Write the equation of a rational function that meets the following conditions: VA: x=-4, x=3 RD: x coordinate of -7. HA: y=4/3

Question

Keep equation in factored form

Answer 1

The question asks for an equation with vertical asymptotes of -4 and 3; a horizontal asymptote of 4/3, and a removable discontinuity with an x coordinate of -7.

In order to answer this question, we need to know how factors of rational functions cause asymptotic and other behaviors. Take, for example, y = 1/(x+5). If the denominator is equal to zero, then the function is undefined and a vertical asymptote will result. Since x = -5 will make the denominator =0, there is a vertical asymptote at x = -5.

What if we are asked to write a function with two vertical asymptotes? There will be two factors in the denominator. If either one of the factors = 0, then the entire denominator = 0 and the function becomes undefined. For example, y = 1/[(x + 5)(x - 10)] would have a vertical asymptote at x = -5 and another at x = 10.

If you are trying to write a function that has a horizontal asymptote, one way is to make the the degree of the denominator greater than the degree of the numerator. For example, y=1/(x+5) will have a horizontal asymptote at y = 0. To move the horizontal asymptote up or down, simply add or subtract a constant to the fractional portion. For example, 1/(x+5) + 3 will have a horizontal asymptote at y = 3.

Finally, to create a removable discontinuity (or hole) in the function, just have the same factor in the numerator and the denominator. Putting it all together in an example, (x+6)/[(x-3)(x+6)(x-4)] -5 will have a removable discontinuity at x = -6, a horizontal asymptote at y = -5, and vertical asymptotes at x = 3 and x = 4. Note that you would not have both a hole and a vertical asymptote at x = -6, only a hole.