1. ## Graphing helpppp...

Find the intercepts, symmetries, domain, range and (vertical and horizontal) asymptotes. Also Sketch the graph:

$1.) y = \frac{1}{8}(x^3-4x)$

$2.) x^2(y^2-4) = 4$

Thanks!

Find the intercepts, symmetries, domain, range and (vertical and horizontal) asymptotes. Also Sketch the graph:

$1.) y = \frac{1}{8}(x^3-4x)$

Thanks!
Intercepts:
y-intecept: This is the point (0,y) on the graph, so:
$y = \frac{1}{8}(0^3-4 \cdot 0) = 0$

So the y-intercept is (0,0).

x-intercepts: These are the points (x,0) on the graph, so:
$0 = \frac{1}{8}(x^3-4x)$

$0 = x(x^2 - 4) = x(x + 2)(x - 2)$

So the x-intercepts are (0,0), (0, -2), and (0, 2).

Symmetries:
Parity: What is y(-x)?
$y(-x) = \frac{1}{8}((-x)^3-4(-x)) = \frac{1}{8}(-x^3 + 4x) = -y(x)$

Since y(-x) = -y(x) we know that y is an odd function.

Reflections:
Reflecting y over the x-axis takes y -> -y:
$-y = \frac{1}{8}(x^3-4x)$ --> $y = -\frac{1}{8}(x^3-4x)$
Since this is not equal to the original y, there is no reflection symmetry.

Reflecting over the y-axis takes x -> -x
As this is the same as the parity symmetry, we already know that y(-x) is not equal to y(x). So there is no reflection symmetry.

Inversion:
This takes (x,y) -> (-x,-y):
$-y = \frac{1}{8}((-x)^3-4(-x)) = -\frac{1}{8}(-x^3 + 4x)$
This gives the original function back again, so y(x) is symmetric under inversion.

Domain:
$y = \frac{1}{8}(x^3-4x)$
We look for values of x that makes the function undefined. There are no fractions, square roots, log functions, etc. so the domain is unrestricted. Thus the domain of y is $(-\infty, \infty)$.

Range:
We look for all possible values the function can take on its domain. The function is continuous. Note that y -> $-\infty$ for x -> $-\infty$ and y -> $\infty$ for x -> $\infty$. So the range of y is $(-\infty, \infty)$.

Asymptotes:
Horizontal asymptotes: These occur when $\lim_{x \to \pm \infty} y(x) = constant$. In this case $\lim_{x \to \pm \infty} y(x) \to \pm \infty$ so there are no horizontal asymptotes.

Vertical asymptotes: These occur when the denominator of a function becomes 0. There are no denominators in this function so there are no vertical asymptotes.

Slant asymptotes: These occur when $\lim_{x \to \pm \infty} y(x) = (constant)x$. In this case $\lim_{x \to \pm \infty} y(x) \to \frac{1}{8}x^3$ so there are no slant asymptotes.

You should be able to sketch this yourself. However for completeness I have attached one at that bottom of this post.

-Dan

3. So the x-intercepts are (0,0), (0, -2), and (0, 2).
isnt that (0,0), (-2,0) & (2,0)?

And thank you very much for the help mr top quark!

So you need to find the coordinates of the equation :
$y = -\frac{1}{8}(-x^3+4x)$
and
$y = \frac{1}{8}(x^3-4x)$
since its symmetrical w/ respect to the origin ryt?

isnt that (0,0), (-2,0) & (2,0)?

And thank you very much for the help mr top quark!

So you need to find the coordinates of the equation :
$y = -\frac{1}{8}(-x^3+4x)$
and
$y = \frac{1}{8}(x^3-4x)$
since its symmetrical w/ respect to the origin ryt?
Yeah, I guess I wrote it backward, didn't I?

I'm not sure what you are trying to confirm for the second part. Coordinates for what?

-Dan

isnt that (0,0), (-2,0) & (2,0)?

And thank you very much for the help mr top quark!

So you need to find the coordinates of the equation :
$y = -\frac{1}{8}(-x^3+4x)$
and
$y = \frac{1}{8}(x^3-4x)$
since its symmetrical w/ respect to the origin ryt?
note that: $-\frac{1}{8}(-x^3+4x) = \frac{1}{8}(x^3-4x)$

6. Isee thanks!

$x^2 (y^2 - 4) = 4$

Isee thanks!

$x^2 (y^2 - 4) = 4$
First observe that $|y| \ge 2$, as otherwise $x$ is imaginary.

also as $y \to \pm \infty, \ x \to 0$.

$x \to \infty,\ y \to \pm 2$, and

$x \to -\infty,\ y \to \pm 2$.

Now differentiating wrt $x$ shows that there are no maxima or minima for finite $x$.

This should be enough to allow us to sketch the curve: