Question

For the following exercise (question 45), find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptotes of the function. use that information to sketch graph.

Answer 1

Rational functions

A rational function is given by the quotient between two polynomials. Assuming that we have two polynomials p(x) and q(x) then f(x) is a rational function:

`f(x)=(p(x))/(q(x))`

The interceptions and asymptotes of this type of functions meet the following:

- The horizontal intercepts or x-intercepts of f(x) are located at the x values for which p(x)=0 and q(x)≠0.

- The vertical intercept or y-intercept is located at the y value equal to f(0): (0,f(0)).

- The vertical asymptotes and blank points are located at the x values where q(x)=0.

- The horizontal asymptote depends on the degrees of p(x) and q(x). If the degree of q(x) is greater than that of p(x) then there's a horizontal asymptote at y=0. If they have the same degree then there's a horizontal asymptote at y=H where H is the quotient between the leading coefficient of p(x) and the leading coefficient of q(x).

- If the degree of p(x) is greater than that of q(x) then there's a slant asymptote given by the quotient of the long division between p(x) and q(x).

Analyzing a(x)

Now that we know how we can find the asymptotes and intercepts of a(x). First of all we should factor both the numerator and the denominator of a(x). Let's start with the numerator that we'll call p(x). Since it's a quadratic expression its factored form looks like this:

`p(x)=a(x-r_1)(x-r_2)`

Where a is the leading coefficient and the r's are its zeros i.e. the x values for which the numerator is 0. The leading coefficient is the number multiplying the squared x that in this case is 1. For the zeros we can use the following formula for a quadratic expression of the form ax²+bx+c:

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

In this case a=1, b=2 and c=-3 so we get:

`\begin{gathered} x=\frac{-2\pm\sqrt[]{2^2-4\cdot1\cdot(-3)}}{2\cdot1}=\frac{-2\pm\sqrt[]{4+12}}{2}=\frac{-2\pm\sqrt[]{16}}{2}=(-2\pm4)/(2) \\ x=1\text{ and }x=-3 \end{gathered}`

Then the numerator can be written like this:

`p(x)=(x+3)(x-1)`

Now let's find the factored form of the denominator. We can equalize it to 0 to find its zeros:

`x^2-1=0`

And we add 1 to both sides

`\begin{gathered} x^2-1+1=0+1 \\ x^2=1 \end{gathered}`

And we apply a square root to both sides:

`\begin{gathered} \sqrt[]{x^2}=\sqrt[]{1} \\ |x|=1 \\ x=1\text{ and }x=-1 \end{gathered}`

So the denominator can be written like:

`q(x)=(x+1)(x-1)`

Then a(x) is equal to:

`a(x)=((x+3)(x-1))/((x+1)(x-1))`

As you can see, the denominator of this expression is equal to 0 at x=1 and x=-1. This means that in those values we can have two things blank points and/or vertical asymptotes. If we can simplify the factors associated to any of those values and make them dissapear then in that value the function has a blank point. If we can't make the term dissapear then at that point we have a vertical asymptote. We can't simplify the term associated to x=-1 (this is x+1) but we can simplify that associated to x=1 (this one is x-1):

`a(x)=((x+3)(x-1))/((x+1)(x-1))=(x+3)/(x+1)`

Then there is avertical asymptote at x=-1. For the blank point we can evaluate the expression after simplifying at x=1:

`(x+3)/(x+1)\rightarrow(1+3)/(1+1)=(4)/(2)=2`

So there's a blank point at (1,2).

Since both p and q have a degree of 2 the horizontal asymptote is given by the quotient between their leading coefficients. since these are equal to 1 the horizontal asymptote is y=1.

Now let's find the intercepts. For the vertical intercept we find a(0):

`a(0)=((0+3)(0-1))/((0+1)(0-1))=-(3)/(-1)=3`

So the vertical intercept is (0,3).

The horizontal intercepts are the points (c,a(c))=(c,0). These are given by the zeros of the numerator i.e the zeros of p(x). These were x=-3 and x=1, however, in x=-1 we had a blank point so the only horizontal intercept is (-3,0).

In order to sketch the graph of a(x) we just need to marked the two intercepts, the blank point as an empty dot and the two asymptotes (these can be dashed lines). Then the graph should look like this:

The interceptions are the blue dots, the blank point is the empty one, the asymptotes are the dashed green lines and the graph of a(x) is the red one.