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