# Homework Help: Graph Problem

1. May 10, 2004

### mustang

Problem 12.
Sketch the graph of k(x)=x^4-64x^2.

2. May 11, 2004

### Muon12

Hmmm. Just graph the function of k? Well, to start off, you should make a chart of respective values for x such as:k(x) when x= 0,1,2,3,4,5,6,...n to as many values as you need to. For this problem, I recommend that you at least go from k(-8) to k(8), because there are zeros at those k values. Other than that, just work the problem out... set maximum horizontal and vertical values for the graph. The values are rather large in this problem, so make the vertical limits large enough to show the overall shape. That's all I suppose. Oh, and use a calculator if you can. This won't be fun if you try to do it by hand :yuck:.

Last edited: May 11, 2004
3. May 11, 2004

### Parth Dave

how much calculus do you know? There is a process using the derivatives of functions to allow you to sketch any function.

4. May 11, 2004

### arildno

We have:
$$x^{4}-64x^{2}=x^{2}(x^{2}-64)=x^{2}(x+8)(x-8)$$

Note the following:
$$\begin{itemize} \item -\infty\leq{x}\leq{-8}\rightarrow{k}(x)\geq{0}\\ \item -8\leq{x}\leq{8}\rightarrow{k}(x)\leq{0} \item 8\leq{x}\leq{\infty}\rightarrow{k}(x)\geq{0} \end{itemize}$$

5. May 11, 2004

### dlgoff

If you know any computer programing languages (perhaps Basic), you could write a simple loop program to give you some values or even plot it for you.

Regards

6. May 12, 2004

### Chen

It's been a while since I was asked to sketch function graphs, but here's what I think we used to do. First option is to follow these steps:

1) Find for which X the function is defined.

2) Find the intersection points of the function with the X and Y axes.

3) Find all minimum, maximum and "twist" (not sure what the English term is) points of the function.

4) Find for which X the function is ascending and for which X it is descending.

5) Find the asymptotes of the function if it has any.

So for your function: $$f(x) = x^4 - 64x^2 = x^2(x^2 - 64) = x^2(x + 8)(x - 8)$$

1) Any X.

2) (0, 0); (8, 0); (-8, 0).

3) $$f'(x) = 4x^3 - 128x = 4x(x^2 - 32) = 4x(x + \sqrt{32})(x - \sqrt{32})$$
$$f''(x) = 12x^2 - 128$$
Minimums: (-$\sqrt{32}$, -1024); ($\sqrt{32}$, -1024).
Maximum: (0, 0).

4) Descending: x < -$\sqrt{32}$; 0 < x < $\sqrt{32}$.
Ascending: -$\sqrt{32}$ < x < 0; $\sqrt{32}$ < x.

5) The function has no asymptotes.

Now draw your axes, mark the meaningful points we found, and considert the descending/ascending regions to complete the graph.

The second option is to buy a graphic calculator.

Good luck,