SOLUTION: Find the asymptotes, intercepts, and graph: T(w)= wˆ2+2w+4 over w-1 ??

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Question 687897: Find the asymptotes, intercepts, and graph: T(w)= wˆ2+2w+4 over w-1 ??
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
T(w)= %28w%5E2%2B2w%2B4%29%2F%28w-1%29 

Change Y(w) to y and w to x

y = %28x%5E2%2B2x%2B4%29%2F%28x-1%29

That has a vertical asymptote.  Set denominator = 0

x-1 = 0
  x = 1

So here is the vertical asymptote:


It has no horizontal asymptote because the numerator's largest
power of x (its degree) which is 2 is larger that the denominator's largest
power of x (its degree) with is 1.  

However since the degree of the numerator is exactly one more
than the degree of the denominator, it has a slanted or "oblique" asymptote.

We find that by long division:

        x+3
x-1)x²+2x+4
    x²- x
       3x+4
       3x-3
          7

So y = x+3 + 7%2F%28x-1%29

And the fraction 7%2F%28x-1%29 gets smaller and smaller as x gets
larger and larger either positively or negatively, so the curve
approaches just being like the line y = x+3 ignoring the remainder
or denominator.  So we draw the slant asymptote y = x+3

   

The y-intercept is gotten by substituting x=0

y = %28x%5E2%2B2x%2B4%29%2F%28x-1%29
y = %280%5E2%2B2%2A0%2B4%29%2F%280-1%29
y = 4%2F%28-1%29
y = -4

So the y-intercept is (0,-4). There is no x-intercept because the numerator
is never zero. So we plot the y-intercept and get a few more points, maybe
(8,12),(-6,-4), (2,12), (-4,-2.4), (3,9.5), (-1,-1.5). and draw the graph:


Edwin