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3.1 3.2 3.3 3.4 3.5 3.6 3.1 Quadratic Functions and Models Quadratic Functions ▪ Graphing Techniques ▪ Completing the Square ▪ The Vertex Formula ▪ Quadratic Models and Curve Fitting Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-1 3.1 Example 1(a) Graphing Quadratic Functions (page 304) Graph f(x) = x2 + 4x – 4 by plotting points. Give the domain and range. X-coordinate of vertex -b/(2a) Vertex: X = -4/2 = -2 X Y -3 -2 -1 -7 -8 -7 Y = (-2)2 + 4(-2) – 4 (-2,-8) Domain: Copyright © 2008 Pearson Addison-Wesley. All rights reserved. Range: 3-2 3.1 add on: Quadratic graph w/ fraction value for a Graph f(x) = 2/3 (x – 5)2 + 4 by plotting points. Give the domain and range. Vertex: X Y (5,4) 2 5 8 10 4 10 Since the value of a = 2/3 choose x-values around the vertex which are 3 units larger and smaller than the x-coor of the vertex Domain: Copyright © 2008 Pearson Addison-Wesley. All rights reserved. Range: [4, ∞ ) 3-3 Add on: Find the equation from the graph Given the graph of a parabola with a vertex at (1,3) passing through the point (-1,-2). Find the equation of the parabola in vertex form: y = a(x – h)2 + k h=1 k=3 y = a (x – 1)2 + 3 Substitute (-1,-2) for (x,y) solve for a -2 = a (-1 – 1)2 + 3 -5 = a (4) - 5/4 = a y = -5/4 (x – 1)2 + 3 Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-4 Add on Find equation of parabola with x-intercepts of 2 and 5 and a y-intercept of 4. Simplify your answer. Roots: Factors: Equation: Substitute x = 2 and x = 5 (x – 2)(x – 5) y = a(x – 2)(x – 5) (0,4) for the x and y and solve for a 4 = a(0 – 2)(0 – 5) 4 = a(10) 2/ = a 5 Equation: Answer: y = 2/5 (x – 2)(x – 5) y = 2/5 x2 – 2/5 x + 4 Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-5 3.1 Example 2 Graphing a Parabola by Completing the Square (page 305) Graph f(x) = x2 + 2x – 5 by completing the square and locating the vertex. X Y I prefer: find vertex 1st X coor = –2/2 = -1 Y coor = (-1)2 + 2(-1) – 5 = -6 Vertex (–1, –6) Axis of Sym: x = –1 -2 -1 0 -5 -6 -5 3.1 Example 3 Graphing a Parabola by Completing the Square (page 305) Graph f(x) = 2x2 + x – 6 by completing the square X and locating the vertex. I prefer: find vertex 1st X coor = –1/4 Y coor = 2( –1/4)2 + (–1/4) – 6 = –49/8 -1 - 1/ 4 0 Y -3 - 49/8 -6 Vertex is (–1/4, –49/8) Axis of sym: X = –1/4 Their way: 3-7 3.1 Example 4 Finding the Axis and the Vertex of a Parabola Using the Vertex Formula (page 308) Find the axis and vertex of the parabola f(x) = –3x2 + 12x – 8 using the vertex formula. a = –3, b = 12, c = –8 Axis: x = 2 Vertex: (2, f(2)) Vertex: (2, 4) Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-8 3.1 Example 5(a) Solving a Problem Involving Projectile Motion (page 308) A ball is thrown directly upward from an initial height of 75 ft with an initial velocity of 112 ft per sec. Give the function that describes the height of the ball in terms of time t. The projectile height function is After how many seconds does the ball reach its maximum height? What is the maximum height? The maximum height occurs at the vertex. (3.5sec,271ft) The ball reaches its maximum height, 271 ft, after 3.5 seconds. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-9 Additional Notes Discriminant: b2 – 4ac • Positive: then the graph crosses the x-axis twice • Zero: then the graph is tangent to the x-axis (meaning it touches the x-axis) • Negative: then the graph never crosses the x-axis because there are two imaginary roots Disc: positive a: positive Disc: zero a: positive Copyright © 2008 Pearson Addison-Wesley. All rights reserved. Disc: positive Disc: negative a: negative a: negative 3-10 3.1 additional notes The number of mosquitos, M(x), in millions, in a certain area of Florida depends on the June rainfall, x, in inches. The function M(x) = 10x – x2 models this phenomenon. Find the amount of rainfall that will maximize the number of mosquitos. What is the maximum number of mosquitos? Vertex: x = -b/2a