1. ## solving a function

Here is a rational function:

$f(x)= \frac {3x^2}{4-x^2}$

Find:
domain: (- infinity, infinity)
Range: (- infinity, -2)U(-2,2)U(2, infinity)
x-intercept:
y-intercept: none
roots:
Asymptotes:

For y-intercept:
Let x=0
$f(0)= \frac {3(0)^2}{4-0^2}$
$f(x)= 0 so, Y=0 (No Y-intercept)$

For x-intercept:

$f(x)= \frac {3x^2}{4-x^2}=0$
I factor the denominator:
$f(x)= \frac {3x^2}{(x-2)(x+2)}$

What is next ? >.<

2. there is actually an intercept at the point (0,0)
as u said:

therefore f(0)= 0/4 =0

3. Originally Posted by chrizzle
there is actually an intercept at the point (0,0)
as u said:

therefore f(0)= 0/4 =0

I found the Vertical Asymptotes X= 2 and X= -2

Horizontal Asymptotes is y=3
thanks!

4. Originally Posted by Anemori
Here is a rational function:

$f(x)= \frac {3x^2}{4-x^2}$

Find:
domain: (- infinity, infinity)
Range: (- infinity, -2)U(-2,2)U(2, infinity)
x-intercept:
y-intercept: none
roots:
Asymptotes:

For y-intercept:
Let x=0
$f(0)= \frac {3(0)^2}{4-0^2}$
$f(x)= 0 so, Y=0 (No Y-intercept)$

For x-intercept:

$f(x)= \frac {3x^2}{4-x^2}=0$
I factor the denominator:
$f(x)= \frac {3x^2}{(x-2)(x+2)}$
There is no point in factoring the denominator. A fraction, a/b, is equal to 0 if and only if the [b]numerator, a, is equal to 0. For this function, both x and y intercepts are at (0, 0).

[/quote]What is next ? >.<[/QUOTE]
Yes, there are vertical asymptotes at x= -2 and x= 2 but the horizontal asymptote is at y= -3, not y= 3.