# solving a function

• Mar 16th 2010, 11:11 PM
Anemori
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 ? >.<
• Mar 16th 2010, 11:19 PM
chrizzle
there is actually an intercept at the point (0,0)
as u said:

http://www.mathhelpforum.com/math-he...94e81c0f-1.gif

therefore f(0)= 0/4 =0
• Mar 16th 2010, 11:41 PM
Anemori
Quote:

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

http://www.mathhelpforum.com/math-he...94e81c0f-1.gif

therefore f(0)= 0/4 =0

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

Horizontal Asymptotes is y=3
thanks!
• Mar 17th 2010, 01:37 AM
HallsofIvy
Quote:

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.