splash screen. lesson menu five-minute check (over lesson 2-4) then/now new vocabulary key...
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Five-Minute Check (over Lesson 2-4)
Then/Now
New Vocabulary
Key Concept: Vertical and Horizontal Asymptotes
Example 1: Find Vertical and Horizontal Asymptotes
Key Concept: Graphs of Rational Functions
Example 2: Graph Rational Functions: n < m and n > m
Example 3: Graph a Rational Function: n = m
Key Concept: Oblique Asymptotes
Example 4: Graph a Rational Function: n = m + 1
Example 5: Graph a Rational Function with Common Factors
Example 6: Solve a Rational Equation
Example 7: Solve a Rational Equation with Extraneous Solutions
Example 8: Real-World Example: Solve a Rational Equation
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Over Lesson 2-4
List all possible rational zeros of f (x) = 2x
4 – x 3 – 3x
2 – 31x – 15. Then determine which, if any, are zeros.
A.
B.
C.
D.
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Over Lesson 2-4
List all possible rational zeros of g (x) = x
4 – x 3 + 2x
2 – 4x – 8. Then determine which, if any, are zeros.
A.
B.
C.
D.
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Over Lesson 2-4
Describe the possible real zeros of f (x) = 2x
5 – x 4 – 8x
3 + 8x 2 – 9x + 9.
A. 4 positive zeros; 1 negative zero
B. 4, 2, or 0 positive zeros; 1 negative zero
C. 3 or 1 positive zeros; 1 negative zero
D. 2 or 0 positive zeros; 2 or 0 negative zeros
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Over Lesson 2-4
Describe the possible real zeros of g (x) = 3x
4 + 16x 3 – 7x
2 – 64x – 20.
A. 1 positive zero; 3 or 1 negative zeros
B. 1 positive zero; 3 negative zeros
C. 1 or 0 positive zeros; 3 or 1 negative zeros
D. 1 positive zero; 2 or 0 negative zeros
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Over Lesson 2-4
Write a polynomial function of least degree with real coefficients in standard form that has 2, –5, and 3 + i as zeros.
A. f (x) = x 4 + 3x
3 – x 2 + 27x – 90
B. f (x) = x 4 – 3x
3 – 20x 2 + 84x – 80
C. f (x) = x 4 – 9x
3 + 18x 2 + 30x – 100
D. f (x) = x 4 – 3x
3 – 18x 2 + 90x – 100
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You identified points of discontinuity and end behavior of graphs of functions using limits. (Lesson 1-3)
• Analyze and graph rational functions.
• Solve rational equations.
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• rational function
• asymptote
• vertical asymptote
• horizontal asymptote
• oblique asymptote
• holes
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Find Vertical and Horizontal Asymptotes
Step 1 Find the domain.
The function is undefined at the real zero of the denominator b (x) = x – 1. The real zero of b (x) is 1. Therefore, the domain of f is all real numbers except x = 1.
A. Find the domain of and the equations
of the vertical or horizontal asymptotes, if any.
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Find Vertical and Horizontal Asymptotes
Step 2 Find the asymptotes, if any.
Check for vertical asymptotes.
Determine whether x = 1 is a point of infinite discontinuity. Find the limit as x approaches 1 from the left and the right.
Because , you
know that x = 1 is a vertical asymptote of f.
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Find Vertical and Horizontal Asymptotes
Check for horizontal asymptotes.
Use a table to examine the end behavior of f (x).
The table suggests that 1.
Therefore, you know that y = 1 is a horizontal
asymptote of f.
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Find Vertical and Horizontal Asymptotes
CHECK The graph of shown supports
each of these findings.
Answer: D = {x | x ≠ 1, x }; vertical asymptote at x = 1; horizontal asymptote at y = 1
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Find Vertical and Horizontal Asymptotes
Step 1 The zeros of the denominator b (x) = 2x2 + 1 are imaginary, so the domain of f is all real numbers.
B. Find the domain of and the
equations of the vertical or horizontal asymptotes,
if any.
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Find Vertical and Horizontal Asymptotes
Step 2 Because the domain of f is all real numbers,
the function has no vertical asymptotes.
Using division, you can determine that
.
As the value of | x | increases, 2x 2 + 1 becomes an
increasing large positive number and
decreases, approaching 0. Therefore,
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Find Vertical and Horizontal Asymptotes
CHECK You can use a table of values to support
this reasoning. The graph of
shown also supports each of these
findings.
Answer: D = {x | x }; no vertical asymptotes; horizontal asymptote at y = 2
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Find the domain of and the equations
of the vertical or horizontal asymptotes, if any.
A. D = {x | x ≠ 4, x }; vertical asymptote at x = 4; horizontal asymptote at y = –10
B. D = {x | x ≠ 5, x }; vertical asymptote at x = 5; horizontal asymptote at y = 4
C. D = {x | x ≠ 4, x }; vertical asymptote at x = 4; horizontal asymptote at y = 5
D. D = {x | x ≠ 4, 4, x }; vertical asymptote at x = 4; horizontal asymptote at y = –2
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Graph Rational Functions: n < m and n > m
A. For , determine any vertical and
horizontal asymptotes and intercepts. Then graph
the function and state its domain.
