SOLUTION: write a rational function that has a vertical asymptote at x=3, a hole at x=-2 and a horizontal asymptote at y=2/3

Question 1199790: write a rational function that has a vertical asymptote at x=3, a hole at x=-2 and a horizontal asymptote at y=2/3
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

(1) vertical asymptote at x=3: factor of (x-3) in the denominator, without a factor of (x-3) in the numerator

(2) hole at x=-2: factors of (x+2) in both numerator and denominator

(3) horizontal asymptote at y=2/3: leading coefficient 2 in the numerator; leading coefficient 3 in the denominator; and numerator and denominator of same degree

(1) and (2) together require two linear factors in the denominator and one in the numerator. Since the degrees have to be the same, we need another linear factor in the numerator. Since no zeros of the function are defined, we can choose any linear factor we want -- except another factor of either (x-3) or (x+2). For simplicity we can choose a factor of x in the numerator.

And lastly we need the required leading coefficients.

ANSWER (one of many possible!):

A graph, showing the horizontal asymptote....

The graph doesn't SHOW the vertical asymptote at x=3; but you can tell it is there.

The graph is not detailed enough to show the hole at x=-2; if you graph the function on a very narrow range like x=-2.01 to x=-1.99 with a good graphing utility, you will see the hole.

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