# Thread: Function Domain, Range etc.

1. ## Function Domain, Range etc.

For this function:

I need to know how to get the:

- Domain
- Range
- X-Intercept(s)
- Y-Intercept(s)
- Vertical Asymptote(s)
- Positive Intervals
- Negative Intervals
- Intervals of Increase
- Intervals of Decrease

2. what have you attempted?

have you graphed the function?

are you saying you can't find anything on that list?

3. Yes, I have graphed the function, and can find the domain and range with the graph. I want to know how to find everything without graphing it, and by doing it with just the equation.

4. Originally Posted by Coolman
For this function:

I need to know how to get the:

- Domain
- Range
- X-Intercept(s)
- Y-Intercept(s)
- Vertical Asymptote(s)
- Positive Intervals
- Negative Intervals
- Intervals of Increase
- Intervals of Decrease
for the domain and the range

http://www.mathhelpforum.com/math-help/pre-algebra-algebra/92992-domain-range.html

x-intercept find the point where f(x) equal zero
y-intercept find f(0)

Positive intervals and negative intervals I will present an example

$\displaystyle f(x)=\frac{(-x^2+5x-6)(x-4)}{(1-x)(3x-2)}=\frac{(x-2)(3-x)(x-4)}{(1-x)3(x-\frac{2}{3})}$ first find criticals points where the function equal zero and the zero of the denominator you have $\displaystyle 2,3,4,1,\frac{2}{3}$ take the points and study each term then multiply them like this

intervals of increasing and decreasing

derive the function then find the positive interval of the derivative this equal to increasing interval in the function f and find the interval when the derivative is negative this is the decreasing intervals

Vertical Asymptotes take the zero of the nominator (root) in my question x=1 and x=2/3 is the Vertical Asymptotes

x-intercept for my question
$\displaystyle f(x)=\frac{(x-2)(3-x)(x-4)}{(1-x)3(x-\frac{2}{3})}$

$\displaystyle \frac{(x-2)(3-x)(x-4)}{(1-x)3(x-\frac{2}{3})}=0$ this function equal zero when x=2,3,4 so the points (2,0),(3,0)(4,0) is the x-intercept

y-intercept find f(0)
$\displaystyle f(0)=\frac{(0-2)(3-0)(0-4)}{(1-0)3(0-\frac{2}{3})}=\frac{(-2)(3)(-4)}{(-2)}=-12$ so (0,-12) y-intercept