Step 2 There is a vertical asymptote at x = –5.
The degree of the polynomial in the numerator is 0, and the degree of the polynomial in the denominator is 1. Because 0 < 1, the graph of k has a horizontal asymptote at y = 0.
Step 1 The function is undefined at b (x) = 0, so the
domain is {x | x ≠ –5, x }
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Graph Rational Functions: n < m and n > m
Step 3 The function in the numerator has no real zeros, so k has no x-intercepts. Because k(0) = 1.4, the y-intercept is 1.4.
Step 4 Graph the asymptotes and intercepts. Then choose x-values that fall in the test intervals determined by the vertical asymptote to find additional points to plot on the graph. Use smooth curves to complete the graph.
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Graph Rational Functions: n < m and n > m
Answer: vertical asymptote at x = –5; horizontal asymptote at y = 0; y-intercept: 1.4; D = {x | x≠ –5, x };
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Graph Rational Functions: n < m and n > m
B. For , determine any vertical and
horizontal asymptotes and intercepts. Then graph
the function and state its domain.
Factoring the denominator yields .
Notice that the numerator and denominator have no
common factors, so the expression is in simplest
form.
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Graph Rational Functions: n < m and n > m
Step 2 There are vertical asymptotes at x = 2 and x = –2.
Compare the degrees of the numerator and denominator. Because 1 < 2, there is a horizontal asymptote at y = 0.
Step 3 The numerator has a zero at x = –1, so the x-intercept is –1. f(0) = 0.25, so The y-intercept is –0.25.
Step 1 The function is undefined at b(x) = 0, so the domain is {x | x ≠ 2, –2, x }.
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Graph Rational Functions: n < m and n > m
Step 4 Graph the asymptotes and intercepts. Then find and plot points in the test intervals determined by the intercepts and vertical asymptotes: (–∞, –2), (–2, –1), (–1, 2), (2, ∞). Use smooth curves to complete the graph.
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Graph Rational Functions: n < m and n > m
Answer: vertical asymptotes at x = 2 and x = –2; horizontal asymptote at y = 0. x-intercept: –1; y-intercept: –0.25; D = {x | x ≠ 2, –2, x }
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A. vertical asymptotes x = –4 and x = 3; horizontal asymptote y = 0; y-intercept: –0.0833
B. vertical asymptotes x = –4 and x = 3; horizontal asymptote y = 1; intercept: 0
C. vertical asymptotes x = 4 and x = 3; horizontal asymptote y = 0; intercept: 0
D. vertical asymptotes x = 4 and x = –3; horizontal asymptote y = 1; y-intercept: –0.0833
Determine any vertical and horizontal asymptotes
and intercepts for .
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Graph a Rational Function: n = m
Determine any vertical and horizontal asymptotes
and intercepts for . Then graph the
function, and state its domain.
Factoring both numerator and denominator yields
with no common factors.
Step 1 The function is undefined at b (x) = 0, so the domain is {x | x ≠ –2, 2, x }.
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Graph a Rational Function: n = m
Step 2 There are vertical asymptotes at x = –2 and
x = 2. There is a horizontal asymptote at
or y = 0.5, the ratio of the leading coefficients
of the numerator and denominator, because
the degrees of the polynomials are equal.
Step 3 The x-intercepts are –3 and 4, the zeros of the numerator. The y-intercept is 1.5 because f(0) = 1.5.
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Graph a Rational Function: n = m
Step 4 Graph the asymptotes and intercepts. Then find and plot points in the test intervals (–∞, –3), (–3, –2), (–2, 2), (2, 4), (4, ∞).
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Graph a Rational Function: n = m
Answer: vertical asymptotes at x = –2 and x = 2; horizontal asymptote at y = 0.5; x-intercepts: 4 and –3; y-intercept: 1.5;
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A. vertical asymptote x = 2; horizontal asymptote y = 6; x-intercept: –0.833; y-intercept: –2.5
B. vertical asymptote x = 2; horizontal asymptote y = 6; x-intercept: –2.5; y-intercept: –0.833
C. vertical asymptote x = 6; horizontal asymptote y = 2; x-intercepts: –3 and 0; y-intercept: 0
D. vertical asymptote x = 6, horizontal asymptote y = 2; x-intercept: –2.5; y-intercept: –0.833
Determine any vertical and horizontal asymptotes
and intercepts for .
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Graph a Rational Function: n = m + 1
Determine any asymptotes and intercepts for
. Then graph the function, and
state its domain.
Step 2 There is a vertical asymptote at x = –3.
The degree of the numerator is greater than the degree of the denominator, so there is no horizontal asymptote.
Step 1 The function is undefined at b (x) = 0, so the domain is D = {x | x ≠ –3, x }.
