Precalculus 1, 161 Fall 018 CRN 4066 Section 010 Time: Saturday, 9:00 a.m. 1:05 p.m. Room BR-11
SYLLABUS Catalog description Functions and relations and their graphs, transformations and symmetries; composition of functions; one-to-one functions and their inverses; polynomial functions; complex numbers; rational functions; conic sections. Prerequisite: MATH 118 with a grade of C or better Learning outcomes Upon successful completion of this course, students will be able to: 1. Determine basic properties of functions. Perform operations on functions 3. Graph polynomial and rational functions. 4. Perform operations on complex numbers 5. Find real and complex roots of quadratic functions. 6. Graph transformations of functions 7. Graph and determine properties of conic sections Book: Instructor: Stewart, Redlin, Watson Math 161 Custom Edition for Community College of Philadelphia. The 7-th edition. Cengage Learning. Dr. Arkady Kitover. Office: Main campus, B-5J, NE campus, 37. Office hours: Email: Web Page: NE campus, TR 5 pm - 6:30 pm, W 5 pm 6 pm. Main campus, S, 4:40 p.m. 5:40 p.m. (by appointment only) akitover@ccp.edu or akitover@hotmail.com (Email is the best way to contact me). http://faculty.ccp.edu/faculty/akitover
The web page contains the syllabus, recommended homework, and the reviews with complete solutions. COURSE OUTLINE Part 1. Linear functions and straight lines. Sections 1.10 and.5 (a) The domain and the range of a linear function y = ax + b. (b) Compositions of linear functions. (c) Inverses of linear functions( a 0). The symmetry between the graphs of a function and its inverse. (d) Slope and intercept. Slope intercept and point slope forms. (e) Horizontal and vertical lines. (f) Parallel and perpendicular lines. Review 1 Test 1 (50 points) Part. Quadratic functions. Section 3.1 (a) The standard form of a quadratic function. The vertex. (b) The domain and the range of a quadratic function. (c) The quadratic formula and the discriminant. (d) The inverse of a quadratic function with restricted domain. (e) Graphing quadratic functions and their inverses. (f) Applications of quadratic functions. Review Test (50 points) Part 3. Polynomial equations. Sections 1.5, 1.6, 3.3, 3.4, 3.5. (a) Complex numbers. Addition and subtraction of complex numbers. (b) Multiplication of complex numbers. Conjugate complex numbers. Division of complex numbers. (c) The main theorem of algebra.
(d) Solving quadratic equations and equations reducible to quadratic. (e) The remainder and the factor theorems. (f) Synthetic division (g) The rational roots test. (h) Some additional topics on finding roots of polynomial equations (if time allows). Review 3 Test 3 (50 points) Part 4. Relations and Functions. Sections.1,.,.5 -.8. (a) Relations. The domain and the range of a relation. Graphs of relations. Inverse relations. (b) Functions. What relations are functions? Vertical line test. (c) Compositions of functions. The domain and the range of a composition. (d) One-to-one functions and their inverses. Horizontal line test. Review 4 Example: radical functions n x as inverses to power functions Test 4 (50 points) Part 5. Polynomial and rational functions. Sections 3. and 3.6 (a) How to graph a completely factored polynomial? (b) Polynomials of quadratic type their graphs. (c) Linear fractions y = ax + b cx + d n n ax bx c + +, their range and, their inverses, and their graphs. Horizontal and vertical asymptotes. When does the graph of a general rational function have a horizontal asymptote? n x.
(d) Rational functions y = ax + bx + c dx + f and their graphs. Slant asymptotes. When does the graph of a general rational function have a slant asymptote? (e) How to graph a rational function when both its numerator and its denominator are completely factored? Review 5 Test 5 (50 points) Part 6. Analytic Geometry. Sections 1.9 and 11.1 11.4 (a) The distance formula. Circles. Standard equations of circles. (b) General equations of circles. Bringing general equations of circles to the standard form. (c) Ellipse. The geometric definition, major and minor axes, foci, vertices, eccentricity. (d) Standard equation of an ellipse. (e) Parabola. The geometric definition. Focus, vertex, and directrix. (f) Standard equation of a parabola. (g) Hyperbola. The geometric definition. Foci, vertices, and asymptotes. (h) Standard equation of a hyperbola. (i) Circle, ellipse, parabola, and hyperbola as conic sections. General equations of conic sections. Shift of coordinate axes. Review 6 Test 6 (50 points) Review for the final exam Final exam (100 points) Grading I will drop the lowest grade of the grades for the six 50-points tests. The grade for the final will not be dropped. If your attendance is satisfactory (not more than two unexcused absences) you can make up each of the regular class tests; one make up for every test; no make up for the final.
