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January 31, 2015

January 31, 2015

Posted by **Karen** on Thursday, July 31, 2014 at 8:43am.

Has vertical asymptotes located at x-2 and x=-1

Has a horizontal asymptote located at y=0(x-axis)

y-intercept:(0,2) x-intercept (4,0)

R(x)= ???

what I did is,

R(x)= ax+b/c(x-2)(x+1)

then plug in (0,2), b=-4c

plug in(4,0), a=c

then what to do next?? Please help me with this I will appreciate that!!

- math(precalculus) -
**Steve**, Thursday, July 31, 2014 at 11:28amYou don't need any c. It can be absorbed into a and b.

Actually, you have solved the problem. Pick any value for c, and plug it in. As long as a=c and b=-4c, R(x) will work. So, make things easy. Let c=1.

As I worked it out,

R(x) = ???/(x+1)(x-2)

Since y=0 is the asymptote, you know that the degree of the numerator is less than the denominator. So,

R(x) = (ax+b) / (x+1)(x-2)

Now, with the intercepts, you know that

b/(1)(-2) = 2, so b = -4

a(4)+b = 0, so 4a-4=0, so a=1

R(x) = (x-4)/(x+1)(x-2)

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