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December 19, 2014

December 19, 2014

Posted by **Rosie** on Wednesday, May 15, 2013 at 10:19pm.

f(x)=(2x^2+x-3)/(2x^2-7x)

1) Determine the symmetry of the function

2) Find the y-intercept

3) Find the x-intercept

4) Find the vertical asymptotes

5) Find the horizontal asymptotes

6) Plot points between and beyond the x-intercept and the vertical asymptotes.

Type the y-coordinate for each of the following points:

-5 -2 -1 2 5

7) What should the graph look like

- College Algebra HELP -
**Steve**, Thursday, May 16, 2013 at 12:11amf(x) = (2x+3)(x-1) / x(2x-7)

So, the denominator is zero at x=0,7/2 making vertical asymptotes

Top and bottom are both quadratics with x^2 coefficient=2, so there is a horizontal asymptote at y = 2/2 = 1

There is no y-intercept, because f(x) is not defined for x=0

It's easy to see where y=0

f(x) is neither even nor odd, so there is no symmetry

visit wolframalpha.com to check out the graph. Just type in the function

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