A: 90 100 % (315 350 points) B: 80 89% (80 314 points) C: 70 79 % (45 79 points) D: 60 69 % (15 44 points) F: 0 59 % (0 14 points) If a student misses a test without a valid excuse I will consider it as the lowest grade test. All the students must take the final exam. Students who miss the final without a valid excuse will be assigned the grade of F. Students who miss the final with a valid excuse will be assigned the grade of incomplete and will have to take the final at the beginning of the next semester. Class Rules The students are required to attend all classes. Students missing an equivalent of three weeks without a valid excuse will be dropped from the class No food in the class room. Cell phones must be put in the vibration mode BEFORE the class starts. You may not use electronic devices in the classroom for purposes not related to the class material. (Texting, surfing the web, et cetera). I will subtract 10 points from your total sum for each violation of this rule. You may not use cell phones, otherwise as calculators during a test.
Recommended homework The numbers of sections, pages, and problems correspond to the book by Stewart, Redlin and Watson Math 161 Custom Edition for Community College of Philadelphia. Edition 7. Section Pages Problems 1.10 114-115 19, 5, 7, 9, 31, 51, 75, 77.5 196 19, 7, 31, 47, 75, 79 3.1 5 54 9, 15, 35, 37, 43, 63, 67 1.5 56 57 59, 65, 81, 85, 89 1.6 64 9, 15, 19, 5, 9, 33, 39, 41, 45, 49, 57, 63, 71 3.5 93 94 13, 9, 35, 41, 59, 69 3.3 74 7, 33, 43, 47, 53, 55 3.4 83 84 19, 33, 45, 83.1 156 157 15, 3, 43, 57. 159 16 39, 49, 57, 67, 75, 77.6 06-07 1, 5, 33, 37, 43, 53.7 16 17 19, 7, 37, 41, 53.8 6-7 37, 43, 45, 53, 57, 61, 69 3. 66 67 1, 9, 35, 63 3.6 308 309 35, 45, 59, 61 1.9 19 104 11, 33, 85, 89, 93 11. 1 788 789 11, 17, 7, 9, 35 11. 796 797 11, 1, 35, 39 11.3 805 806 13, 1, 31, 37, 41 11.4 814 815 5, 9, 19, 5, 31
161 Precalculus 1 Review 1 Problem 1. Let u and v be linear functions defined as u( x) x 3, v( x) 3x. Find the compositions u vandv u. Graph the linear functions ( u v)( x) and ( v u)( x) in the same coordinate system. Solution. ( u v)( x) u( v( x)) v( x) 3 (3x ) 3 6x 1, ( v u)( x) v( u( x) 3 u( x) 3(x 3) 6x 7. The graphs of these two compositions are parallel lines shown below.
