Objectives Lesson 5 Graphs of Rational Functions the table. near 0, we evaluate fix) to the left and right of x = 0 as shown in Because the denominator is zero when x = 0, the domain off is all real numbers except x 0. To determine the behaviour of f Solution Sketch the graph of f(x) =!. When you graphed polynomial functions, the domain was always the set of real numbers. However, rational domains are When sketching the graph of a rational function, you should Example 1 that are not in the domain. give special attention to the shape of the graph near x-values In fix), you cannot have x = 2 or you will have division by zero. valüës of which will make the denominator 0. h(x), the domain is the set of reals because there are no real In g(x), you cannot have x = 1 or x = 2 or zero will result. In often restricted to avoid division by zero. 12, g(x)=( f(x)= 21 2),orh(x)= Some emmples of rational functions are: q(x) 0 where p(x) and q(x) are polynomials in x. q(x) A rational function is a function of the form f(x) = p(x) where asymptotes. 2. You will learn how to deal with vertical and horizontal 3. You will learn how to sketch the graph of a rational function. 1. You will learn the definition of rational functions. Senior 3 Pre-alculus Mathematics Module 8, Lesson 5 59
I f(x) * as x As x approaches 0 from the right. As x approaches 0 from the left. - 60 Module 8, Lesson 5 SenIor 3 Pre-alculus Mathematics - I I -- _I_ The behaviour of 1(x) = means x = 0 is denoted by As x approaches 0 from the right, f(x) increases without bound. f(x) > as x, The behaviour of f(x) = means x = 0 is denoted by As x approaches 0 from the left, fix) decreases without bound. 1 1 I x 1 (F 1
line y = 0 is the horizontal asymptote. The line x = 0 is a vertical asymptote of the graph of/tx). The - V Senior 3 Pre-alculus Mathematics Module 8, Lesson 5 61 leading coefficient ofp(x) and b is the leading coefficient of graph has no horizontal asymptote. V - the line y = 0 is a horizontal asymptote. line If the degree ofp(x) is less than the degree of q(x), then the If the degree ofp(x) is equal to the degree of q(x), then the is If the degree ofp(x) is greater than the degree of q(x), then q(x). a horizontal asymptote where a is the a 2. The graph off has, at most, one horizontal asymptote 1. The graph off has a vertical asymptote at each real zero of determined as follows: q(x). Let f(x) = where p(x) and q(x) have no common factors. The following guidelines pertain to the asymptotes. The line y = b is a horizontal asymptote of the graph offif or ftx) -, asx -oo orx -s. oo. as x -. a, either from the right or from the left. The line x a is a vertical asymptote of the graph offifflx) - Definition of Vertical Asymptote Definition of Horizontal Asymptote asymptote horizontal I I 411 x I vertical asymptote y
. x2 c) x3+1 2 b) Note: The degree of the numerator is zero as x = 1. and the equation isy = 0. the denominator, then the x-axis is the horizontal asymptote Since the degree of the numerator is less than the degree of of the vertical asymptote. q(x). Therefore, x + 2 = 0, x = 2 is the zero and the equation 2 Find the vertical and horizontal asymptotes of Example 2 H 62 Module 8, Lesson 5 SenIor 3 Pre-alculus Mathematics Because the degree of the numerator is larger than the so the equation is x =0. degree of the denominator, there is no horizontal asymptote denominator. Therefore, x2 =0, x =0. This is the y-axis and The vertical asymptote occurs at the zeros of the y = is a horizontal asymptote denominator. ratio of the leading coefficients of the numerator and the denominator, the horizontal asymptote is given by the Since the degree of the numerator is equal to the degree of function has no vertical asymptote. denominator q(x). since 3x 2 + 1 has no real zeros, then the The vertical asymptote occurs at the zeros of the f(x)= 3x 2 1 The vertical asymptote(s) occurs at the zeros of denominator (x+2) a) 1
The above procedure for finding the asymptotes becomes a part Remember that most rational functions are discontinuous because of the nature of the asymptotes. Sketch the graph of f(x) = x 2 Example 3 beyond the vertical asymptotes. p(x) Let f(x) where p(x) and q(x) are polynomials with no q (x) 2. Find the zeros of the numerator (if any) by spiving the 3. Sketch the corresponding vertical asymptotes by solving the 5. Use the sign analysis to show where the function portion is 6. Use smooth curves to complete the graph between and 4. Find and sketch the horizontal asymptote (if any) by using at the zero location. equation of the denominator q(x) = 0 to find the zeros of the denominator. Remember the vertical asymptotes are drawn equation p(x) =0. Then plot the corresponding x-intercepts. xis 0. 1. Find and plot the y-intercept (if any) by evaluating /(O). common factors. Guidelines find the key features. Remember the y-intercept is at a point on the y-value and so points. Identifying discontinuities before you graph can help you negative. Rational functions can be difficult to graph by merely plotting the rule for finding the horizontal asymptote given earlier. of the following guidelines to graph a rational function. Senior 3 Pie-alculus Mathematics Module 8, Lesson 5 63
Solution There are no common factors to the numerator or denominator. Sketch the graph of f(x) = : L x 2 x 4 Example 4 you curve 3 x=2 x 2=0 E :. the point is at 1 less than the degree of the denominator, the horizontal asymptote is the x-axis and its equation is y = 0 help with the sketch. Draw in the curve may wish to locate additional points to x <2 is below the x-axis x> 2 + curve is above the x-axis sign nlysis: fory = vertical asymptote: find the zero by letting the expression in there is no x-intercept the denominator equal 0 x-intercept: since there is no variable in the denominator, horizontal asymptote: since the degree of the numerator is 1 11 i I I I I> 02 =_.. f(o)=: Letf(x)=O 64 Module 8, Lesson 5 SenIor 3 Pre-alculus Mathematics
Solution The numerator and denominator have no common factors. (x 2)(x+2)=0 x=o the ratio of the leading coefficient of the numerator and it goes to positive infinity zeros, and draw the vertical lines through those zeros vertical asymptote: Let the denominator x2 4=0, find the x-intercept: Let 2x2 = 0 20 =O 02_4 f(o)= x = 2 or 2 x = 2 from the right it goes to negative infinity x < 2 + as the curve approaches x = 2 from the left sign analysis: it goes to negative infinity and as it approaches it goes to positive infinity. x > 2 + as the curve approaches x = 2 from the right denominator are the same, the horizontal asymptote equation is denominator as the curve approaches x = 2 from the left horizontal asymptote: Since the degree of the numerator and y-intercept: Find flo).<hiliih> Senior 3 Pre-alculus Mathematics Module 8, Lesson 5 65
numerator. y-mtercept: Find /(0) x+4 r Solution in the region to help. x< 2. 66 Module 8, Lesson 5 SeniorS Pre-alculus Mathematics unsure of the slope. f(0)= 02±4=2 Sketch the graph of 1(x) = 2 Example 5 sign analysis: The sign analysis yields a positive region less than the degree of the denominator, the y-axis is the horizontal asymptote. throughout. Determine points that lie on the curve in areas where you are vertical asymptote: None, since x2 + 4 has no real zeros. horizontal asymptote: Since the degree of the numerator is x-mtercept: None, since there is only a constant in the There are no common factors to the numerator or denominator. horizontal asymptote Note: You can always determine the coordinates of some points the horizontal asymptote from above the line where x> 2 and The x- and y-intercepts are at the origin. The graph approaches F r
x+2 Sketch the graph of f(x) = X 4 Example 6 can see the break in the curve. When the cursor comes to x = 2, the y-value is undefined. You Graph and use trace. 1 = y-intercept: fto) =0 2 =2 x 2 Entery 4 mode to dot. However, x 2 so on the graph offix) = x 2, there is a point of The graphing calculator portrays this very well. hange the the discontinuity. discontinuity at x = 2. this is shown by an open circle to show horizontal asymptote: None. vertical asymptote: None x-intercept: x + 2 = 0, x = 2 / I I I I I I I > adds the restriction that cancellation of x +2 f(x)=x 2 xs2 x+2 f(x_2)+2) This function has a factor in common. Solution SenIor 3 Pre-alculus Mathematics Module 8, LessonS 87
L L E 2x+4 x 3 x+3 x+2 I 2. Match the function with its graph. x+4 L 4 E E x 1 68 ModuleS, Lesson 5 SenIor 3 Pre-alculus Mathematics r --- : ce -4- :z:: i) y d) f(x)= c) f(x)= = :24 b) f(x) a) f(x)= c) y= 2 29 b) Y= (x-2)(x-3) a) y equation of any vertical or horizontal asymptotes: 1. Identify the 1) x-intercepts, ii) y-intercept, and iii) the Assignment 5
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i) f(x)=: 2X1 21 x +x 2 I x+2 x+2 x-intercept(s), y-intercept(s), equation of horizontal 3. Sketch the graph of each rational function. Label the e) k) f(x)= j) f(x)= 9 h) f(x)= x 1 - g) f(x)= 2 f(x)= 2_1 d) f(x) = c) f(x)= b) f(x)= 3x a) f(x)= x 4 asymptote, equation of vertical asymptotes, if they exist. r 70 Module 8, Lesson 5 SenIor 3 Pre-alculus Mathematics -
Senior 3 Pre-alculus Mathematics Module 8, Lesson 5, Answer Key 31 Answer Key Lesson 5 1. a) i) x-intercept is the zero of the numerator x 1O,x1 ii) y-intercept is the value of x 1 1 1 = (x 2)(x 3) ( 2)( 3) 6 iii) Equation of vertical asymptote are vertical lines through the zero of the denominator (x 2)(x 3)= 0 x 2=0 x 3=O x=2 x=3 Equation of horizontal asymtote ince the degree of the numerator is less than the degree of the denominator, the x-axis is the horizontal asymptote which has the equation y =0. b) i) x-intercept is the zero of the numerator x0 ii) y-intercept is the value of iii) Equation of vertical asymptote are vertical lines drawn through the zero of the denominator x 3=0 x+3=0 x=3 x= 3 The equation of horizontal asymptote since the degree of the numerator is less than the degree of the denominator, the x-axis is the horizontal asymptote. Its equation is y = 0.
32 ModuleS, Lesson 5, Answer Key Senior 3 Pre-alculus Mathematics c) i) x-intercept: x ii) y-intercept 2 4 0, x = ±2 iii) 2. a) ii) b) i) c) iv) d) iii) 3.a) vertical asymptotes: since the denominator x2 + 4 has no real zeros, there are no vertical asymptotes horizontal asymptotes: since the degree of the numerator is equal to the degree of the denominator, = the ratio of the leading coefficient of numerator to that of the denominator. Therefore, y = 1. vertical fix) =, asymptote x 4 y.intercept x4 A! Ii F 1 <I I I II I horizontal ( asymptote W y=o There are no factors common to numerator or denominator. x-intercept: None (no variable in numerator) y-intercept: f(0) = 0 4 = vertical asymptote: x 4= 0, x = 4 horizontal asymptote: y = 0 sign analysis: x> 4 curve is below the x-axis x < 4 + curve is above the x-axis L L
