SOLUTION: Stan is so good i had to come back for more. 1)Given the function y=2x-3/x-1 what are the a)x intercept b)y intercept c)vertical asymptote(s) d)horizontal asymptote(s)

Algebra ->  Rational-functions -> SOLUTION: Stan is so good i had to come back for more. 1)Given the function y=2x-3/x-1 what are the a)x intercept b)y intercept c)vertical asymptote(s) d)horizontal asymptote(s)      Log On


   

Question 156953: Stan is so good i had to come back for more.
1)Given the function y=2x-3/x-1 what are the
a)x intercept
b)y intercept
c)vertical asymptote(s)
d)horizontal asymptote(s)
e)slant asymptote
2)For the function f(x)=x^2(x-2)
a)find the X intercepts and for each intercept indicate whether the graph of F crosses (c) or merely touches (t)the X axis at each intercept.
b)find the Y intercept
c)Over what interval is f>0?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first one to get you going in the right direction


1)

a) x intercept


The x-intercept occurs when y=0.


y=%282x-3%29%2F%28x-1%29 Start with the given equation


0=%282x-3%29%2F%28x-1%29 Plug in y=0


0%28x-1%29=2x-3 Multiply both sides by x-1


0=2x-3 Multiply 0 and x-1 to get 0


3=2x Add 3 to both sides.


3%2F2=x Divide both sides by 2


So the answer is x=3%2F2 which means that the x-intercept is , which is (1.5, 0). Note: the x-intercept is in the form of (x, 0)


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b)y intercept


The y-intercept occurs when x=0



y=%282x-3%29%2F%28x-1%29 Start with the given equation


y=%282%280%29-3%29%2F%280-1%29 Plug in x=0


y=%280-3%29%2F%280-1%29 Multiply


y=%28-3%29%2F%28-1%29 Subtract


y=3 Reduce


So the answer is y=3 which means that the y-intercept is Note: the y-intercept is in the form of (0, y)


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c)vertical asymptote(s)


To find the vertical asymptote, just set the denominator equal to zero and solve for x (remember you CANNOT divide by zero)


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


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d)horizontal asymptote(s)


Since the degree of the numerator and the denominator are the same, we can find the horizontal asymptote using this procedure:

To find the horizontal asymptote, first we need to find the leading coefficients of the numerator and the denominator.

Looking at the numerator 2x-3, the leading coefficient is 2

Looking at the denominator x-1, the leading coefficient is 1

So the horizontal asymptote is the ratio of the leading coefficients. In other words, simply divide 2 by 1 to get %282%29%2F%281%29=2


So the horizontal asymptote is y=2


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e) slant asymptote

Since horizontal asymptote exists, there isn't a slant asymptote (you either have one or the other, but not both)


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Notice if we graph y=%282x-3%29%2F%28x-1%29, we can visually verify our answers:


Photobucket - Video and Image Hosting
Graph of y=%282x-3%29%2F%28x-1%29%29 with the x-intercept (or the point (1.5, 0)), the y-intercept (0,3), the horizontal asymptote y=2 (blue line) and the vertical asymptote x=1 (green line)



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With this info, you should have a better understanding on how to approach # 2. If you're still having trouble with # 2, then repost the question.