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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 Lesson Menu

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. 5–Minute Check 1

List all possible rational zeros of g (x) = x 4 – x 3 + 2x 2 – 4x – 8 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. 5–Minute Check 2

A. 4 positive zeros; 1 negative zero 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 5–Minute Check 3

A. 1 positive zero; 3 or 1 negative zeros 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 5–Minute Check 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 5–Minute Check 5

Analyze and graph rational functions. Solve rational equations. 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. Then/Now

rational function asymptote vertical asymptote horizontal asymptote oblique asymptote holes Vocabulary

Key Concept 1

Find Vertical and Horizontal Asymptotes A. Find the domain of and the equations of the vertical or horizontal asymptotes, if any. 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. Example 1

Step 2 Find the asymptotes, if any. Check for vertical asymptotes. 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. Example 1

Check for horizontal asymptotes. 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. Example 1

CHECK The graph of shown supports each of these findings. 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 Example 1

Find Vertical and Horizontal Asymptotes B. Find the domain of and the equations of the vertical or horizontal asymptotes, if any. Step 1 The zeros of the denominator b (x) = 2x2 + 1 are imaginary, so the domain of f is all real numbers. Example 1

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, Example 1

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 Example 1

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 Example 1

Key Concept 2

Step 2 There is a vertical asymptote at x = –5. 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 1 The function is undefined at b (x) = 0, so the domain is {x | x ≠ –5, x  } 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. Example 2

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. Example 2

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 }; Example 2

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. Example 2

Step 2 There are vertical asymptotes at x = 2 and x = –2. Graph Rational Functions: n < m and n > m Step 1 The function is undefined at b(x) = 0, so the domain is {x | x ≠ 2, –2, x  }. 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. Example 2

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. Example 2

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 } Example 2

Determine any vertical and horizontal asymptotes and intercepts for . 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 Example 2

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  }. Example 3

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. Example 3

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, ∞). Example 3

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; Example 3

Determine any vertical and horizontal asymptotes and intercepts for . 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 Example 3

Key Concept 3

Step 2 There is a vertical asymptote at x = –3. Graph a Rational Function: n = m + 1 Determine any asymptotes and intercepts for . Then graph the function, and state its domain. Step 1 The function is undefined at b (x) = 0, so the domain is D = {x | x ≠ –3, x  }. 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. Example 4

Therefore, the equation of the oblique asymptote is y = x – 2. 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. Example 4

Graph a Rational Function: n = m + 1 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) ≈ 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, ∞). Example 4

Graph a Rational Function: n = m + 1 Example 4

Graph a Rational Function: n = m + 1 Answer: vertical asymptote at x = –3; oblique asymptote at y = x – 2; x-intercepts: and ; y-intercept: ; Example 4

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 Example 4

Factoring both the numerator and denominator yields h(x) = 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  }. Example 5

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) = . Example 5

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. Example 5

Graph a Rational Function with Common Factors –4 –2 2 4 Answer: vertical asymptote at x = –2; horizontal asymptote at y = 1; x-intercept: –3 and y-intercept: ; hole: ; ; Example 5

Determine the vertical and horizontal asymptotes and holes of the graph of . 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) Example 5

Solve . Original Equation Multiply by the LCD, x – 6. Simplify. Solve a Rational Equation Solve . Original Equation Multiply by the LCD, x – 6. Simplify. Quadratic Formula Simplify. Example 6

Solve a Rational Equation 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. Answer: Example 6

Solve . A. –22 B. –2 C. 2 D. 8 Example 6

The LCD of the expressions is x – 1. Solve a Rational Equation with Extraneous Solutions Solve . The LCD of the expressions is x – 1. Original Equation Multiply by the LCD. x 2 – x + x = 3x – 2 Simplify. x 2 – 3x + 2 = 0 Subtract 3x – 2 from each side. Example 7

(x – 2)(x – 1) = 0 Factor. x = 2 or x = 1 Solve. Solve a Rational Equation with Extraneous Solutions (x – 2)(x – 1) = 0 Factor. x = 2 or x = 1 Solve. Because the original equation is not defined when x = 1, you can eliminate this extraneous solution. So, the only solution is 2. Answer: 2 Example 7

Solve . A. –2, 1 B. 1 C. –2 D. –2, 5 Example 7

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. Example 8

12r = 2r 2 – 32 Combine like terms. Solve a Rational Equation Original Equation Multiply by the LCD. 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. Example 8

0 = (r – 8)(r + 2) Factor. r = 8 or r = –2 Solve. Solve a Rational Equation 0 = (r – 8)(r + 2) Factor. r = 8 or r = –2 Solve. Because r is the rate of the boat, r cannot be negative. Therefore, r is 8 miles per hour. Answer: 8 Example 8

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? A. 1.7 or 8.3 seconds B. 2 or 7 seconds C. 4.7 seconds D. 12 seconds Example 8

End of the Lesson