M(5) = 10(5) – 52 = 25 Vertex is (5,25) x = -10/-2 = 5 Since the parabola opens down the max # of mosquitos in Flordia is 25 million which will occur when 5 inches of rain has fallen in June Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-11 3.1 Summary Quadratics: y = ax2 + bx + c y = a(x – h)2 + k, vertex (h,k) Axis of symmetry of a quadratic equation: x = -b/2a Vertex of a quadratic equation: (-b/2a, f(-b/2a )) Quadratic formula: Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-12 3.2 Synthetic Division Synthetic Division ▪ Evaluating Polynomial Functions Using the Remainder Theorem ▪ Testing Potential Zeros Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-13 3.2 Example 1 Using Synthetic Division (page 323) Use synthetic division to divide . x – 3 is in the form x – k. Bring down the 1st #, then mult, add, mult, add, … 12 -9 6 -3 2 -4 3.2 Example 2 Applying the Remainder Theorem (page 324) Let theorem to find f(4). . Use the remainder Use synthetic division with k = 4. f(4) = 17 Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-15 3.2 Example 3(a) Deciding Whether a Number is a Zero (page 325) Let . Is k = 2 a zero? Use synthetic division with k = 2. 2 is not a zero. Let . Is k = –3 a zero? –3 is a zero. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. Additional Example Express f(x) in the form (x – k)q(x) + r for the given value of k f(x) = 2x3 + 4x2 – 5x + 2 and k = 4 4 2 4 8 2 12 -5 2 48 172 43 174 Answer: (x – 4)(2x2 + 12x + 43) + 174 Copyright © 2008 Pearson Addison-Wesley. All rights reserved. What if the 2 was missing in the original f(x) equation? We would need to put a zero place holder in the last position of synthetic division 3-17 3.2 Example 3(c) Deciding Whether a Number is a Zero (page 325) Let . Is k = 1 + 3i a zero? Use synthetic division with k = 1 + 3i. 1 + 3i is a zero. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-18 3.2 Summary How to tell if a number is a zero of a polynomial or not Synthetic Division: Always rewrite the original problem in the correct order and fill in zeros for missing exponents The number to the left of the division bar comes from setting the divisor = 0 and solving Bring the first value down, then multiply, add, multiply, add, … When writing the answer, start with one degree less than the original problem Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-19 3.3 Zeros of Polynomial Functions Factor Theorem ▪ Rational Zeros Theorem ▪ Number of Zeros ▪ Conjugate Zeros Theorem ▪ Zeros of a Polynomial Function ▪ Descartes’ Rule of Signs Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-20 3.3 Example 1(a) Deciding Whether x – k is a Factor of f(x) (page 328) Let . Is x + 4 a factor? By the factor theorem x + 4 is a factor of f(x) if and only if f(–4) = 0. Insert 0 as the coefficient for the missing x3-term. Since f(–4) = 0, x + 4 is a factor of f(x). Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-21 3.3 Example 1(b) Deciding Whether x – k is a Factor of f(x) (page 328) Let Is x + 4 a factor? . By the factor theorem x + 4 is a factor of f(x) if and only if f(–4) = 0. Since f(–4) ≠ 0, x + 4 is not a factor of f(x). Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-22 3.3 Example 2 Factoring a Polynomial Given a Zero (page 329) Factor 5 is a zero of f. into linear factors if Since 5 is a zero of f, x – 5 is a factor. Divide f(x) by x – 5. The quotient is Copyright © 2008 Pearson Addison-Wesley. All rights reserved. . Side note: 6x2 – 7x – 3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) + 1(2x – 3) = (3x + 1) (2x – 3) 3-23 Additional example Factor f(x) into linear factors given that k is a zero of f(x). f(x) = x3 + (7 – 3i)x2 + (12 – 21i)x – 36i; k = 3i 3i 1 1 7 – 3i 12 – 21i – 36i 3i 21i 36i 7 12 0 f(x) = (x – 3i)(x2 + 7x + 12) Answer: f(x) = (x – 3i)(x + 3)(x + 4) Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-24 3.3 Example 3(a) Using the Rational Zeros Theorem (page 330) List all possible rational zeros. Note: •from Algebra II “factors of last over factors of first” •from College Alg book “factors of p/q” p must be a factor of a0 = 4: ±1, ±2, ±4 q must be a factor of a4 = 8: ±1, ±2, ±4, ±8 Possible rational zeros, : Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-25 3.3 Example 3(b) Using the Rational Zeros Theorem (page 330) Find all rational zeros and factor f(x) into linear factors. Is x – 1 a factor? Possible rational factors: ± {1, ½, ¼, 1/8, 2, 4, 8} x – 1 is not a factor. Is x + 1 a factor? x + 1 is a factor. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-26 3.3 Example 3(b) Using the Rational Zeros Theorem Possible rational factors: ± {1, ½, ¼, 1/8, 2, 4, 8} (cont.) Use the quotient polynomial find the next factor. Is x – 4 a factor? to x – 4 is a factor. The rational zeros are Copyright © 2008 Pearson Addison-Wesley. All rights reserved. . Side note: 8x2 – 2x – 1 = 8x2 – 4x + 2x – 1 = 4x(2x – 1) + 1(2x – 1) = (4x + 1)(2x – 1) 3-27 3.3 Example 4(a) Finding a Polynomial Function that Satisfies Given Conditions (Real Zeros) (page 332) Find a function f defined by a polynomial of degree 3 that has zeros of –3, –2, and 5, and f(–1) = 6. The three zeros are: x = -3, x = -2, x = 5 The factors are: (x + 3) (x + 2) (x – 5) Since f(–1) = 6, we can solve for a: Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-28 3.3 Example 4(b) Finding a Polynomial Function that Satisfies Given Conditions (Real Zeros) (page 332) Find a function f defined by a polynomial of degree 3 that has zero 4 of multiplicity 3, and f(2) = –24. The function f has form Since f(2) = –24, we can solve for a: Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-29 3.3 Example 5 Finding a Polynomial Function that Satisfies Given Conditions (Complex Zeros) (page 334) Find a polynomial function of least degree having only real coefficients and zeros –4 and 3 – i. By the conjugate zeros theorem, 3 + i is also a zero. Zeros are: x = -4 x = 3 – i x = 3 + i Factors are: (x + 4) (x – 3 + i) (x – 3 – i) F(x) Copyright © 2008 Pearson Addison-Wesley. All rights reserved. Have the students make a 3x3 box to multiply Have the students make a 2x3 box to multiply 3-30 3.3 Example 6 Finding All Zeros of a Polynomial Function Given One Zero (page 335) Find all zeros of given that 2 + i is a zero. Since 2 + i is a zeros of f(x), then 2 – i is also a zero. 2–i 2–i 1 3 6 – 3i -10 -20 + 10i 0 F(x) = (x – 2 – i)(x – 2 + i)(x2 + 3x – 10) F(x) = (x – 2 – i)(x – 2 + i)(x +5)(x – 2) The zeros of f(x) are 2 + i, 2 – i, –5, and 2. 3.3 Example 7 Applying Descartes’ Rule of Signs (page 336) Determine the possible number of positive real zeros and negative real zeros of f(x) = – + – + – Thus, f(x) has 4 sign changes, so f(x) has 4, or 2, or 0 positive real zeros. For negative zeros, consider the variations in signs for f(–x). f(-x)= – – – – – There are no changes in sign, so there are no negative zeros. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-32 3.3 Example 7 Applying Descartes’ Rule of Signs (page 336) Cont. from previous slide to include complex imaginary roots Determine the possible number of positive real zeros and negative real zeros of f(x) = – + – + – Thus, f(x) has 4 sign changes, so f(x) has 4, or 2, or 0 positive real zeros. + – i 4 0 0 2 0 2 0 0 4 f(-x)= – – – – – There are no changes in sign, so there are 0 negative zeros. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. There are 0, 2, or 4 complex imaginary zeros. 3-33 3.3 Example 8 Applying Descartes’ Rule of Signs (page 336) Determine the possible number of positive real zeros and negative real zeros of – f(x) = – – – + – Thus, f(x) has 2 sign changes, so f(x) has 2 or 0 positive real zeros. + – i 2 2 0 0 2 0 2 0 2 4 f(-x)= – + – – – Thus, f(x) has 2 sign changes, so f(x) has 2 or 0 negative real zeros. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. There are 0, 2, or 4 complex imaginary zeros. 