, 2015

I assume that you want to graph this without technology.

The graph is at the end of the solution. (I believe it's more instructive to not have it to start.)

$f \left(x\right) = {x}^{4} - 3 {x}^{2} + 2 x$

$f$ is a polynomial, so, of course its domain is the set of all real numbers.

$y$-intercept $f \left(0\right) = 0$
The $y$-intercept is $0$. (Or $\left(0 , 0\right)$ if you prefer.)

$x$- intercept(s) :
Solve $f \left(x\right) = {x}^{4} - 3 {x}^{2} + 2 x = 0$

Factor out the $x$: to get $x \left({x}^{3} - 3 x + 2\right) = 0$
By inspection or by the Rational Zero Theorem (or "Test"), $1$ is a zero of $\left({x}^{3} - 3 x + 2\right)$.
By the Factor Theorem, $\left(x - 1\right)$ is a factor.
Use division (in some form) to get the quadratic factor:
$\left({x}^{3} - 3 x + 2\right) = \left(x - 1\right) \left({x}^{2} + x - 2\right)$

The quadratic is straightforward to factor: $\left({x}^{2} + x - 2\right) = \left(x + 2\right) \left(x - 1\right)$

So we see that $f \left(x\right) = x \left(x + 2\right) {\left(x - 1\right)}^{2}$ the zeros (the $x$-intercepts are: $- 2 , 0 , 1$ ($1$ is a multiple zero of even multiplicity.)

(Write the intercepts: $\left(- 2 , 0\right) , \left(0 , 0\right)$, and $\left(1 , 0\right)$ if you prefer.)

Analysis of $f '$
Although we can work with $f \left(x\right)$ in any form, I prefer the standard polynomial form over the 3 factor form.

$f \left(x\right) = {x}^{4} - 3 {x}^{2} + 2 x$
So $f ' \left(x\right) = 4 {x}^{3} - 6 x + 2$ Which is never non-existent, so we only need to solve: $f ' \left(x\right) = 4 {x}^{3} - 6 x + 2 = 0$
By inspection or by the Rational Zero Theorem or by observing that multiple zeros of a polynomial are also zeros of the derivative, we see that $1$ is again a zero, so $\left(x - 1\right)$ is a factor and:

$f ' \left(x\right) = 4 {x}^{3} - 6 x + 2 = 2 \left(2 {x}^{3} - 3 x + 1\right) = 2 \left(x - 1\right) \left(2 {x}^{2} + 2 x - 1\right)$

The quadratic factor has irrational zeros: $\frac{- 1 \pm \sqrt{3}}{2}$.

Observe that $1 < \sqrt{3} < 2$, so one of the zeros is negative and the other positive. And $\frac{- 1 + \sqrt{3}}{2} < \frac{- 1 + 2}{2} = \frac{1}{2} < 1$.

The critical numbers for $f$ are Left to right
: $\frac{- 1 - \sqrt{3}}{2} , \frac{- 1 + \sqrt{3}}{2} , \text{ and } 1$.

For ease of reference, let's call the negative critical number ${z}_{1} = \frac{- 1 - \sqrt{3}}{2}$ and the positive one ${z}_{2} = \frac{- 1 + \sqrt{3}}{2}$

We need to investigate the sign of $f '$ on each of the intervals:

$\left(- \infty , {z}_{1}\right)$, $\left({z}_{1} , {z}_{2} ,\right)$, $\left({z}_{2} , 1\right)$, $\left(1 , \infty\right)$

If you like test numbers, I'd suggest: $- 10 , 0 , \frac{1}{2} , 10$.
It may not be clear that $\frac{1}{2}$ is in $\left({z}_{2} , 1\right)$ until it is observed that:

$\sqrt{3} < 2$ $\implies$ $\frac{- 1 + \sqrt{3}}{2} < \frac{- 1 + 2}{2} = \frac{1}{2}$

If you prefer, use a factor table.

Whichever method you use, you'll find that:

Increasing/Decreasing
$f ' \left(x\right) < 0$ on $\left(- \infty , {z}_{1}\right)$, so $f$ is decreasing on $\left(- \infty , {z}_{1}\right)$
$f ' \left(x\right) > 0$ on $\left({z}_{1} , {z}_{2} ,\right)$, so $f$ is increasing on $\left({z}_{1} , {z}_{2} ,\right)$
$f ' \left(x\right) < 0$ on $\left({z}_{2} , 1\right)$, so $f$ is decreasing on $\left({z}_{2} , 1\right)$
$f ' \left(x\right) > 0$ on $\left(1 , \infty\right)$, so $f$ is increasing on $\left(1 , \infty\right)$

Local extrema:
$f \left({z}_{1}\right) = f \left(\frac{- 1 - \sqrt{3}}{2}\right)$ is a local minimum (also global)
$f \left({z}_{2}\right) = f \left(\frac{- 1 + \sqrt{3}}{2}\right)$ is a local maximum
$f \left(1\right) = 0$ is a local minimum.

Analysis of $f ' '$

$f ' ' \left(x\right) = 12 {x}^{2} - 6 = 6 \left(2 {x}^{2} - 1\right)$

Whose zeros are : $\pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$

Investigating the sign of $f ' '$ on the appropriate intervals leads us to:

Concavity:
$f ' ' \left(x\right) > 0$ on $\left(- \infty , - \frac{\sqrt{2}}{2}\right)$ So $f$ is concave up.
$f ' ' \left(x\right) < 0$ on $\left(- \frac{\sqrt{2}}{2} , \frac{\sqrt{2}}{2}\right)$ So $f$ is concave down.
$f ' ' \left(x\right) > 0$ on $\left(\frac{\sqrt{2}}{2} , \infty\right)$ So $f$ is concave up.

There are two inflection points. They are:
((-sqrt2 / 2, f(-sqrt2 / 2) ) which is $\left(- \frac{\sqrt{2}}{2} \text{,} - \frac{5}{4} - \sqrt{2}\right)$
and
((sqrt2 / 2, f(sqrt2 / 2) ) which is $\left(\frac{\sqrt{2}}{2} \text{,} - \frac{5}{4} + \sqrt{2}\right)$

Now sketch the graph (It may take a couple of rough sketches first:

graph{y=x^4-3x^2+2x [-10, 10, -5, 5]}