Question 156953
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=(2x-3)/(x-1)}}} Start with the given equation



{{{0=(2x-3)/(x-1)}}} Plug in {{{y=0}}}



{{{0(x-1)=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.



{{{3/2=x}}} Divide both sides by 2



So the answer is {{{x=3/2}}} which means that the x-intercept is *[Tex \LARGE \left(\frac{3}{2},0\right)], 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=(2x-3)/(x-1)}}} Start with the given equation



{{{y=(2(0)-3)/(0-1)}}} Plug in {{{x=0}}}



{{{y=(0-3)/(0-1)}}} Multiply



{{{y=(-3)/(-1)}}} Subtract



{{{y=3}}} Reduce



So the answer is {{{y=3}}} which means that the y-intercept is *[Tex \LARGE \left(0,3\right)]  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 <b>CANNOT</b> divide by zero)



{{{x-1=0}}} Set the denominator equal to zero



{{{x=0+1}}}Add 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 {{{(2)/(1)=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=(2x-3)/(x-1)}}}, we can visually verify our answers:



<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/rational_graph.png" alt="Photobucket - Video and Image Hosting">

Graph of {{{y=(2x-3)/(x-1))}}} with the x-intercept *[Tex \LARGE \left(\frac{3}{2},0\right)] (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.