Holes & Slant Asymptotes Rational Functions Holes & Slant Asymptotes Holt McDougal Algebra 2 Holt Algebra 2

Now we find out why we have to start out factoring….to find holes in the graph.

In some cases, both the numerator and the denominator of a rational function will equal 0 for a particular value of x. As a result, the function will be undefined at this x-value. If this is the case, the graph of the function may have a hole. A hole is an omitted point in a graph.

Example : Graphing Rational Functions with Holes Identify holes in the graph of f(x) = . Then graph. x2 – 9 x – 3 Factor the numerator. (x – 3)(x + 3) x – 3 f(x) = The expression x – 3 is a factor of both the numerator and the denominator. Set it = to 0 to find the x- coordinate of the hole. There is a hole in the graph at x = 3. (x – 3)(x + 3) (x – 3) For x ≠ 3, f(x) = = x + 3 Divide out common factors.

To find the y-coordinate of the hole plug 3 into the reduced equation. Example Continued The graph of f is the same as the graph of y = x + 3, except for the hole at x = 3. On the graph, indicate the hole with an open circle. The domain of f is all real #’s except 3 Hole at x = (3, 6) To find the y-coordinate of the hole plug 3 into the reduced equation.

Check It Out! Example 5 Identify holes in the graph of f(x) = . Then graph. x2 + x – 6 x – 2 Factor the numerator. (x – 2)(x + 3) x – 2 f(x) = The expression x – 2 is a factor of both the numerator and the denominator. Set it = to 0 to find the x- coordinate of the hole. There is a hole in the graph at x = 2. For x ≠ 2, f(x) = = x + 3 (x – 2)(x + 3) (x – 2) Divide out common factors.

Check It Out! Example 5 Continued The graph of f is the same as the graph of y = x + 3, except for the hole at x = 2. On the graph, indicate the hole with an open circle. The domain of f is {x|x ≠ 2}. Hole at x = 2

Domain: Vertical Asymptotes: Horizontal Asymptotes: Holes: Intercepts:

Domain: Vertical Asymptotes: Horizontal Asymptotes: Holes: Intercepts:

Domain: Vertical Asymptotes: Horizontal Asymptotes: Holes: Intercepts:

Let’s go back and look at the worksheet and find the problems with holes.

Slant Asymptotes Slant asymptotes occur when the degree of the numerator is exactly one bigger than the degree of the denominator. In this case a slanted line (not horizontal and not vertical) is the function’s asymptote. To find the equation of the asymptote we need to use long division – dividing the numerator by the denominator.

When dividing to find slant asymptotes: Do synthetic division (if possible); if not, do long division! The resulting polynomial (ignoring the remainder) is the equation of the slant asymptote.

EXAMPLE: Finding the Slant Asymptote of a Rational Function Rational Functions and Their Graphs EXAMPLE: Finding the Slant Asymptote of a Rational Function Find the slant asymptotes of f (x) = Solution Because the degree of the numerator, 2, is exactly one more than the degree of the denominator, 1, the graph of f has a slant asymptote. To find the equation of the slant asymptote, divide x - 3 into x2 - 4x - 5: 1 -4 -5 3 -3 1 -1 -8 3 Remainder more

EXAMPLE: Finding the Slant Asymptote of a Rational Function 3.6: Rational Functions and Their Graphs EXAMPLE: Finding the Slant Asymptote of a Rational Function Find the slant asymptotes of f (x) = Solution The equation of the slant asymptote is y = x - 1. Using our strategy for graphing rational functions, the graph of f (x) = is shown. -2 -1 4 5 6 7 8 3 2 1 -3 Vertical asymptote: x = 3 Slant asymptote: y = x - 1

Graph: Notice that in this function, the degree of the numerator is larger than the denominator. Thus n>m and there is no horizontal asymptote. However, if n is one more than m, the rational function will have a slant asymptote. To find the slant asymptote, divide the numerator by the denominator: The result is . We ignore the remainder and the line is a slant asymptote.

1st, find the vertical asymptote. Graph: 1st, find the vertical asymptote. 2nd , find the x-intercepts: and 3rd , find the y-intercept: 4th , find the horizontal asymptote. none 5th , find the slant asymptote: 6th , sketch the graph.

A Rational Function with a Slant Asymptote Graph the rational function Factoring:

A Rational Function with a Slant Asymptote Finding the x-intercepts: –1 and 5 (from x + 1 = 0 and x – 5 = 0) Finding the y-intercepts: 5/3 (because )

A Rational Function with a Slant Asymptote Finding the horizontal asymptote: None (because degree of numerator is greater than degree of denominator) Finding the vertical asymptote: x = 3 (from the zero of the denominator)

A Rational Function with a Slant Asymptote Finding the slant asymptote: Since the degree of the numerator is one more than the degree of the denominator, the function has a slant asymptote. Dividing, we obtain: Thus, y = x – 1 is the slant asymptote.

A Rational Function with a Slant Asymptote Here are additional values and the graph.

Slant Asymptotes and End Behavior So far, we have considered only horizontal and slant asymptotes as end behaviors for rational functions. In the next example, we graph a function whose end behavior is like that of a parabola.

Finding a Slant Asymptote If There will be a slant asymptote because the degree of the numerator (3) is one bigger than the degree of the denominator (2). Using long division, divide the numerator by the denominator.

Finding a Slant Asymptote Con’t.

Finding a Slant Asymptote Con’t. We can ignore the remainder The answer we are looking for is the quotient and the equation of the slant asymptote is

Graph of Example 7 The slanted line y = x + 3 is the slant asymptote

Graph the rational function which has the following characteristics Vert Asymp at x = 1, x = -3 Horz Asymp at y = 1 Intercepts (-2, 0), (3, 0), (0, 2) Passes through (-5, 2)

Graph the rational function which has the following characteristics Vert Asymp at x = 1, x = -1 Horz Asymp at y = 0 Intercepts (0, 0) Passes through (-0.7, 1), (0.7, -1), (-2, -0.5), (2, 0.5)

Identify the zeros, asymptotes, and holes in the graph of . Then graph. x2 – 3x + 2 x2 – x f(x) = zero: (2,0); asymptotes: x = 0, y = 1; hole at (1, -1)