### Assignments:

Unfinished Assignment Study Questions for Lesson 22

### Lesson Objectives:

- Define a Rational Function
- Vertical Asymptotes
- Horizontal Asymptotes
- Oblique Asymptotes
- Holes A rational function is a quotient of two polynomials, p(x) and q(x).

So it's of the form f(x) = (p(x)/q(x)), where q(x) != 0.

The domain of x includes all values of x for which q(x) != 0. A vertical asymptote for a rational function is an imaginary vertical line, x=a, where a is the zero of the denominator, but not the numerator. The function approaches but never crosses this line. The horizontal asymptote is an imaginary horizontal line which the graph of a function approaches and may cross.

Given f(x) = (p(x)/q(x)), if our degrees are the same, then y = a/b, where a is the coefficient of our numerator and b is the coefficient of denominator.

And if the degree of our numerator is less than our degree of our denominator, then y = 0 is our asymptote.

And then if the degree of our numerator is greater than the degree of our denominator, then a horizontal asymptote does not exist. Determine the vertical and horizontal asymptotes of h(x).

h(x) = (x+7)/(3-x). So to find our vertical asymptote, we need to find the zero of our denominator. So set 3-x = 0, and solve for x. We get x = 3.

Now let's find our horizontal asymptote. So for our horizontal asymptote, we have to look at the degree of our numerator and denominator. And they're both degree 1. Our horizontal asymptote is y equals the coefficient of our numerator, 1, over the coefficient of our denominator, -1.

So if we draw this on a graph, x = 3 is an imaginary vertical line over here, and y = -1 is an imaginary horizontal line over here. And try plugging this function into your calculator, and you'll get a graph that looks like this. Oblique Asymptote or Slant Asymptote.

It's a slanted line which the graph of a function approaches and may cross.

For the rational function f(x) = (p(x)/q(x)), where p(x) represents a polynomial function in the numerator, and q(x) represents a polynomial function in the denominator, where q(x) != 0; if the degree of your numerator is 1 more than the degree of your denominator, then there exists an oblique asymptote, which is equal to the quotient of (p(x)/q(x)).

So in order to find your oblique asymptote, divide (p(x)/q(x)), and throw out the remainder.

Remember that the graph of a function cannot have an oblique asymptote if it already has a horizontal asymptote. Find all the asymptotes, the domain and the x and y intercepts and then graph the function.

We're given f(x) = (x^2-5x-9)/(x+1).

Let's find the zero of our denominator. x+1 = 0 when x = -1. This means that our domain is (-oo, -1) cup (-1, oo) in interval notation, or you can also write this in set notation, {x| x != 1}. x = -1 can be our vertical asymptote, since it's only a zero for our denominator, and not our numerator.

Now, for our horizontal asymptote, we need to compare the degree of our numerator and denominator. We have a degree of 2 and a degree of 1. Since the degree of our numerator is greater than the degree of our denominator, that means that there is no horizontal asymptote.

But since the degree of our numerator is greater than the degree of our numerator, by 1, that means that we have an oblique asymptote, and we can find that by dividing our numerator by our denominator.

So let's use long division. x^2-5x-9 divide by x+1. So we have x(x+1) gives us x^2+x, and then we subtract, and we get -6x and bring down the -9. And then we have -6 here, because when we multiply that by x+1, we get -6x-6. And then we subtract, and then we get -3 for our remainder. Our oblique asymptote is y = x-6.

For our y-intercept, we need to find f(0). That's (0^2-5(0)-9)/(0+1), or -9. So the y-intercept is at the point (0, -9).

And for our x-intercept, we need to solve f(x) = 0, or 0 = (x^2-5x-9)/(x+1). If we start by multiplying everything by our denominator, x+1, x+1 cancels, and (x+1)0 = 0, and we're left with 0 = x^2-5x-9. And we can solve this using the quadratic formula, where a = 1, b = -5," and "c = -9.

So x is equal to (-b+-sqrt(b^2-4ac))/(2a). x is about 6.4 and -1.4. So our x-intercepts are at (6.4, 0) and (-1.4, 0).

So now let's graph our function using all of the information we just gathered. We can draw the vertical asymptote at x = -1, and there is no horizontal asymptote, and then we have an oblique asymptote at y = x-6. This is the same as the graph of y = x, which goes straight through the origin, except it's shifted down 6 units.

The x-intercept of our oblique asymptote is at (6, 0). And then our y-intercept is at the point (0, -9) and our x-intercept is at (6.4, 0), and our other x-intercept is at (-1.4, 0). So now we can draw our graph. Holes.

For any rational function, f(x) = (p(x)/q(x)), where denominator is not equal to zero.

There is a hole or open circle at the point (a, f(a)), where 'a' is both a zero of the numerator and denominator. Find all of the asymptotes, the domain and the x and y intercepts and then graph the function.

f(x) = (2x^2+7x-4)/(x^2+2x-8).

So since the degree of our numerator is the same as the degree of our denominator, we know that we have a horizontal asymptote at y = 2/1.

Now before we can find our vertical asymptote, we have to factor our numerator and denominator. So we get ((2x-1)(x+4))/((x-2)(x+4)). So the zeros of our denominator are 2 and -4, but -4 is also a zero of our numerator, so our vertical asymptote only includes the 2. So our vertical asymptote is x = 2.

Now when the numerator and the denominator share a factor, or share a zero, of -4, that means we have a hole at the point (-4, f(-4)). We'll need to cancel out the x+4's, so f(-4) = (2(-4)-1)/(-4-2), which is equal to -8-1, which is -9, over -4-2, which is -6. Let's cancel out our negative terms and divide the numerator and denominator by 3. So we have 3/2 or 1.5.

So our hole is at the point (-4, 1.5).

Now the domain cannot include any values of x for which the denominator is zero. So it doesn't include the value 2 and the value -4. So in set notation, this is written {x| x != -4, x != -2}. Or in interval notation, this is (-oo, -4) cup (-4, 2) cup (2, oo), where we have parentheses at 2 and -4, since our graph does not include values of x that are 2 and -4.

Now let's find our x-intercept. We have to set our function equal to zero. Then multiply both sides by the denominator and factor 2x^2+7x-4, so we get 0 = (2x-1)(x+4). So 2x-1 = 0 or x+4 = 0, or x = 1/2 or x = -4. So our x-intercept is (1/2, 0). It doesn't include the point x = -4, because that point is not included in our domain. Now let's find our y-intercept. We have to plug 0 in for x. And we get (2(0)^2+7(0)-4)/((0)^2+2(0)-8). So our zero terms cancel, our negative signs cancel, and we're left with f(0) = 1/2. So our y-intercept is at the point (0, 1/2). Now all that's left to do is to use all of our information to graph our function.
So we have a horizontal asymptote at y = 2, and a vertical asymptote at x = 2, and a hole at (-4, 1.5), and an x-intercept at (1/2, 0), and a y-intercept at (0, 1/2).

Our graph looks like this.