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Graph a Rational Function: n = m + 1
Because the degree of the numerator is exactly one more than the degree of the denominator, f has an oblique asymptote. Using polynomial division, you can write the following.
f(x) =
Therefore, the equation of the oblique asymptote is y = x – 2.
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Graph a Rational Function: n = m + 1
Step 4 Graph the asymptotes and intercepts. Then find and plot points in the test intervals (–∞, –3.37), (–3.37, –3), (–3, 2.37), (2.37, ∞).
Step 3 The x-intercepts are the zeros of the
numerator, and , or
about 2.37 and –3.37. The y-intercept is
about –2.67 because f(0) ≈
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Graph a Rational Function: n = m + 1
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Graph a Rational Function: n = m + 1
Answer: vertical asymptote at x = –3;
oblique asymptote at y = x – 2;
x-intercepts: and ;
y-intercept: ;
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Determine any asymptotes and intercepts for
.
A. vertical asymptote at x = –2; oblique asymptote at y = x; x-intercepts: 2.5 and 0.5; y-intercept: 0.5
B. vertical asymptote at x = –2; oblique asymptote at y = x – 5; x-intercepts at ; y-intercept: 0.5
C. vertical asymptote at x = 2; oblique asymptote at y = x – 5; x-intercepts: ; y-intercept: 0
D. vertical asymptote at x = –2; oblique asymptote at y = x2– 5x + 11; x-intercepts: 0 and 3; y-intercept: 0
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Graph a Rational Function with Common Factors
Determine any vertical and horizontal asymptotes,
holes, and intercepts for . Then
graph the function and state its domain.
Factoring both the numerator and denominator yields
h(x) =
Step 1 The function is undefined at b (x) = 0, so the domain is D = {x | x ≠ –2, 3, x }.
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Graph a Rational Function with Common Factors
Step 2 There is a vertical asymptote at the real zero of the simplified denominator x = –2. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.
Step 3 The x-intercept is –3, the zero of the
simplified numerator. The y-intercept is
because h(0) = .
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Graph a Rational Function with Common Factors
Step 4 Graph the asymptotes and intercepts. Then
find and plot points in the test intervals
There is a hole at
because the original function is
undefined when x = 3.
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Graph a Rational Function with Common Factors
Answer: vertical asymptote at x = –2; horizontal
asymptote at y = 1; x-intercept: –3
and y-intercept: ; hole: ;
;
–4 –2 2 4
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A. vertical asymptote at x = –2, horizontal asymptote at y = –2; no holes
B. vertical asymptotes at x = –5 and x = –2; horizontal asymptote at y = 1; hole at (–5, 3)
C. vertical asymptotes at x = –5 and x = –2; horizontal asymptote at y = 1; hole at (–5, 0)
D. vertical asymptote at x = –2; horizontal asymptote at y = 1; hole at (–5, 3)
Determine the vertical and horizontal asymptotes
and holes of the graph of .
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Original Equation
Solve a Rational Equation
Solve .
Multiply by the LCD, x – 6.
Simplify.
Quadratic Formula
Simplify.
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Solve a Rational Equation
Answer:
CHECK Because the zeros of the graph of the related
function appear to be at
about x = 6.6 and x = –0.6, this solution is
reasonable.
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A. –22
B. –2
C. 2
D. 8
Solve .
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Solve a Rational Equation with Extraneous Solutions
The LCD of the expressions is x – 1.
Solve .
x 2 – x + x = 3x – 2
Simplify.
x 2 – 3x + 2 = 0
Subtract 3x – 2 from each side.
Multiply by the LCD.
Original Equation
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Solve a Rational Equation with Extraneous Solutions
(x – 2)(x – 1) = 0Factor.
x = 2 or x = 1Solve.
Because the original equation is not defined when x = 1, you can eliminate this extraneous solution. So, the only solution is 2.
Answer: 2
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A. –2, 1
B. 1
C. –2
D. –2, 5
Solve .
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Solve a Rational Equation
WATER CURRENT The rate of the water current in a
river is 4 miles per hour. In 2 hours, a boat travels
6 miles with the current to one end of the river and
6 miles back. If r is the rate of the boat in still water,
r + 4 is its rate with the current, r – 4 is its rate
against the current, and , find r.
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Solve a Rational Equation
6r + 24 + 6r – 24 = 2r 2
– 32 Simplify.
12r = 2r 2 – 32
Combine like terms.
0= 2r 2 – 12r – 32
Subtract 12r from each side.
0 = r 2 – 6r – 16
Divide each side by 2.
Multiply by the LCD.
Original Equation
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Answer: 8
Solve a Rational Equation
0 = (r – 8)(r + 2)Factor.
r = 8 or r = –2Solve.
Because r is the rate of the boat, r cannot be negative. Therefore, r is 8 miles per hour.
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A. 1.7 or 8.3 seconds
B. 2 or 7 seconds
C. 4.7 seconds
D. 12 seconds
ELECTRONICS Suppose the current I, in amps, in
an electric circuit is given by the formula ,
where t is time in seconds. At what time is the
current 2 amps?
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