Problem. Let 1 1 y( x) x. Find the inverse function 3 both functions in the same coordinate system. y x and graph 1 ( ) 1 1 y x for x. Multiplying both parts 3 3 3 3y x whence x 3y. Interchanging the input x Solution. First we solve the equation of it by 3 we get and the output y in the last formula we get y ( x) 3x. 1 3 The graphs of the functions yx ( ) and y x are shown below. 1 ( )
Problem 3. Find an equation of the straight line through the points (-3, 5) and (, -3) in the slope-intercept form. Solution. (a) We will find the slope of the line according to the formula a y x y x 1 1 3 5 8 ( 3) 5. (b) Using one of the given points we will write an equation of the line in the point-slope form. If we use e.g. the second point then such an equation will be written as Or y y a( x x ). 8 y ( 3) ( x ) 5 (c)solving the last equation for y we obtain 8 1 y x. 5 5
Problem 4. Find an equation of the line parallel to the line you found in Problem 3 and through the point (-1, 0). Write this equation in the slope-intercept form and graph both lines in the same coordinate system. Solution. Parallel lines have the same slope whence the slope of the line we are looking for is 8 y 0 [ x ( 1)], or 5 8. An equation in the point slope form is 5 8 8 y x. The graphs are shown below. 5 5
Problem 5. Find an equation of the line perpendicular to the line you found in Problem 3 and through the point (, -). Write this equation in the slopeintercept form and graph both lines in the same coordinate system. Solution. Slopes of perpendicular lines are negative reciprocals whence the slope of the line we are looking for is 5 8. An equation in the point slope form is 5 5 13 y ( ) ( x ), or y x. The graphs are shown below. 8 8 4
161 Precalculus 1 Review In problems 1 4 consider the following quadratic function y x x = 3 + 7 + 1. Problem 1 Write the function in the standard form and find the coordinates of the vertex. y= ax ( xv) + yv Solution. We first factor out the coefficient by x, 7 1 y= 3 x x 3 3. Next we complete the square inside the parentheses using the formula In our case we have x bx x b b + = +. 1 7 49 x x= x 3 6 36 expression into the formula for y we get. After we plug in this y 7 61 = 3 x + 6 1 This is the standard form of the function y x x = 3 + 7 + 1. From the standard form we see that the coordinates of the vertex are x v 7 61 =, yv =. 6 1
Problem Find the x -intercepts of the graph of y (if any). Graph the function. Solution. The x -intercepts can be found by quadratic formula x 7 7 4 ( 3) 1 b ± b 4ac ± 7 ± 61 = = = a ( 3) 6 The x -intercepts are located approximately at -0.13 and.47. The graph of the function is shown below.
Problem 3 Consider function y over the restricted domain to the right from the vertex. Find the inverse function for this restriction of y. State the domain and the range of the inverse function. Solution The restricted domain of y is the interval 7, 6. To find an expression for the inverse function we have to solve the equation y x x = 3 + 7 + 1for x. Writing it as the quadratic formula we get 3x 7x y 1 0 + = and applying 7 ± 49 4 3 ( y 1) 7 ± 61 1y x = = 6 6 7 Because according to our choice of the restricted domain x 6 in the above formula is +. Interchanging the input x and the output y we get the inverse function in the form the correct sign y = 7 + 61 1x 6 Notice now that that the original function has the domain range 61, 1 7, 6 and the. Because for the inverse function we have to interchange the domain and the range we see that the domain of the inverse function is 61, 7 1 and its range is, 6.
Problem 4. Graph the function y with the restricted domain and its inverse in the same coordinate system. Solution Notice that the graph of y is the part of the graph from problem to the right of the vertex. The graph of the inverse function can be obtained by interchanging the x and y coordinates, in other words by reflecting the graph of y about the line y= x. The graphs are shown below.
Problem 5 A penny is thrown up from the observation deck of the Empire State building (1050ft) with the velocity 50 ft/sec. Answer the following questions (a) After what time will the penny reach its maximum height? (b) What is the maximum height? (c) After what time will the penny hit the ground? Solution. In a problem like this the height of the object aftert seconds is given by the formula ht ( ) = 16t + vt 0 + h0 wherev 0 is the initial velocity of the object in ft/sec and h0 is its initial height. In our case v 0 = 50 and h 0 = 1050whence ht t t ( ) = 16 + 50 + 1050 Because the leading coefficient of the quadratic function h is negative the function takes it greatest value at the vertex. (a) The t coordinate of the vertex is given by the formula t v b 50 = = = 1.565 s a ( 16) This answers question (a). (b) The h coordinate of the vertex is given by b 50 hv = c = 1050 = 1089.065 ft 4a 4( 16) This answers question (b). To answer question (c) we have to find the positive solution of the equation
ht t t ( ) = 16 + 50 + 1050 + + = 16t 50t 1050 0 By the quadratic formula 50 50 4( 16)1050 t = 9.81 ( 16) s
161 Precalculus 1 Review 3 Problem 1 Perform computations and present the result as a complex number in the standard forma bi. Solution (3 i)(4 3 i) (6 5 i)(5 6 i) (3 i)(4 3 i) 1 9i 8i 6i (6 5 i)(5 6 i) 30 36i 5i 30i. Recalling that i 60 11 i 1 18 we see that the above expression is equal to i. Finally we perform the division 18 i (18 i)(60 11 i) 1091 138i 1091 138 60 11 i (60 11 i)(60 11 i) 60 11 371 371 i.
Problem Find all the solutions of the equation x 6 x 3 6 0. Solution The left part of the equation is a trinomial of quadratic type and it can be factored as 3 3 ( x 3)( x ) Therefore we have to solve two equations of degree 3 Recall that all the solutions of the equation 3 x 3 x 3and 3 x. awhere a is a real number can be obtained by multiplying all the solutions of the equation As we discussed in class the solutions of the last equation are 1 3i 1 3i 1,,. Therefore all six solutions of our original equation can be obtained as 3 3 1 3i 3 1 3i x1 3, x 3, x3 3, 3 3 1 3i 3 1 3i x4, x5, x6. 3 x 1by 3 a.
Problem 3 Solve the radical equation x x 1 x. Solution. Squaring both parts of the equation we get Or equivalently Squaring both parts again we obtain x x( x 1) x 1 x. x( x 1) 1 x. 4 x( x 1) x x 1, which is equivalent to the quadratic equation By quadratic formula x 3x 6x 1 0 6 6 4 3 ( 1) 6 48 3 1 3 6 3. Clearly, the negative value of x is not a solution of the original equation, but the positive value is (e.g. we can use our calculators to see that 1 ( / 3) 3 0.1547 and that 0.1547 1.1547.1547 ). Finally 3 x 1. 3
Problem 4 Using your calculator and the method of halving the interval locate the positive solution of the equation Solution Let 3 x x 1 0with accuracy 0.01. 3 f ( x) x x 1. Then f (0) 1 0 and f (1) 1 0. Therefore the solution is in the interval[0,1]. The successive steps are shown in the table below. Interval Midpoint Sign of the value of f at the midpoint [0, 1] 1/ - [1/, 1] 3/4 - [3/4, 1] 7/8 + [3/4, 7/8] 13/16 + [3/4, 13/16] 5/3 + [3/4, 5/3] 49/64 + [3/4, 49/64] 97/18 + [3/4, 97/18] The length of the last interval is1/18 solution of our equation in the interval 0.01. We have located the positive 3 4,97 18 [0.75,0.757815]
Problem 5 Using the rational roots test, the factor theorem, and the synthetic division find all the solutions of the polynomial equation and factor the polynomial completely. 4 3 6x x 5x 11x 6 0 Solution. By the rational roots test if there are rational solutions of the equation above then their numerators can be only the numbers 1,, 3, 6, and their denominators can be only from the list1,, 3, 4, 6. Therefore the list of all possible rational roots is 1,, 3, 6, 1, 3, 1 3, 3, 1 4, 3 4, 1 6. Plugging in the numbers from the list into the original equation (either by using the synthetic division or a calculator) we successively eliminate the numbers 1,, 3, 6, 1, 3. But 3is a solution of the original equation. Indeed the synthetic division goes as follows -3/ 6-1 -5 11-6 -9 15-15 6 6-10 10-4 0 whence 3is indeed a root of the original equation and 4 3 6x x 5x 11x 6 3 3 ( x 3 )(6x 10x 10x 4) (x 3)(3x 5x 5x ) We have reduced the problem to solving the cubic equation 3 3x 5x 5x 0. The list of possible rational roots for this equation is much shorter : 1 3, 3. We easily eliminate 13but when we come to 3we hit another root.
/3 3-5 5 - - 3-3 3 0 Therefore 3 3x 5x 5x ( x 3)(3x 3x 3) (3x )( x x 1). It remains to solve the quadratic equation x x 1 0. By the quadratic formula x ( 1) ( 1) 4 1 1 1 3 1 of solutions of the original equation is The polynomial can be factored as 3 1 3i 1 3i,,, 3 i. Finally, the complete list 4 3 1 3i 1 3i 6x x 5x 11x 6 (3x )(x 3) x x
161 Precalculus 1 Review 4 Problem 1 (a) A relation is given as the set of pairs R {(1,0),(, 1),(3,1),(1,)} Find the domain and the range of this relation, graph the relation, and decide whether this relation is a function. Solution. The domain is the set of inputs{1,,3}. The range is the set of outputs { 1,0,1,}. The graph of the relation is shown below. The relation is not a function because to the same input 1 correspond two different outputs 0 and. Also we see that the graph does not pass the vertical line test.
(b) Describe the relation inverse to the relation from Problem 1. Is this relation a function? Graph it. Solution We get the inverse relation if we interchange the input and the output in each pair. 1 R {(0,1),( 1,),(1,3),(,1)}. This relation is a function: to each input corresponds only one output. The graph below passes the vertical line test.
Problem Consider the following two functions Describe the composition f composition. Graph the composition. Solution f ( x) x, g( x) x x g. Find the domain and the range of this f ( g( x)) g( x) x x. To find the domain of this function notice that the expression under the square root cannot be negative; thus we have to solve the inequality x x 0. To solve it we factor the left part and write the equation as( x 1)( x ) 0. The left part will be nonnegative either on the interval (, ] or on the interval[1, ). Therefore the domain is(, ] [1, ). On this domain the quadratic function g takes all nonnegative values whence the composition f g g also takes all nonnegative values and its range is[0, ).
Problem 3 Consider the composition g f where g and f are functions from Problem. Describe the composition; find its domain and its range; graph the composition function. g( f ( x) ( x) x x x. Clearly this expression Solution is defined if and only if x 0and therefore the domain of the composition is [0, ). The question about the range can be reduced to the following: what u u 1take ifu 0 values does the quadratic trinomial? To answer this question notice that the vertex of this quadratic trinomial is at u v b 1 a. To the right of the vertex the quadratic function is increasing because a 0and therefore the range of the composition is [ g( f (0), ) [, ). The graph of the composition is shown below
Problem 4 Consider the function 3 f ( x) x 1. Find its domain and its range. Prove that the function f is one-to-one and find its inverse Solution The function f is defined if inequality we write it as x 3 1 0 ( 1)( 1) 0 x x x. Because 1 f.. Factoring the left part of this 1 3 x x 1 x 0the last inequality is equivalent to x 1 0 4 x. Thus the domain of f is[ 1, ). The range of f is clearly[0, ). or 1 To prove that f is one-to-one and to find its inverse let that y 0. By squaring both parts we obtain y x 3 1, and 3 y x 1. Notice 3 x y 1. Thus, if the output y is given the input x is defined in the unique way which proves that f is one-to-one. Finally we get an expression for interchanging the input and the output. 1 3 f ( x) x 1, x 0. f 1 by
Problem 5 Graph the functions f and coordinate system. f 1 from Problem 4 in the same Solution
161 Precalculus 1 Review 5 Problem 1 Graph the polynomial function P( x) ( x ) x ( x 1). Solution The polynomial is of degree 4 and therefore it is positive to the left of its smallest real root and to the right of its largest real root. The roots of the polynomial are,0, and 1. The sign of the polynomial changes at and at 1, because they are simple roots, and does not change at0 because 0 is a root of multiplicity two. The sign of the polynomial is shown in the table below Interval (, ) (,0) (0,1) (1, ) Sign + - - + The graph of the polynomial is shown below.
Problem Consider the polynomial (a) Find the x -intercepts. (b) Find the critical points and the range. (c) Graph the polynomial 4 P( x) x 3x 1. Solution (a) to find the x -intercepts we have to find the real solutions of the x 3x 1 0. This equation is of quadratic type and applying the equation 4 quadratic formula we get x ( 3) ( 3) 4 1 ( 1) 3 13 a Sign minus does not provide any real solutions and taking sign plus we obtain x 3 13 1.8 (b) To find the range and the position of critical points we have to solve the equation that 4 y x 3x 1for x. Applying again the quadratic formula we see 3 9 4( y 1) x y. The expression under square root cannot be negative whence4 13 0 13 Therefore y and the range of Pis[ 13 / 4, ). 4
There are critical points corresponding to the value y 13 4. Respectively x and x 3 1. there is one more critical point at(0, 1). The list of critical points is 3. Because polynomial Pis an even function ( 3, 13 4),(0, 1),( 3, 13 4) (c) The graph of Pis shown below.
Problem 3 Consider the linear fraction Rx ( ) x 3 3x 4. (a) Find the x and y -intercepts (b) Find equations of the horizontal and the vertical asymptote (c) Graph the function together with its horizontal and vertical asymptotes (d) Find an equation of the inverse function and graph it together with its horizontal and vertical asymptotes Solution (a) to find the x -intercept we solve the equation y 0 equivalent to x 3 0 whence x 3 and the x intercept is which is ( 3,0). To find the y -intercept we just notice that if x 0 then y 34and the y -intercept is(0, 3 4). x (b) Looking at the ratio of leading terms 3x 3 horizontal asymptote is y. 3 we see that an equation of the The function is undefined if x 43whence an equation of the vertical asymptote is x 43.
(c) The graph is shown below (d) solving the equation y x 3 3x 4 whence (3y ) x 4y 3and for x we get 3xy 4y x 3 x 1 4x 3 R ( x) 3x 4y 3 3y. The inverse function is The x and y -intercepts are, respectively, ( 3 4,0) and(0, horizontal asymptote is 43 graph is shown below. 3 ). The y, the vertical asymptote is x 3. The
Problem 4 Consider the rational function (a) Find the x and the y -intercepts, if any Rx ( ) (b) Find an equation of the horizontal and the slant asymptote (c) Find the critical points, if any, and the range of the function (d) Graph the function together with its horizontal and slant asymptotes x x x 1 1. Solution (a) the equation no x -intercepts. The y -intercept i (0,1). x x 1 0has no real solutions whence there are (b) The vertical asymptote is clearly x 1. To find an equation of the slant asymptote we will divide x x 1by x 1 using e.g. the synthetic division -1 1 1 1-1 0 1 0 1 The quotient is x whence an equation of the slant asymptote is y (c)to find the range and the critical points (if they exist) we have to solve the x x 1 equation y x 1 quadratic equation x (1 y) x (1 y ) 0. The quadratic formula provides x x for x. This equation is equivalent to the following y y y y y y 1 ( 1) 4( 1) 1 ( 1)( 3) The expression is defined if and only if ( y 1)( y 3) 0which happens when either y 1or y 3. The range of Ris therefore
(, 3] [1, ) To find the critical points we plug in into the expression for x the values y 1and y 3getting x 0 and x are located at (, 3)and at(0,1). (d) The graph of R and its asymptotes is shown below, respectively. The critical points
Problem 5 Consider the function R x x 4 1 (a) Find the x -intercepts (b) Find the vertical asymptotes (c) Find the horizontal or slant asymptote, if any (d) Find the sign of the function (e) Graph the function and its asymptotes Solution (a) the x -intercepts are (,0)and(,0) (b) the vertical asymptotes are x 1and x 1 (c) the degree of the numerator equals the degree of the denominator; the ratio of the leading terms is 1; therefore there is the horizontal asymptote with the equation y 1 (d) we see from (c) that the sign of Ris positive far right and far left. Because Rx ( ) ( x )( x ) ( x 1)( x 1) every root of the numerator and of the denominator is simple and the sign of the function changes at points, 1,1 and. Interval (, ) (, 1) ( 1,1) (1,) (, ) Sign of R + - + - +
(f) The graph is shown below
161 Precalculus 1 Review 6 Problem 1 The equation of a circle is given as x x y y 8 + + 4 = 5 (a) Bring this equation to the standard form (b) Find the center and the radius of the circle (c) Graph the circle Solution (a) we will complete squares using the formula Then x bx x b b + = + x 8 x= ( x 4) 16and y + 4 y= ( y+ ) 4. Thus the standard form of the original equation is ( x 4) + ( y+ ) = 5 (b) From the standard equation we see that the center of the circle is (4, ) and its radius is 5 (c) The graph of the circle is shown below
Problem the vertices of an ellipse are at ( 5,0) of the ellipse is ε = 0.4 (a) Find the standard equation of the ellipse (b) Find the position of foci and of the ends of the minor axis (c) Graph the ellipse and at(5,0); the eccentricity Solution (a) from the information provided we see that the center of the ellipse is at the origin and thata = 5. Next we see that 1 b a = ε = 0.16 whenceb = a (1 0.16) = 5 0.84 = 1. The standard equation thus is ( ) (b) Foci are located at a b,0 x y + = 1 5 1 ± or at(,0) and at(,0).
The ends of the minor axis are located at ( 0, ± 1) or approximately at (0, ± 4.6) The graph of the ellipse is shown below
Problem 3 the equation of an ellipse is x + 8x+ 3y 18y= 1 (a) Write this equation in the standard form (b) Find the center, vertices, ends of the minor, axes, foci, and the eccentricity of this ellipse (c) Graph the ellipse Solution we transform the right part of the equation in the following way using the completion of squares x 8x 3y 18y ( x 4 x) 3( y 6 y) + + = + + = = [( x+ ) 4] + 3[( y 3) 9] = = + + ( x ) 3( y 3) 35 Therefore ( x ) 3( y 3) 36 + + = and finally the standard equation is ( x+ ) ( y 3) + = 1 18 1 (b)notice that a = 18 = 3, b = 1 = 3, and = = 6. The center of the ellipse is at (,3) f a b, the vertices are at ( ± 3,3), the ends of the minor axis are at(,3± 3), the foci b 1 are at( ± 6,3) and the eccentricity is ε = 1 = 0.58 a 3
(d) The graph of the ellipse is shown below Problem 4 consider the following equation of a parabola y y x + 4 + 4 = 8 (a) Write this equation in the standard form (b) Find the position of the vertex and the focus of the parabola. Find an equation of the directrix (c) Graph the parabola and the directrix Solution completing the squares we rewrite the equation as ( y ) 4x 4 8 + + =. The standard form is
( y + ) x 3 = 4 (b) From the standard equation we see that the vertex is at(3, ). The sign minus in the right part shows that the graph of the parabola is to the left of the vertex. The distance from the vertex to the focus and to the directrix is 1 4 4 = 1. The focus is at (47 / 16, ). An equation of the directrix is 16 x = 49 16 (c) The graph is shown below (the distances between the focus, the directrix, and the vertex are shown greater then in reality).
Problem 5 consider the following equation of a hyperbola x x y y + + 8 = 9 (a) Bring the equation to the standard form (b) Find the location of vertices and foci and equations of asymptotes (c) Graph the hyperbola and its asymptotes Solution (a) x x y y x x y y + + 8 = + ( 4 ) = = ( x+ 1) 1 [( y ) 4] = = + + ( x 1) ( y ) 7 Whence ( x 1) ( y 1) + = and the standard equation is (x +1) ( y ) = 1 The center of the hyperbola is at ( 1,). The axis of the hyperbola is horizontal. Notice that a = and b = 1whence f = a + b = 3. The vertices are at ( 1±,)and the foci are at ( 1± 3,). The asymptotes are given by equations y = ± b a (x +1) = ± (x +1).
(d) The graph is shown below 3.4 y 3. 3.8.6.4. 1.8 1.6 1.4 1. 1 0.8 0.6 0.4 0. 0 x 5.4-5. -5-4.8-4.6-4.4-4. -4-3.8-3.6-3.4-3. -3 -.8 -.6 -.4 -. - -1.8-1.6-1.4-1. -1-0.8-0.6-0.4-0. 0 0. 0.4 0.6 0.8 1 1. 1.4 1.6 1.8..4.6.8 3 3. 3.4 3.6 3.8 4 4. 4.4-0. -0.4-0.6-0.8-1 -1. -1.4-1.6
161 Precalculus 1 Review for the Final Exam 1. Consider two points on the coordinate plane P (-, 3) and Q (4,-1). (a) Find the distance between P and Q. ( points) d ( x x ) ( y y ) [4 ( )] ( 1 3) 1 1 6 4 5 13 (b) Find the coordinates of the midpoint. ( points) x1 x 4 xm 1. y1 y 3 ( 1) ym 1. (c) Find the slope of the line through P and Q. ( points) y y1 1 3 m. x x 4 ( ) 3 1 (d) Find the "slope-intercept" equation of this line. ( points) First we will find a point-slope equation using, for example, the second point y y m( x x ), y ( 1) [ x 3 4], y 1 8 x. 3 3 Finally, y 5 x+ is the slope-intercept equation. 3 3. Consider the quadratic function y x 3x 1 (a) Find the coordinates of the vertex. ( points) b 3 3 xv. a ( 1) y v b 9 13 c 1 4a 4 4
(b) Find the x-intercepts (if any). ( points) x x 1 b b 4ac 3 13 3 3 4 ( 1) 1 a ( 1).3, 3 13 x 3.3. (c) Find the range of the function. ( points) The coefficient a is negative, therefore the range is from the vertex down: (-, 13/4]. (d) Graph the function. (4 points). 3. A rectangular plot of land on the edge of a river is to be enclosed with fence on three sides. Find the dimensions of the rectangular enclosure of the greatest area if the side that goes along the river does not require fencing and the total length of the fence is 00 m. (10 points) Let x be the side along the river and y be the perpendicular side. Then x y 00. whence x 00 y. The area of the rectangle is xy (00 y) y y 00 y. The greatest value of this quadratic function will be at the vertex
Then x 00 50 100. y 00 ( ) 50. 4. Graph the polynomial function (10 points) P( x) x ( x 4) The x-intercepts are at -, 0, and. The function is positive to the left of - (because the leading term is x 4 ). It changes sign at - and at (because the exponent in each case is 1), but does not change sign at 0 (because the exponent is ). 5. Find the rational roots of the polynomial 3 x 6x 11x 6 and factor the polynomial completely. (10 points) Possible rational roots are factors of 6: 1,-1,,-,3,-3,6,-6. The solutions cannot be negative (every term becomes negative) so we try only 1,,3,6. We can see at once that 1 is a root and therefore x-1 is a factor. From synthetic (or long) division we see that 3 x 6x 11x 6 ( x 1)( x 5x 6) ( x 1)( x )( x 3). The roots are 1,, 3. 6. For the function f ( x) x 1 (a) Find the domain. (3 points)
The function is defined when x 1 0, whence x 1. The domain is the interval [1, ) (b) Find the range. (3 points) All values from and up. The range is the interval [, ). (c) Find the inverse function, its domain and range. (4 points) y x x y x y 1, 1, 1 ( ), x y y y ( ) 1 4 5. The inverse g( x) x 4x 5. The domain of the inverse function is the range of the original function in our case the interval[, ). The range equals to the domain of the original function, [1, ). 7. For the rational function x 1 f( x) x 4 (a) Find the horizontal or slant asymptote, if any. (3 points) The ratio of leading terms is x / x 1/ x and it is close to 0 when x is a large positive or a large negative number. Therefore the graph has the horizontal asymptote y=0 (the x- axis). (b) Find the vertical asymptotes, if any. (3 points) The denominator is 0 when x= or x=-. The vertical asymptotes go through points and - on the x-axis. (c) Find the x and y intercepts, if any. (3 points) The x-intercept: x=1. The y-intercept (when x=0) is ¼.
(d) Graph the function. (5 points)
8. For the functions f x x g x x (a) Find the compositions (4 points) f ( g( x)) and g( f ( x )) ( ) and ( ) 1 f ( g( x)) x 1, g( f ( x)) ( x) 1 x 1. (b) Find the domain of each composition function. (8 points) 1. (-, ).. [0, ). 9. Function f is defined in the following way f( x) 0, x 0 x, 0 x 1 x,1 x 0, x (a) Graph the function. (4 points)
(b) Graph the function g( x) f ( x 1) 3. (4 points) (c) How are the two graphs related? ( points) We obtain the second graph if we move the first one by one unit to the left, stretch it twice along the y-axis, and move by three units down. 10. Consider the following functions. f x x x 4 ( ) 3 1, g x x x x 5 3 ( ) 3, h( x) x x 4 3 (a) Which of these functions are even? ( points) f ( x) f ( x), f is an even function. (b) Which are odd? ( points) g( x) g( x), g is an odd function. (c) What kind of symmetry do their graphs have? ( points) Graph of f is symmetric about the y-axis. Graph of g is symmetric about the origin. 11. Consider the following equation of a conic. 8x y x 5y 0. a. Identify the conic ( points). The conic is a hyperbola (or a degenerate conic) because the coefficients by x and y have opposite signs. b. Write an equation of the conic in the standard form (5 points).
1 1 1 1 1 8x x 8( x x) 8[( x ) ] 8( x ) ; 4 8 64 8 8 5 5 y 5 y ( y ). 4 After we plug these expressions into original equation it becomes ( y 5/ ) 8( x 1/8) 49 /8, or ( y 5/ ) ( x 1/8) 1. 49/8 49/ 64 c. Find the center, the vertices and the foci (5 points). The center is at (-1/8, -5/). The axis of the hyperbola is vertical, a 7 /8, b 7 / 8.5, and f 49 49 1 8 64 8 vertices are at ( 1/ 8, 5 / 7 / 8) and ( 1/ 8, 5 / 7 / 8). The foci are at (-1/8, - 41/8) and at (-1/8, 1/8). d. Graph the conic (5 points).. The