Introduction to Rational Functions

Presentation

•

Mathematics

•

12th Grade

•

Hard

Joseph Anderson

FREE Resource

10 Slides • 20 Questions

Rational Functions: Finding asymptotes

Multiple Choice

Factor the rational function:

$f\left(x\right)\ =\ \frac{x+3}{x^2-8x+12}$

$\frac{x+3}{\left(x-3\right)\left(x-4\right)}$

$\frac{x+3}{\left(x-6\right)\left(x-2\right)}$

$\frac{x+3}{\left(x+3\right)\left(x+4\right)}$

$\frac{x+3}{\left(x+6\right)\left(x+2\right)}$

(x-2) and (x-6) will determine the domain of the function because we can NOT divide by zero.

Notice the factors in the denominator.

Multiple Choice

What values of x would make the denominator of f(x) zero?

$f\left(x\right)\ =\ \frac{x+3}{\left(x-6\right)\left(x-2\right)}$

x = -6 only

x = 2 only

x = 2 and x = 6

x = -2 and x = -6

Domain of f(x)

Since x = 2 and x = 6 make the denominator of f(x) zero, they cannot be in the domain of the function. All other real numbers will be included. So the domain of f(x) can be written:

(-∞,2)U(2,6)U(6,∞)

Notice that we use parenthesis here since the values 2 and 6 are NOT included.

Multiple Choice

What is the domain of h(x)?

$h\left(x\right)\ =\ \frac{x+2}{x^2-4}$

(Remember to ALWAYS factor first)

$\left(-\infty,-2\right)\cup\left(2,\infty\right)$

$\left(-\infty,-2\right]\cup\left[-2,2\right]\cup\left[2,\infty\right)$

$\left(-\infty,-2\right]\cup\left[2,\infty\right)$

$\left(-\infty,-2\right)\cup\left(-2,2\right)\cup\left(2,\infty\right)$

Vertical Asymptotes and Holes

Any number NOT in the domain of a rational function is either a hole or a vertical asymptote.
HOLES happen when factors cancel.
VERTICAL ASYMPTOTES happen when you set factors in the denominator equal to zero and solve.

Multiple Choice

Find the vertical asymptotes and holes of the following function:

$f\left(x\right)\ =\ \frac{x^2-3x}{3x^2+6x}$

Remember to factor first!

No Hole.

VA at x = -2

Hole at x = 0,

VA at x = -2

Hole at x = 3,

VA at x = 2

Hole at x = -2,

VA at x = 3

Finding the hole

Multiple Choice

What is the value of f(0)?

$f\left(0\right)\ =\ \frac{\left(0-3\right)}{3\left(0+2\right)}$ .

y = -3/0

y = -1/2

y = 0

y = -3/5

Holes as ordered pairs

So f(x) has a hole at (0,-1/2).

Zeros of f(x)

The zeros of a function come from the numerator.

First, factor the function. Any factors that cancel from the numerator and denominator become holes (excluded from the domain but not counted as zeros).
Set each factor in the numerator equal to zero and solve. Those are the x-intercepts or zeros of your function.

Multiple Choice

Find the zeros of j(x):

$j\left(x\right)\ =\ \frac{3x^2-8x-3}{x^2-9}$

x = 1/3

x = 3

x = -1/3

x = -3

y-intercepts

y-intercepts are found when you plug in zero for x. For example in the function j(x),
j(0) = -3/-9. So the y-intercept is at (0, 1/3).

Multiple Choice

What is the y-intercept of the following function?

$g\left(x\right)\ =\ \frac{x^2-x-12}{x^2-2x-8}$

(0, 3/2)

(0, 3)

(0, -3/2)

(0, -2)

Horizontal Asymptotes

Horizontal asymptotes are a special ratio of leading coefficients based on the degree of the polynomial in the numerator and the polynomial in the denominator.
Use the chart on the next slide to find the equation for the horizontal asymptotes of a few functions.

Multiple Choice

What is the horizontal asymptote of the function

$f\left(x\right)\ =\ \frac{3x^2-8x-3}{x^2-9}$

There is no horizontal asymptote

y = 3

y = 0

y = -3

Multiple Choice

What is the horizontal asymptote of the function

$h\left(x\right)\ =\ \frac{x^2-3x}{3x^2+6x}$

There is no horizontal asymptote

y = -1/3

y = 0

y = 1/3

Multiple Choice

What is the horizontal asymptote of the function

$j\left(x\right)\ =\ \frac{x^3-2x+7}{3x^2+6x}$

There is no horizontal asymptote

y = -1/3

y = 0

y = 1/3

Multiple Choice

What is the horizontal asymptote of the function

$k\left(x\right)\ =\ \frac{4x^2-5x}{3x^3+2x}$

There is no horizontal asymptote

y = -1/3

y = 0

y = 1/3

Use your "Graphing Rational Functions" Cheat sheet to answer the following quesitons.

Let's practice!

Rational Functions Overview

Multiple Choice

Factor the following function:

$g\left(x\right)\ =\ \frac{2x^2-4x-6}{x^2-3x-4}$

$g\left(x\right)\ =\ \frac{\left(2x-3\right)\left(x+1\right)}{\left(x-4\right)\left(x+1\right)}$

$g\left(x\right)\ =\ \frac{2\left(x+3\right)\left(x-1\right)}{\left(x+4\right)\left(x+1\right)}$

$g\left(x\right)\ =\ \frac{2\left(x-3\right)\left(x+1\right)}{\left(x-2\right)\left(x-2\right)}$

$g\left(x\right)\ =\ \frac{2\left(x-3\right)\left(x+1\right)}{\left(x-4\right)\left(x+1\right)}$

Multiple Choice

g(x) has a hole at:

x = 1

x = 3

x = -1

x = 4

Multiple Choice

The y-value of the hole is

y = 8/5

y = 3/4

y = 3/2

y = 0

Multiple Choice

g(x) has a vertical asymptote at

x = 4

x = -4

x = 3

x = -1

Multiple Choice

g(x) has an x-intercept at

x = 4

x = -3

x = 3

x = -1

Multiple Choice

g(x) has a y-intercept at

(0, 3/2)

(0, 3/4)

(0, -3/2)

(0, 1/2)

Multiple Choice

g(x) has a horizontal asymptote at

y = 0

y = 2

y = 1/2

There is no horizontal asymptote

Poll

After this lesson, I am feeling ____ about rational functions.

great

better but not great

not good

horrible

Open Ended

What is one thing you want Ms. Beck to clarify about rational functions when she gets back tomorrow?

Type response here

Rational Functions: Finding asymptotes

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