Help finding turning points to plot quartic and cubic functions

Question

My teacher assigned us some graphing homework. I know how to graph all of them manually all except these two

1. $f_1(x)=x^4-x^3-4x^2$

2. $f_2(x)=\frac{1}{2}x^3+2x^2-8x-2$

When i asked him how he said it was above our grade level and it was a calculus question so we should just use graphing calculators. So since I'm still curious, I'm asking here. Detailed answers with steps would be nice. The main thing I need to know is how to find the exact location of turning points.

Answer 1

Let's just take $f_1(x) = x^4-x^3-4x^2 = x^2\ (x^2 - x - 4)$. Start by finding the roots (the values of $x$ where $f_1(x) = 0$). A quartic equation may have up to four real roots. From the $x^2$ factor, we see that two of them are $x = 0$ and $x = 0$ — a "double" root.

What about the other two roots? To solve $x^2 - x - 4 = 0$, we can use the quadratic equation:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{17}}{2} $$

So, $y = x^2 - x - 4$ would describe a parabola that crosses the $x$-axis at -1.562… and 2.562…. It is centered at $x = \frac{1}{2}$ (the midpoint between the roots), with a minimum at (0.5, -4.25).

Now let's consider what happens when we multiply that by $x^2$. That turns the parabola into a lopsided W-like curve (but smoother than a W), shifted to the right of the origin, and touching the $x$-axis at -1.562…, 0, 0, and 2.562….

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To find interesting points such as the minima and inflection points, though, you would need calculus. To give you a taste of what you haven't learned… we take the first derivative ("$\frac{d}{dx}\ f_1(x)$"), which tells you the slope of the curve for any $x$.

$$\frac{d}{dx}\ f_1(x) = 4x^3 - 3x^2 - 4 \cdot 2 \cdot x = \frac{x}{16} (8x - (3 - \sqrt{137}))(8x - (3 + \sqrt{137}))$$

We find that the slope is 0 at $x = 0$, $x \approx -1.088$, and $x \approx 1.838$. We conclude that $f_1(x)$ has a local maximum at (0, 0), a local minimum at (-1.088, -2.0458), and a local minimum at (1.838, -8.3097).

To find the inflection points, we would take the derivative of the derivative. This second derivative is

$$\frac{d^2}{dx^2}\ f_1(x) = 4 \cdot 3 \cdot x^2 - 3 \cdot 2 \cdot x - 8 = \frac{1}{12} (12x - (3 - \sqrt{105}))(12x - (3 + \sqrt{105}))$$

The inflection points, where the second derivative is zero, are at $x \approx -0.604$ and $x \approx 1.104$. That is, $f_1(x)$ is $\cap$-shaped in the interval $\frac{3-\sqrt{105}}{12} \lt x \lt \frac{3+\sqrt{105}}{12}$, and $\cup$-shaped everywhere else.

All of the information above can be seen in the plot. The roots are the red points, the extrema are green, and the inflection points are purple.

Answer 2

In the calculus classes you would be introduced to differentiation and next you will know how to use those derivatives to get turning points.

Chart out your road at first by calculating and plotting on a graph. Then and next later on you will appreciate those techniques of turning.

Answer 3

I don't see how this can be of any use to you, but for what it's worth: Turning points of graphs come from places where the derivative is zero, because the derivative of the function gives the slope of the tangent line. In calculus you would learn to compute the first derivative here as $4x^3-3x^2-8x$, so you'd find its zeroes and then check in any of several ways which of them give turning points.

Answer 4

Well, since you are so enthusiastic, I give just one step of differentiation. $$ f(x) = x^4 -x^3 -4 x^2 $$

$$ f'(x) = 4 x^3 - 3 x^2 - 8 x = 0, i.e., x\, ( 4 x^2 - 3 x -8 ) = 0 $$

It gives one root x = 0 at origin and you get two more roots after solving above quadratic.