SOLUTION: f(x)=(5x+1)/(x^2-5x+6).... find vertical asymptotes

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Question 611123: f(x)=(5x+1)/(x^2-5x+6).... find vertical asymptotes

Found 2 solutions by jsmallt9, ewatrrr:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
The vertical asymptotes occurs for x values that make a denominator zero, if any. So all you have to do is find out what x values make the denominator zero. In other "words", solve:
x%5E2-5x%2B6+=+0

Since this is a quadratic equation we want one side to be zero. The right side is already zero so we can proceed directly to the next step: Factor or use the Quadratic Formula. This factors quite easily:
(x-2)(x-3) = 0
From the Zero Product Property we know that a product can be zero onlyif one (or more of the factors is zero. So:
x-2 = 0 or x-3 = 0
Solving these we get:
x = 2 or x = 3
These are the equations of the vertical asymptotes of f(x).

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi,
f(x)= (5x+1)/(x^2-5x+6)
f(x)=%285x%2B1%29%2F%28%28x-3%29%28x-2%29%29 Note: denominator CANNOT = 0
x = 3 and x = 2 are the vertical asymptotes
Might recommend the FREE graph software at http://www.padowan.dk.com
See Graph: