Find an equation of a rational function f that satisfies the given conditions.

Question

Given Conditions:

 

vertical asymptotes: x = −1, x = 0 horizontal asymptote: y = 0 x-intercept: 3; f(

Answer 1

The vertical asymptotes are the x-values in which the graph gets closer to but never meets. This is because those x values are undefined.  When we set our denominator to zero in a rational function or expression, it makes the whole thing undefined.

 

Since the vertical asymptotes are x = -1 and x = 0, our factors of the denominator are x and (x + 1).  If we multiply them out, we will have x² + x.  The denominator of the rational function is  x² + x.

 

Horizontal asymptotes are y-values in which the graph get closer to but never meets.  There are two ways to find the horizontal asymptotes.  We can evaluate the limit of the function as x approaches infinity from the left or right.  You will learn this in calculus. Another way is use the degree of the polynomial in the numerator and denominator. 

 

If the degree of the numerator equals the degree of the denominator, the horizontal is the coefficient of the numerator's degree divided by the coefficient of the denominator's degree.

 

If the degree of the numerator is less than the degree of the denominator, the horizontal is zero.

If the degree of the numerator is more than the degree of the denominator, there is no steady horizontal.

 

Since the degree of the denominator is 2 because the leading term is x², and the horizontal asymptote is y=0, then the leading term of the numerator must be x.  The degree is 1.   1<2.

 

So far, the function will look like this:

 

f(x) = x / (x² + x)

 

The x-intercept is value of x when the numerator is set to zero.  This means f(x) = 0. x-intercept = 3.

 

x = 3

 

Subtract 3 on both sides of equation to make the right side equal zero.

 

x - 3 = 0

 

The numerator is (x - 3).

 

Our function final function is

 

f(x) = (x - 3) / (x² + x)