Question
Write the equation of a Rational Function satisfying given conditions.
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!!
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!!
Answers
Steve
You 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)
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)