3-34 3.3 Summary Given a polynomial, determine if a binomial is a factor or not Factor a polynomial given a factor or a zero List the possible rational zeros of a polynomial Find all the rational zeros of a polynomial Given a zeros of a polynomial find the polynomial Multiplicity is the number of times the zeros is counted as a factor If a polynomials with integral coefficients has an imaginary root, then its conjugate must also be a root. To find the number of possible positive and negative zeros count the number of sign changes for f(x) and f(-x) respectfully, and subtract 2 at a time Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-35 3.4 Polynomial Functions: Graphs, Applications, and Models Graphs of f(x) = axn ▪ Graphs of General Polynomial Functions ▪ Turning Points and End Behavior ▪ Graphing Techniques ▪ Intermediate Value and Boundedness Theorems ▪ Approximating Real Zeros ▪ Polynomial Models and Curve Fitting Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-36 3.4 Example 2(a) Examining Vertical and Horizontal Translations (page 341) Graph X Y -1 0 0 1 1 2 2 9 The graph of is the same as the graph of , but translated 1 unit up. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-37 3.4 Example 2(b) Examining Vertical and Horizontal Translations (page 341) Graph X Y 1 -1 2 0 3 1 4 32 The graph of is the same as the graph of , but translated 2 units right. The multiplicity of 5 for the zero at 2 makes the graph flatter at that point Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-38 3.4 Example 2(c) Examining Vertical and Horizontal Translations (page 341) Graph X Y -5 -3 -3 5 -1 -3 1 -123 The graph of f(x) = -½ (x + 3)4 + 5 is the same as the graph of f(x) = x4,but translated 3 units left, reflected across the x-axis, stretched vertically by a factor of ½, and then translated 5 units up. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-39 3.4 Example 3 Determining End Behavior Given The Defining Polynomial (page 344) Use the symbols for end behavior to describe the end behavior of the graph of each function. (a) Since a = −1 < 0 and f has even degree, the end behavior is shaped (b) Since a = 1 > 0 and g has odd degree, the end behavior is shaped Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-40 3.4 Example 3 Determining End Behavior Given The Defining Polynomial (cont.) (c) Since a = −1 < 0 and h has odd degree, the end behavior is shaped (d) Since a = 1 > 0 and f has even degree, the end behavior is shaped Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-41 3.4 Example 4 Graphing a Polynomial Function (page 345) Graph X Y -1 6 0 -6 1 -12 2 0 Have girls do synthetic division for -1 and boys synthetic division for 2 Setting each factor equal to zero gives the zeros of f as 2, -1/2, and –3. Add on Graph (for the notes we will just complete the table but the homework you would graph these) y = 2x(x – 5)(x + 2) y = 1/5 (x – 2)2 + 3 X Y X Y 0 0 -3 8 5 0 2 3 -2 0 7 8 1 -24 12 23 This equation is in factored form so I would start with the zeros of the polynomial Copyright © 2008 Pearson Addison-Wesley. All rights reserved. The coefficient has a 5 in the denominator so I choose numbers + 5 units from the x-coordinate of the vertex 3-43 3.4 Example 5 Locating a Zero (page 346) Use synthetic division and a graph to show that has a real zero between 1 and 2 Use synthetic division to find f(1) and f(2). f(1) = 2 f(2) = –3 By the Intermediate Value Theorem, there is a real zero between 1 and 2. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3.4 Example 6(a) Using the Boundedness Theorem (page 348) Show that greater than 1. has no real zero f(x) has real coefficients and the leading coefficient, 1, is positive, so the Boundedness Theorem applies. Divide f(x) synthetically by x – 1. Since 1 is positive and all the numbers in the last row of the synthetic division are nonnegative, f(x) has no real zeros greater than 1. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-45 3.4 Example 6(b) Using the Boundedness Theorem (page 348) Show that less than –2. has no real zero f(x) has real coefficients and the leading coefficient, 1, is positive, so the Boundedness Theorem applies. Divide f(x) synthetically by x + 2. Since the 2 is negative and the numbers in the last row of the synthetic division alternate in sign, f(x) has no real zeros less than –2. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-46 Additional example Given f(x) = x2(x – 4)(x + 5)2 Find the zeros, state the multiplicity for each, state if graph crosses or is tangent at each zero, the degree of the polynomial, the end behavior and then make a rough sketch of the graph without the use of a graphing calculator Zeros: Mult: Cross/tangent: X=0 2 tangent X = 4 X = -5 1 2 cross tangent Degree of poly: 5 End behavior: Down, Up Copyright © 2008 Pearson Addison-Wesley. All rights reserved. The graph should have a smooth curve NOT a sharp point at (-5,0) 3-47 Additional example Find the polynomial function with the given graph. Assume the leading coefficient is 1. Zeros: Mult: (0,-128) X = -4 X = -1 X = 2 1 2 2 X=4 1 F(x) = a(x + 4)2(x + 1)(x – 2)2(x – 4) F(0) = a(0 + 4)2(0 + 1)(0 – 2)2(0 – 4) -128 = a(16)(1)(4)(–4) -1/2 = a F(x) = -1/2(x + 4)2(x + 1)(x – 2)2(x – 4) Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-48 Additional example: Make a table points used to graph the following: y = 2x(x – 1)(x + 2) Degree = 3 End behavior is down,up The zeros of the function are 0, 1, -2 Copyright © 2008 Pearson Addison-Wesley. All rights reserved. X Y 0 1 -2 0 0 0 -1 4 You will need one more point than the degree for the homework system to graph the polynomial for you 3-49 Additional example: Make a table points used to graph the following: y = x3 + 9x2 + 11x – 21 Degree = 3 End behavior is down,up X Y 0 1 -1 -21 0 -24 2 45 You will need one more point than the degree for the homework system to graph the polynomial for you Use synthetic division to obtain each yvalue of the table after x = 1 … I will show you on the whiteboard if needed Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-50 3.4 Summary End behavior of polynomials Steepness and shifting of graphs from parent functions Reflecting polynomials over x- or y-axis Multiplicity: Odd multiplicity, the graph crosses the x-axis at that zero Even multiplicity, the graph is tangent to the x-axis at that zero Larger the value of the multiplicity the flatter the graph at that zero Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-51 3.5 Rational Functions: Graphs, Applications, and Models The Reciprocal Function ▪ The Function Asympototes ▪ Steps for Graphing Rational Functions ▪ Rational Function Models Copyright © 2008 Pearson Addison-Wesley. All rights reserved. ▪ 3-52 3.5 Example 1 Graphing a Rational Function (page 360) Graph . Give the domain and range. The expression can be written as , indicating that the graph can be obtained by stretching the graph of vertically by a factor of 4. Domain: Range: Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-53 3.5 Example 2 Graphing a Rational Function (page 361) Graph . Give the domain and range. Shift the graph of three units to the right, then reflect the graph across the x-axis to obtain the graph of . The horizontal asymptote is y = 0. The vertical asymptote is x = 3. Domain: Range: Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-54 3.5 Example 3 Graphing a Rational Function (page 363) Graph . Give the domain and range. To obtain the graph shift the graph of two units to the right, stretch the graph vertically by a factor of two, and then shift the graph four units down. The vertical asymptote is x = 2. The horizontal asymptote is y = –4. Domain: Range: Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-55 3.5 Example 4(a) Finding Asymptotes of Rational Functions (page 364) Find all asymptotes of the function To find the vertical asymptotes, set the denominator equal to 0 and solve. The equations of the vertical asymptotes are x = –4 and x = 4. The denominator has a bigger degree than the numerator, so the horizontal asymptote is y = 0. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-56 3.5 Example 4(b) Finding Asymptotes of Rational Functions (page 364) Find all asymptotes of the function To find the vertical asymptotes, set the denominator equal to 0 and solve. The equation of the vertical asymptote is . The numerator and the denominator have the equal degrees, so the equation of the horizontal asymptote is Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-57 3.5 Example 4(c) Finding Asymptotes of Rational Functions (page 364) Find all asymptotes of the function To find the vertical asymptotes, set the den = 0 and solve. Vert asy is x = 3. Since the degree of the numerator is exactly one more than the denominator, there is an oblique asymptote. Divide the numerator by the denominator and, disregarding the remainder, set the rest of the quotient equal to y to obtain the equation of the asymptote. I would have used Synthetic Division Oblique asy is y = 2x + 6. 3-58 3.5 Example 5 Graphing a Rational Function with the x-axis as Horizontal Asymptote (page 366) Graph To find the vertical asymptotes, set the den = 0 and solve. Find the horizontal asymptote. The degree of den > degree of num, so the horizontal asy is y = 0. Find the y-intercept Find the x-intercept 3-59 3.5 Example 6 Graphing a Rational Function That Does Not Intersect its Horizontal Asymptote (page 367) Graph To find the vertical asymptotes, set the den = 0 and solve. Find the horizontal asymptote. The degree of num = degree of den, so the horizontal asymptote is y = 4. Find the y-int Find the x-int 3-60 3.5 Example 8 Graphing a Rational Function With an Oblique Asymptote (page 369) Find the y-int Graph Find the x-int Find the vertical asy x(x+1) = 0 Find the horizontal asy. Since the deg num > deg den there isn’t a horiz asy. Divide to find oblique asy. I would have used Synthetic Division oblique asy is y=x+2 x = 0 or -1 3.5 Example 9 Graphing a Rational Function Defined by an Expression That is not in Lowest Terms (page 370) Graph The domain of f cannot include 3. In lowest terms, the function becomes The graph of f is the same as the graph of y = x + 2, with the exception of the point with x-value 3. A “hole” appears in the graph at (3, 5). Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-62 3.5 Ex 7 Graphing a Rational Function That Intersects its Horizontal Asymptote (page 368) Calculus bound students only Graph Find the x-int Find the vertical asy Find the horizontal asy. Degree of num = degree of denom, so the horizontal asy is y = 2. Find the y-int 3.5 Summary Rational Graphs X-intercept: solve for x, when y = 0 Y-intercept: solve for y, when x = 0 Vertical asymptote: set the denominator = 0 Horizontal asymptote: degree of bottom = top, y = coef (BETC) degree is bigger on bottom, y=0 (BOBO) Oblique asymptote: (BOTD) degree bigger on top, divide asy: y = quotient Holes occur in the graph when a factor reduces from the numerator and denominator Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-64 3.6 Variation Direct Variation ▪ Inverse Variation ▪ Combined and Joint Variation Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-65 3.6 Example 1 Solving a Direct Variation Problem (page 380) At a given average speed, the distance traveled by a vehicle varies directly as the time. If a vehicle travels 156 miles in 3 hours, find the distance it will travel in 5 hours at the same average speed. Equation: d = kt. Find k, substitute d = 156 and t = 3 Final equation: d = 52t. Solve the equation for d with t = 5. The vehicle will travel 260 miles in 5 hours. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-66 3.6 Example 2 Solving an Inverse Variation Problem (page 380) In a certain manufacturing process, the cost of producing a single item varies inversely as the square of the number of items produced. If 100 items are produced, each costs $1.50. Find the cost per item if 250 items are produced. Let n = # of items produced and c = cost per item Solve for c with n = 250 Find k, sub c = 1.50 and n = 100 The cost per item will be $0.24. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3.6 Example 3 Solving a Joint Variation Problem (page 382) The volume of a cylinder varies directly as the height and the square of the radius. A cylinder with radius 5 cm and height 10 cm has volume 785 cm3. Find the volume of a cylinder with radius 10 cm and height 15 cm. V = khr2 Find k, sub V = 785, h =10 and r = 5 Copyright © 2008 Pearson Addison-Wesley. All rights reserved. Solve for V with h = 15 and r = 10. The volume of the cylinder is 4710 cm3 3.6 Example 4 Solving a Combined Variation Problem (page 383) Vanessa Jones has determined that the number of children enrolled in her day care center varies directly with the amount she spends on advertising per year and inversely with her weekly charge for each child. This year she spent $900 on advertising, charged $150 per child per week, and had 51 children enrolled. Next year, if she spend $1000 on advertising and charges $170 per week, how many children should she expect to enroll? Let n = the number of children, a = $ spent on advertising/yr, and c = the weekly charge. She should expect to enroll 50 children. 3-69 3.6 Summary Varies Directly: y = kx Varies Inversely: y = k/x Varies jointly: y = kxw Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-70 Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-71 3.1 Ex 5(c) Solving a Problem Involving Projectile Motion (pg 308) For what interval of time is the height of the ball greater than 200 ft? Use the quadratic formula a = –16, b = 112, c = –125 The ball will be greater than 200 ft above ground level between 1.39 and 5.61 seconds after it is thrown. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3.1 Example 5(d) Solving a Problem Involving Projectile Motion (page 308) After how many seconds will the ball hit the ground? Use the quadratic formula to find the positive solution of Reject The ball hits the ground after about 7.62 sec. Verify with a graphing calculator. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-73 3.1 Additional Story Problem A farmer has 500 feet of fencing and is planning to enclose a rectangular lot using the barn for one side of the lot. What should the dimensions of the lot be for the lamb to have the maximum amount of area to graze on? A = x(500 – 2x) A = 500x – 2x2 A = -2x2 + 500x barn x x 500-2x Max area occurs when x = – 500/–4 = 125 Max area is 500(125) – 2(125)2 = 31250 Maximum area of 31250 ft2 occurs when the dimensions of the rectangle are 125 ft by 250 ft Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-74 3.1 Additional Story Problem (like Hw 3.1G#5) A piece of sheet metal is 2.9 times as long as it is wide. It is to be made into a box with an open top by cutting 3-inch squares from each corner and folding up the sides. L = 2.9x Restrictions for x: If we are to cut away 3 inch squares then x > 6 x Equation for the volume of the box: V = 3(x – 6)(2.9x – 6) The problem wants us to find the value(s) for x, in order for the box to have a volume between 600 and 800 inches In the calculator Y1 = 3(x – 6)(2.9x – 6) and Y2 = 600 calculate the intersection In the calculator Y1 = 3(x – 6)(2.9x – 6) and Y2 = 800 calculate the intersection Copyright © 2008 Pearson Addison-Wesley. All rights reserved. x must be between 12.6 & 13.8 inches 3-75 3.4 Example 7 Approximating Real Zeros of a Polynomial Function (page 349) Approximate the real zeros of The greatest degree term is have end behavior similar to , so the graph will . There are at most 3 real zeros. f(0) = 10, so the y-intercept is 10. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 3-76 3.4 Example 7 Approximating Real Zeros of a Polynomial Function (cont.) The graph shows that there are 3 real zeros. Use the calculator to calculate each zero. The zeros are approximately –8.33594, –.9401088, and 1.2760488. 3-77