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Rational Functions
4-2
A Rational Function is an equation in the form of f(x) = p(x)/q(x), where p(x) and q(x) are polynomial functions, and q(x) does not equal zero.
The parent rational function is f(x) = . Its graph is a hyperbola, which has two separate branches.
1x
Rational functions may
have asymptotes. The
function f(x) = has a
vertical asymptote at
x = 0 and a horizontal
asymptote at y = 0.
1x
The rational function f(x) = can be transformed by using methods similar to those used to transform other types of functions.
1x
Using the graph of f(x) = as a guide, describe the transformation and graph each function.
A. g(x) =
Because h = –2, translate f 2 units left.
1 x + 2
1x
B. g(x) =
Because k = –3, translate f 3 units down.
1x
– 3
Identify the zeros and vertical asymptotes of f(x) = . (x2 + 3x – 4)
x + 3
Factor the numerator. (x + 4)(x – 1) x + 3
f(x) =
Step 1 Find the zeros and vertical asymptotes.
Zeros: –4 and 1 The numerator is 0 when x = –4 or x = 1.
Vertical asymptote: x = –3 The denominator is 0 when x = –3.
Identify the zeros and vertical asymptotes of
f(x) = . (x2 + 7x + 6) x + 3
Factor the numerator. (x + 6)(x + 1) x + 3
f(x) =
Step 1 Find the zeros and vertical asymptotes.
Zeros: –6 and –1 The numerator is 0 when x = –6 or x = –1 .
Vertical asymptote: x = –3 The denominator is 0 when x = –3.
Some rational functions, including those whose graphs are hyperbolas, have a horizontal asymptote. The existence and location of a horizontal asymptote depends on the degrees of the polynomials that make up the rational function.
Note that the graph of a rational function can sometimes cross a horizontal asymptote. However, the graph will approach the asymptote when |x| is large.
To find Horizontal Asymptotes, compare the degrees of the numerator and the denominator. (3 scenarios):
xx
xxf
34
18)(
2
25
9)(
x
xxf
54
63)(
2
x
xxxf
Since degree of numerator is less than the degree of the denominatorA horizontal asymptote occurs at y = 0
Since the degree of numerator is larger thanthe degree of the denominator No horizontal asymptote
Since degrees of the numerator and denominator are equal, divide the coefficients of the highest degree terms.
A horizontal asymptote occurs at y = 9/5
Factor the denominator.
Identify the zeros and asymptotes of the function. Then graph.
x – 2 x2
– 1f(x) =
x – 2 (x – 1)(x + 1)
f(x) =
The numerator is 0 when x = 2.
The denominator is 0 when x = ±1.
Degree of p < degree of q.
Zero: 2
Vertical asymptote: x = 1, x = –1Horizontal asymptote: y = 0
Identify the zeros and asymptotes of the function. Then graph.
Graph with a graphing calculator or by using a table of values.
Factor the numerator.
Identify the zeros and asymptotes of the function. Then graph.
4x – 12 x – 1
f(x) =
4(x – 3) x – 1
f(x) =
The numerator is 0 when x = 3.The denominator is 0 when x = 1.
Zero: 3
Vertical asymptote: x = 1
Horizontal asymptote: y = 4The horizontal asymptote is
y =
= = 4. 4 1
leading coefficient of p leading coefficient of q
Identify the zeros and asymptotes of the function. Then graph.
Graph with a graphing calculator or by using a table of values.
Factor the numerator.
Identify the zeros and asymptotes of the function. Then graph.
x2 + 2x – 15 x – 1
f(x) =
(x – 3)(x + 5) x – 1
f(x) =
The numerator is 0 when x = 3 or x = –5.
The denominator is 0 when x = 1.
Degree of p > degree of q.
Zeros: 3 and –5
Vertical asymptote: x = 1
Horizontal asymptote: none
A Hole occurs at x = a whenever there is a common factor (x – a) in the numerator and denominator of the function.
Example of a function with a hole:4
16)(
2
x
xxf
(x – 3)(x + 3)x – 3
f(x) =
Identify holes in the graph of f(x) = . Then graph.
x2 – 9 x – 3
Factor the numerator.
The expression x – 3 is a factor of both the numerator and the denominator.
There is a hole in the graph at x = 3.
Divide out common factors.
(x – 3)(x + 3)(x – 3)
For x ≠ 3,
f(x) = = x + 3
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 {x|x ≠ 3}.
Hole at x = 3
(x – 2)(x + 3)x – 2
f(x) =
Identify holes in the graph of f(x) = . Then graph.
x2 + x – 6 x – 2
Factor the numerator.
The expression x – 2 is a factor of both the numerator and the denominator.
There is a hole in the graph at x = 2.
Divide out common factors.
For x ≠ 2,
f(x) = = x + 3(x – 2)(x + 3)(x – 2)
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
2.
g is f translated 4 units right.
asymptotes: x = 1, y = 2; D:{x|x ≠ 1}; R:{y|y ≠ 2}
1. Using the graph of f(x) = as a guide, describe the transformation and graph the function g(x) = .
1x 1
x – 4
Identify the asymptotes, domain, and range of the function g(x) = + 2. 5
x – 1
zero: 2; asymptotes: x = 0, y = 1; hole at x = 1
3. Identify the zeros, asymptotes, and holes in the graph of . Then graph. x2 – 3x + 2
x2 – x f(x) =
Example: Create a function of the form y = f(x) that satisfies the given set of conditions:
a) Vertical asymptote at x = 2, hole at x = -3
2 2
5
x xf ( x )
x
If we were looking for horizontal asymptotes, we would use scenario 3 where the degree of the numerator is greater than the degree of the denominator, and therefore no horizontal asymptote.
However, when the degree of the numerator is exactly one greater
than the degree of the denominator it is possible
that there is a Oblique or Slant ASYMPTOTE.
25 2x x x
Use polynomial long division
184
5f x x
x
What happens as x
So the slant asymptote is the line 4f x x
5
22
x
xxxf )(Find the equation of the slant
asymptote for the function .
To determine the equation of the oblique asymptote, divide the numerator by the denominator,( use polynomial long division or synthetic division) and then see what happens as x
The equation of the slant asymptote for this example is y = x.
3
2
1xf ( x )
x
Example: Determine the slant asymptote for the following function:
3
143)(
2
x
xxxf
22 5 2
2 7
x x
x
22 7 2 5 2x x x
5-Minute Check Lesson 3-8A
5-Minute Check Lesson 3-8B
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