SOLUTION: Find an equation of a rational function that satisfies the following conditions: • Vertical asymptotes: x = −3 • Horizontal asymptote: y=3/2 • x -intercept: 5 • Hole

Algebra ->  Rational-functions -> SOLUTION: Find an equation of a rational function that satisfies the following conditions: • Vertical asymptotes: x = −3 • Horizontal asymptote: y=3/2 • x -intercept: 5 • Hole       Log On


   

Question 1153467: Find an equation of a rational function that satisfies the following conditions:
• Vertical asymptotes: x = −3
• Horizontal asymptote: y=3/2
• x -intercept: 5
• Hole at x =2
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


vertical asymptote at x = -3:
This requires a factor of (x+3) in the denominator, without a like factor in the numerator:
1%2F%28x%2B3%29

horizontal asymptote at y = 3/2:
(We will take care of this last)

x-intercept at x=5:
This requires a factor of (x-5) in the numerator, without a like factor in the denominator:
%28x-5%29%2F%28x%2B3%29

Hole at x=2:
This requires factors of (x-2) in BOTH numerator and denominator:
%28%28x-5%29%28x-2%29%29%2F%28%28x%2B3%29%28x-2%29%29

horizontal asymptote at y = 3/2:
This requires the numerator and denominator to be the same degree, with the ratio of leading coefficients 3:2. The degrees of the numerator and denominator are the same at this point; we just need to add constant factors to make the ratio of the leading coefficients equal to 3/2.
%283%28x-5%29%28x-2%29%29%2F%282%28x%2B3%29%28x-2%29%29

A graph showing the vertical asymptote at x = -3 and the x-intercept at x=5:



A graph showing the horizontal asymptote at y = 3/2:



The graphing utility used on this site won't show the hole in the graph at x=2. A good graphing calculator like the TI-83 will show it if you graph the function on a very small range of values of x either side of x=2.