What function has 2 vertical asymptotes at x=-3/2 and x=3/2 and a horizontal asymptote at 3/4? I know that the coefficients of the rational function are 3 and 4 but I don't know how to figure anything else out.
2 answers
It should look like a corner in the II quadrant and IV quadrant and a -x^3 function in the middle.
vertical asymptotes: y = 1/((x+3/2)(x-3/2))
for a nonzero horizontal asymptote, the top and bottom must have the same degree, so
horizontal asymptote: y = 3x^2/(4(x^2 - 9/4))
If you want corners in QII and QIV, you'll have to adjust things a bit.
for a nonzero horizontal asymptote, the top and bottom must have the same degree, so
horizontal asymptote: y = 3x^2/(4(x^2 - 9/4))
If you want corners in QII and QIV, you'll have to adjust things a bit.
1 answer