50 ! ! construct a rational function that has a vertical asymptote at x=3 and a removable discontinuity at x=-2. explain how you determined your answer.

The vertical asymptote means that you need an equation that equals 0, so when x=3 you would have x - 3 at that value it zeroes the denominator.

The removable discontinuity means you need an equation that equals 0, so when x=-2, the equation would be x + 2 at that value it zeroes both the numerator and denominator.

So now you have x-3 as part of the denominator and x+2 as part of both the numerator and denominator.

The rational function then becomes:

f(x) = (x+2) / (x+2) (x-3)  

or

Simplifying the denominator you have f(x) = (x+2) / (x^2 -x -6)

The vertical asymptote means that you need an equation that equals 0, so when x=3 you would have x - 3 at that value it zeroes the denominator.

The removable discontinuity means you need an equation that equals 0, so when x=-2, the equation would be x + 2 at that value it zeroes both the numerator and denominator.

So now you have x-3 as part of the denominator and x+2 as part of both the numerator and denominator.

The rational function then becomes:

f(x) = (x+2) / (x+2) (x-3)  

or

Simplifying the denominator you have f(x) = (x+2) / (x^2 -x -6)

i’m here for points.

step-by-step explanation: