Find the intercepts, symmetries, domain, range and (vertical and horizontal) asymptotes. Also Sketch the graph:
$\displaystyle 1.) y = \frac{1}{8}(x^3-4x)$
$\displaystyle 2.) x^2(y^2-4) = 4$
Thanks!
Intercepts:
y-intecept: This is the point (0,y) on the graph, so:
$\displaystyle y = \frac{1}{8}(0^3-4 \cdot 0) = 0$
So the y-intercept is (0,0).
x-intercepts: These are the points (x,0) on the graph, so:
$\displaystyle 0 = \frac{1}{8}(x^3-4x)$
$\displaystyle 0 = x(x^2 - 4) = x(x + 2)(x - 2)$
So the x-intercepts are (0,0), (0, -2), and (0, 2).
Symmetries:
Parity: What is y(-x)?
$\displaystyle y(-x) = \frac{1}{8}((-x)^3-4(-x)) = \frac{1}{8}(-x^3 + 4x) = -y(x)$
Since y(-x) = -y(x) we know that y is an odd function.
Reflections:
Reflecting y over the x-axis takes y -> -y:
$\displaystyle -y = \frac{1}{8}(x^3-4x)$ --> $\displaystyle y = -\frac{1}{8}(x^3-4x)$
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):
$\displaystyle -y = \frac{1}{8}((-x)^3-4(-x)) = -\frac{1}{8}(-x^3 + 4x)$
This gives the original function back again, so y(x) is symmetric under inversion.
Domain:
$\displaystyle y = \frac{1}{8}(x^3-4x)$
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 $\displaystyle (-\infty, \infty)$.
Range:
We look for all possible values the function can take on its domain. The function is continuous. Note that y -> $\displaystyle -\infty$ for x -> $\displaystyle -\infty$ and y -> $\displaystyle \infty$ for x -> $\displaystyle \infty$. So the range of y is $\displaystyle (-\infty, \infty)$.
Asymptotes:
Horizontal asymptotes: These occur when $\displaystyle \lim_{x \to \pm \infty} y(x) = constant$. In this case $\displaystyle \lim_{x \to \pm \infty} y(x) \to \pm \infty$ 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 $\displaystyle \lim_{x \to \pm \infty} y(x) = (constant)x$. In this case $\displaystyle \lim_{x \to \pm \infty} y(x) \to \frac{1}{8}x^3$ 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
isnt that (0,0), (-2,0) & (2,0)?So the x-intercepts are (0,0), (0, -2), and (0, 2).
And thank you very much for the help mr top quark!
So you need to find the coordinates of the equation :
$\displaystyle y = -\frac{1}{8}(-x^3+4x)$
and
$\displaystyle y = \frac{1}{8}(x^3-4x)$
since its symmetrical w/ respect to the origin ryt?
First observe that $\displaystyle |y| \ge 2$, as otherwise $\displaystyle x$ is imaginary.
also as $\displaystyle y \to \pm \infty, \ x \to 0$.
$\displaystyle x \to \infty,\ y \to \pm 2$, and
$\displaystyle x \to -\infty,\ y \to \pm 2$.
Now differentiating wrt $\displaystyle x$ shows that there are no maxima or minima for finite $\displaystyle x$.
This should be enough to allow us to sketch the curve: