Graph Rational Functions

Rational Functions - How it Works - Video

Rational Function Definition

Rational Function Definition:

Remember the definition of a rational number: is number that can be written as ratio or mathematically t/v where t and v can be any integer and v ≠ 0.

Except this time we don't have individual numbers we have functions, polynomials, or equations or however you might to define it.

One thing is that we have to be careful when writing the domain and range especially with h(x) since we can't have zeros in the denominator.

Notation

Notation:

Here we have some notation.

The key take away is that if we have x and the arrow symbol we move left to right (negative sign) or right to left (positive sign). And, if we have f(x), we move up (positive ∞) or down (negative ∞).

Definition of Vertical Asymptote

Definition of Vertical Asymptote:

We have four scenarios with vertical asymptotes. We graph a dashed vertical line (x = some number) to represent that.

1 - The graph approaches the dashed line from the left and goes downward or f(x) → -∞ as x → w-.

2 - The graph approaches the dashed line from the right and goes downward or f(x) → -∞ as x → w+.

3 - The graph approaches the dashed line from the right and goes upward or f(x) → +∞ as x → w+.

4 - The graph approaches the dashed line from the left and goes upward or f(x) → +∞ as x → w-.

Definition of Horizontal Asymptote

Definition of Horizontal Asymptote:

We have four scenarios with horizontal asymptotes. We graph a dashed horizontal line (y = some number) to represent that. The first two have the same notation as well as the the last two.

1 - The graph approaches the dashed line from the bottom and goes to the right or f(x) → q as x → +∞.

2 - The graph approaches the dashed line from the top and goes to the right or f(x) → q as x → -∞.

3 - The graph approaches the dashed line from the top and goes to the left or f(x) → q as x → -∞.

4 - The graph approaches the dashed line from the bottom and goes to the left or f(x) → q as x → +∞.

Theorem on Horizontal Asymptotes

Theorem on Horizontal Asymptotes:

We have 3 scenarios when we have f(x) = g(x)/h(x).

1 - The biggest power of g(x) is less than the biggest power of h(x) or n < k which means we have a horizontal asymptote at y = 0.

2 - The power of g(x) is equal to the power of h(x) or n = k which means we have a horizontal asymptote at y = [the leading coefficient of g(x)] divided by [the leading coefficient of h(x)].

3 - The biggest power of g(x) is greater than the biggest power of h(x) or n > k which means we do not have a horizontal asymptote.

Examples on Theorem on Horizontal Asymptotes

Examples on Theorem on Horizontal Asymptotes:

We have 3 examples on the scenarios when we have f(x) = g(x)/h(x).

1 - The biggest power of g(x) is less than the biggest power of h(x) or n < k which means we have a horizontal asymptote at y = 0.

2 - The power of g(x) is equal to the power of h(x) or n = k which means we have a horizontal asymptote at y = [the leading coefficient of g(x)] divided by [the leading coefficient of h(x)].

3 - The biggest power of g(x) is greater than the biggest power of h(x) or n > k which means we do not have a horizontal asymptote.

Slant (Oblique) Asymptotes - How to Find Them

Slant (Oblique) Asymptotes - How to Find Them:

In scenario 3 of horizontal asymptotes sometimes, we can get slant (oblique) asymptotes. That occurs when n is greater than k by 1. In this scenario we 2 for n and 1 for k. We don't have a horizontal asymptote, but we do have a slant (oblique) asymptote.

To find the slant (oblique) asymptote we have to use long division or synthetic division when applicable. We can write the f(x) = g(x)/h(x) as f(x) = (ax + b) + r(x)/h(x). The (ax + b) is our slant (oblique) asymptote.

Guidelines for Sketching the Graph

Guidelines for Sketching a Rational Function:

Here are the first 4 guidelines.

Guidelines for Sketching a Rational Function:

Here are the last 4 guidelines.

Example 1

Example 1:

Guidelines 1 - We want to simplify and find any holes.

We have f(x) = g(x)/h(x) where g(x) = 3x - 4 and h(x) = 2x + 5.

We cannot simplify here as we don't have any factors that are the same in the numerator and denominator. Since we didn't cancel anything, we don't have any holes or points of discontinuity.

Example 1:

Guidelines 2 and 3 - We want to find the vertical and horizontal asymptotes.

We have f(x) = g(x)/h(x) where g(x) = 3x - 4 and h(x) = 2x + 5.

We set h(x) = 3x + 5 = 0 and solve for x and we get x = -3/5 as our vertical asymptote.

Since we have highest power in the numerator, 3x1, is the same the denominator, 2x1, we have n = k. This means that we have to find the coefficients of each g(x) and h(x) and divide them so we can find our horizontal asymptote, which is y = 3/2.

We don't have any slant (oblique) asymptotes.

Example 1:

Guidelines 4 and 5 - We want to find the x-intercept(s) and the y-intercept if there are any.

We have f(x) = g(x)/h(x) where g(x) = 3x - 4 and h(x) = 2x + 5.

We set f(x) = (3x - 4)/(2x + 5) = 0 and solve for x. When we cross multiply the denominator will always be multiplied by 0 so we could just set g(x) equal to 0 to find our x-intercept(s). After solving for x, we get x = 4/3. So our point is (4/3, 0).

Now we need to find the y-intercept and to do that we set our input to 0. After solving for f(0), we get f(0) = -4/5, So our point is (0, -4/5).

Example 1:

Guidelines 6 - We determine if the graph intersects any of the asymptotes.

We have f(x) = g(x)/h(x) where g(x) = 3x - 4 and h(x) = 2x + 5.

We set f(x) = (3x - 4)/(2x + 5) = 3/2 our horizontal asymptote and solve for x. After we cross multiply and solve for x, we get x 15 ≠ -8. Since that is not true, our graph does not cross the horizontal asymptote.

Since we don't have a slant (oblique) asymptote, we don't have to worry about that one.

Example 1:

Guidelines 7 and 8 - We create an XY table and graph.

Our vertical asymptote is x =-5/2 = -2.5. So we used that point as our middle point and we went 3 points to the left and 3 points to the right. We skipped -2 and -3 because those points are not on the graph, and as for last we picked 4/3 instead of 1 because we found it during step/guideline 4.

After finding each point, by substituting each input value, we created our points. Then we plotted and graphed our function.

Example 2

Example 2:

Guidelines 1 - We want to simplify and find any holes.

We have f(x) = g(x)/h(x) where g(x) = x - 3 and h(x) = x2 - 5x + 6.

We can simplify. Now when we have a degree 2 polynomial like we do with h(x), we should factor it so that we can see whats with our graph easier. It will find the vertical asymptotes.

h(x) = x2 - 5x + 6 can be factored as (x - 2)(x - 3). Wooh! One of these match g(x). That means we can cancel them out and make our f(x) easier to work with.

Since we can cancel those factors out, our new f(x) is 1/(x - 2). When we canceled, we "lost" a problem since we cannot have zero in the denominator. So we substitute 3 into our new function and our result is 1. So our point is (3, 1). This point is a point of discontinuity since we cannot have 3 in our original f(x). So we have found a hole, and we can the point (3, 1) as an empty/closed circle.

Example 2:

Guidelines 2 and 3 - We want to find the vertical and horizontal asymptotes.

We have f(x) = g(x)/h(x) where g(x) = x - 3 and h(x) = x2 - 5x + 6.

We set our new h(x) = x - 2 = 0 and solve for x and we get x = 2 as our vertical asymptote.

Since our highest power in the numerator is x0 and the highest power in the denominator is x1, we have n < k, which means we have a horizontal asymptote at y = 0 (the x-axis).

We don't have any slant (oblique) asymptotes.

Example 2:

Guidelines 4 and 5 - We want to find the x-intercept(s) and the y-intercept if there are any.

We have f(x) = g(x)/h(x) where g(x) = x - 3 and h(x) = x2 - 5x + 6.

We set our new f(x) = 1/(x - 2) = 0 and solve for x. When we cross multiply the denominator will always be multiplied by 0 so we could just set our new g(x) equal to 0 to find our x-intercept(s). Our result is 0 ≠ 1, which means we don't have any x-intercepts. This makes sense since we have a horizontal asymptote at y = 0 (the x-axis).

Now we need to find the y-intercept and to do that we set our input to 0. After solving for f(0), we get f(0) = -1/2, So our point is (0, -1/2).

Example 2:

Guidelines 6 - We determine if the graph intersects any of the asymptotes.

We have f(x) = g(x)/h(x) where g(x) = x - 3 and h(x) = x2 - 5x + 6.

We set our new f(x) = 1/(x - 2) = 0 our horizontal asymptote and solve for x. After we cross multiply and solve for x, we get x 01. Since that is not true, our graph does not cross the horizontal asymptote.

Since we don't have a slant (oblique) asymptote, we don't have to worry about that one.

Example 2:

Guidelines 7 and 8 - We create an XY table and graph.

We have f(x) = g(x)/h(x) where g(x) = x - 3 and h(x) = x2 - 5x + 6.

Let's use the new f(x), to create our XY table. Our vertical asymptote is x = 2. So we used that point as our middle point and we went 3 points to the left and 3 points to the right. Remember the point (3, 1) is a hole.

After finding each point, by substituting each input value, we created our points. Then we plotted and graphed our function.