# Thread: rational function need help: asymptotes, intercepts, holes

1. ## rational function need help: asymptotes, intercepts, holes

f(x)= 3x^2+3x-6 / x^2-6x+5

1. Determine if the graph of f(x) has any holes.
2. Find all asymptotes for f(x).
3. Find atleast one point between any verticle asymptote(s) and x-intercept(s).

Thank u so much for any help.

2. Originally Posted by tpwkbs

f(x)= 3x^2+3x-6 / x^2-6x+5

1. Determine if the graph of f(x) has any holes.
find the zeros of the numerator and the denominator. if you have a zero that is the same for both, it is a hole

2. Find all asymptotes for f(x).
these occur at x-values (that are not holes) where the denominator is zero.

3. Find atleast one point between any verticle asymptote(s) and x-intercept(s).

3. Originally Posted by tpwkbs

f(x)= 3x^2+3x-6 / x^2-6x+5

1. Determine if the graph of f(x) has any holes.
2. Find all asymptotes for f(x).
3. Find atleast one point between any verticle asymptote(s) and x-intercept(s).

Thank u so much for any help.
Factor the numerator and the denominator of f:

$\displaystyle f(x)=\frac{3x^3+3x-6}{x^2-6x+5}= \frac{3(x-1)(x+2)}{(x-1)(x-5)}$

to #1.: The graph is punctured(?) at x = 1

to #2.: Since $\displaystyle \lim_{|x|\mapsto \infty}f(x) = 3$ there exist a horizontal asymptote $\displaystyle y = 3$

Since f(x) is not defined at x = 5 there exist a vertical asymptote $\displaystyle x = 5$

to #3.: I don't understand your question - but because the x-intercept is at $\displaystyle x = -2$ maybe you mean a point like $\displaystyle (0, 0)$ ?

thank you