SOLUTION: Find all horizontal/vertical asymptotes and all x-intercepts: {{{(x^2-5x+6)/(x-1)}}}

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Question 165403: Find all horizontal/vertical asymptotes and all x-intercepts:
%28x%5E2-5x%2B6%29%2F%28x-1%29
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
x-intercepts:


y=%28x%5E2-5x%2B6%29%2F%28x-1%29%29 Start with the given equation


0=%28x%5E2-5x%2B6%29%2F%28x-1%29%29 Plug in y=0


x%5E2-5x%2B6=0 Set the numerator equal to zero


%28x-3%29%28x-2%29=0 Factor the left side (note: if you need help with factoring, check out this solver)



Now set each factor equal to zero:
x-3=0 or x-2=0

x=3 or x=2 Now solve for x in each case


So our answers are

x=3 or x=2



So the x-intercepts are (3,0) and (2,0)


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Asymptotes:

Looking at the numerator x%5E2-5x%2B6, we can see that the degree is 2 since the highest exponent of the numerator is 2. For the denominator x-1, we can see that the degree is 1 since the highest exponent of the denominator is 1.


Oblique Asymptote:

Since the degree of the numerator (which is 2) is greater than the degree of the denominator (which is 1), there is no horizontal asymptote. In this case, there's an oblique asymptote.


To find the oblique asymptote, simply use polynomial long division to find it.


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So the quotient is x-4, this means that the oblique asymptote is y=x-4




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Vertical Asymptote:
To find the vertical asymptote, just set the denominator equal to zero and solve for x

x-1=0 Set the denominator equal to zero


x=0%2B1Add 1 to both sides


x=1 Combine like terms on the right side


So the vertical asymptote is x=1


Notice if we graph y=%28x%5E2-5x%2B6%29%2F%28x-1%29, we can visually verify our answers:

Graph of y=%28x%5E2-5x%2B6%29%2F%28x-1%29%29 with the x-intercepts (3,0) and (2,0), the oblique asymptote y=x-4 (blue line) and the vertical asymptote x=1 (green line)