2:. b) 3x x2 1 f(x)= x.intercept y.-intercept I I I- I I x3 f(x) x 3 c) x-i-2 choose points or do a sign analysis to complete the graph vertical asymptote: x +2= 0, x = 2 horizontal asymptote equation is y = = y = 3 horizontal asymptote: since the degrees are equal, the 3.0 x-intercept: 3x 0, x =0 denominator. y-intercept: f(0) = 0+2 =0 There are no factors common to numerator or x- and y-intercepts <1 I I T7j1 I I I> Senior 3 Pre-alculus Mathematics Module 8, Lesson 5, Answer Key 33
1 3)(x.1.:: y = 0 X = 3: x= 3 d) 3 0 t r U L 1 since 1Z1 U 34 Module 8, Lesson 5, Answer Key Senior 3 Pre-alculus Mathematics H analysis. choose points that work in each region or use sign less than degree of denominator horizontal asymptote: y 0, since degree of numerator is x 3 or x vertical asymptote: (x +3) = 0 y-intercept: Find f(0) = 0 9 = 3 no x-intercept. denominator. There are no factors common to numerator or x-intercept: Since the numerator has no variable, there is y.jntercept - liii I I. the graph. using selected ordered pairs or sign analysis, complete horizontal asymptote: denominator are equal Y = = 1 degrees of numerator and vertical asymptote: x 3 0, x = 3 x 3 3 y-mtercept: Fmd f(0) = = 0+2 2 x+ 2 = 0, x 2 x-intercept: Let numerator = 0 denominator. There are no factors common to numerator or f(x)= x2 9 r n
x = 3i I, f(x)= 2 x 9 I X 3 e) y 4 1) Senior 3 Pre-alculus Mathematics Module 8, Lesson 5, Answer Key 35 x+1 f(x)=, x1 x+ 2 I - I - I I I I I 1 I x+ 2 x+l y-interept (x+2)(x 1) (x 1)(x+ 1) x = x +x 2 2 = 1 choose ordered pairs in each region or do sign analysis denominator have equal degrees vertical asymptote: (x 3) = 0 (x + 3) = 0 horizontal asymptote: y = I, ince the numerator and = 0 9 = 0 4 4 x = 2 or 2 x = 3 or x = 3 (x 2)(x + 2) = 0 the x-intercept of the function denominator. y-mtercept: Fmd f(0) There are no factors common to numerator or x-intercept: Let numerator = 0 to find its zeros which are x-intercep t x-intercept I I Ix
L 1 36 Module 8, Lesson 5, Answer Key Senior 3 Pre-alculus Mathematics L1.fl use ordered pairs to complete graph or do sign analysis. horizontal asymptote: y = 1 xlorx 1 x 1 0 (x 1) = 0 vertical asymptote: (x 1) (x + 1) = 0 y-mtercept Fmd f(0) = 00 1=0 x-intercept: Let x2 0, :. x =0. There ar no factor common to numerator or denominator. < I I I I I I I 2t N= x ii xl g) y horizontal asymptote: degrees of numerator and denominator are equal vertical asymptote: x + 2 = 0 x = 2 0+1 1 x-intercept: Let numerator = 0.x + 1 = 0, x = 1 y-rntercept: Fmd f(0) = 0+2 = domain. There are common factors so we exclude x = 1 from the
x+1 x---1 h) f(x)= x+1 The numerator and denominator have no common factors. sign analysis. hoose ordered pairs to complete the graphand/or do a y-intercept: f(0) = 0 horizontal asymptote: y = 0 x+1 x 1 x 10 x+10 (x 1)(x+ 1) 0 vertical asymptote: x2 1 = 0 x-intercept: x 0 X.andYiePtS i) f(x)= complete graph 2X horizontal asymptote: y 0 vertical asymptote: x 1 = 0, x = 1 y-intercept: Find f(o) = 1 x-intercept: none The numerator and denominator have common factors. y-intercept f(x)= 1 x 1 (x 1)(x+1) Senior 3 Pre.alculus Mathematics Module 8, Lesson 5 Answer Key 37
complete graph vertical asymptote: x 9=0 vertical asymptote: none horizontal asymptote: y 0 x-intercept: 3x = 0,.. y-intercept: f(0) = 2 x-intercept: none The numerator and denominator have common factors. <ritfr 1 horizontal asymptote: y = 0 x+3 x= 3 2 y-intercept: f(0) =0 = 0 x 30 x+30 x (x 3)(x+ 3)= 0 factors. x 3 x=3 The numerator aed denominator have no common. complete graph 38 Module 8, Lesson 5, Answer Key Senior 3 Pm-alculus Mathematics j) 3x f _I x+1 k) 2 f(x)= 2