SOLUTION: determine the equations of any vertical asymptotes and the values of x for any holes in the graph of each rational function. f(x)=2x^2-x-10/(2x-5)

Algebra ->  Rational-functions -> SOLUTION: determine the equations of any vertical asymptotes and the values of x for any holes in the graph of each rational function. f(x)=2x^2-x-10/(2x-5)      Log On


   

Question 547530: determine the equations of any vertical asymptotes and the values of x for any holes in the graph of each rational function.
f(x)=2x^2-x-10/(2x-5)
Found 2 solutions by Edwin McCravy, KMST:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
f(x) = %282x%5E2-x-10%29%2F%282x-5%29
f(x) = %28%28x%2B2%29%282x-5%29%29%2F%282x-5%29
2x-5 must not be 0 since denominators are never 0.
So
2x - 5 ≠ 0
2x ≠ 5
x ≠ 5%2F2
We can cancel those as long as we require x ≠ 5%2F2
f(x) = %28%28x%2B2%29%28cross%282x-5%29%29%29%2F%28cross%282x-5%29%29, where x ≠ 5%2F2
f(x) = x+2, x ≠ 5%2F2
Since x ≠ 5%2F2, f(x) ≠ 5%2F2%2B2 or 9%2F2
So the graph is this line with a hole:

with a hole at the point H(5%2F2,9%2F2)
There are no asymptotes.
Edwin

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=%282x%5E2-x-10%29%2F%282x-5%29 is a rational function (meaning a function involving at most polynomials and maybe quotients of polynomials). If you thought you would not have to factor and divide polynomials ever again, you were wrong.
With rational functions, you have to factor, or divide often. It is essential figure out what happens when a denominator is zero. For values of x that make a denominator zero, the function is undefined. It could be a hole, or it could be a vertical asymptote.
Factoring, we find:
f%28x%29=%282x%5E2-x-10%29%2F%282x-5%29=%282x-5%29%28x%2B2%29%2F%282x-5%29
2x-5=0 <---> x=5%2F2 and the function does not exist.
For any other x, f%28x%29=x%2B2
So x=5%2F2 is a hole in the line y=x%2B2.
No vertical asymptote. Just a hole.