write an equation for graph given

Question

Write a formula for the function whose graph is shown below 

 

points (-1,-1)and (7,-4) given 

 

vertical and horizontal asymptotes given at y=-1/2x-1 and x=3 

 

I am unsure of where to start and how to formulate this formula

Answer 1

Lauren,

This is a rather long solution, but I'll do my best to make it clear. First, I need to point out that there is not just one equation that fits the given scenario. We just need to find one that does, so we'll keep it as simple as possible.

 

Let's recognize that there is no horizontal asymptote in this problem, but rather one vertical and one oblique (or slant) asymptote. The difference is important. Horizontal asymptotes only occur in rational functions where the degree of the numerator is ≤ the degree of the denominator. Slant asymptotes happen where the degree of the numerator is exactly one more than the degree of the denominator. So we use this fact to set up the simplest rational function we can in general form:

 

          f(x) = ax² + bx + c

h(x) = –––    –––––––––––

          g(x) =   dx + e

 

You can see that I have used a quadratic function in the numerator and a linear function in the denominator to satisfy the slant condition in the simplest form possible. However, five variables does not sound simple. We need to start paring this down. We'll begin with the vertical asymptote at x = 3. That means that the denominator must be zero when x = 3.

 

g(3) = 3d + e = 0

                  e = -3d  substitution gives:

g(x) = dx - 3d

 

Knowing that a slant asymptote is found by the polynomial division, we look at the polynomial division:

             -½x - 1

           ___________

dx - 3d)ax² + bx + c

       -[(-d/2)x² + (3d/2)x]

        =   0   -  x(b - 3d/2) + c     ⇒   a = -d/2

                         -(-dx + 3d)

                  =       0  + c - 3d     ⇒   -d = b - 3d/2 or b = d/2

 

Because there is no x left, c - 3d is a remainder from the division. It is not used for anything

If you look back, you'll see that we now have a, b, and e in terms of d. Only c is outstanding. Because we only need to find an equation that works, I'll pick a value of d and then calculate the corresponding values of a, b, and e. AFter that we solve for c.

 

I pick d = 2 

 

a = -1,   b = 1,    e = -6

 

            -x² + x + c

h(x) = ––––––––––––

               2x - 6

 

Now we use either of the two know points to solve for c:

 

h(7) = (-49 + 7 + c)/(14 - 6) = -4

                               -42 + c = -32

                                        c = 10

 

Final Equation:

 

           -x² + x + 10

h(x) = ––––––––––––

               2x - 6