Math 11. Practice Questions Chapters and 3 Fall 01 1. Find the other endpoint of the line segment that has the given endpoint and midpoint. Endpoint ( 7, ), Midpoint (, ). Solution: Let (, ) denote the other endpoint, then = 7 + and = +. Therefore, = ( ) ( 7) = 1 and = () = 0 and so the other endpoint is ( 1, 0).. A circle has a diameter with endpoints (, ) and (, ). (a) Find the center of the circle (this will be the midpoint of the endpoints of the diameter). (b) Find the radius of the circle (this will be the distance from the center to a point on the circle). (c) Use the information from (a) and (b) to find the equation of the circle in standard form. ( ) + + ( ) Solution: (a) The center is, = (, 5). (b) The radius is the distance from one of the points on the circle, sa (, ) to the center: r = ( ) + ( 5 ( )) = + 1 = 5 (c) The equation of the circle in standard form is thus ( ) + ( + 5) = 5 3. Write the equation of the circle + + + 9 = 0 in standard form. Then find the center and radius of the circle. Solution: First, + + = 9 and completing the squares: + + 3 + + 3 = 9 + 3 + 3 and so the equation in standard form is ( + 3) + ( 3) = 9 Thus the center of the circle is ( 3, 3) and its radius is 9 = 3.
. Given the circle ( ) + ( ) = 3. (a) The center is (, ) and the radius is. (b) The center is (, ) and the radius is. (c) The center is (, ) and the radius is 3. (d) The center is (, ) and the radius is 3. (e) None of these. Solution: (a) 5. Consider the function f defined b f() = 5 + 1 (a) Find the domain of f. Write our answer in interval notation. (b) Is in the range of f? If so, find all so that f() =. (c) Is 1 5 in the range of f? If so, find all so that f() = 1 5. Solution: (a) A number is in the domain of f if 5 + 1 0, that is when 1/5. Thus the domain is (, 1 ) ( 15 ) 5, (b) For this, we solve f() =, if possible. If there is no solution, then is not in the range of f. 5 + 1 = = (5 + 1) = 0 + 19 = = 19 Thus is in the range of f, and f() = when =. 19 (c) For this, we solve f() = 1/5, if possible. If there is no solution, then 1/5 is not in the range of f. 5 + 1 = 1 5 10 = 5 + 1 0 = 11 5 However, 0 11 and so there is no solution. Therefore 1 5 is not in the range of f.. Let = 3. (a) Complete the following table. 1 0 1 3 5 (b) Plot the graph of = 3 and determine its -intercepts and -intercepts. Page
Solution: (a) The completed table is as follows: 1 0 1 3 5 1 0 1 3 1 0-1 (b)the graph is as follows 8 8 From the table of values and graph above, we see the -intercept is (0, 1) and the - intercepts are ( 1, 0) and (5, 0). Without reference to the graph, we find the -intercept b finding the -value when = 0: 3 0 = 3 = 1 and so the -intercept is (0, 1). To find the -intercept, find -values for which the -value is 0: solve 3 = 0 this means = 3 and this means = 3 or = 3 then = + 3 = 5 or = 3 + = 1 and so the -intercepts are ( 1, 0) and (5, 0) 7. An open bo is to be made from a square piece of cardboard having dimensions 88 cm b 88 cm (w = 88) b cutting out squares of area from each corner, as shown in the figure below. (a) Epress the volume V (in cubic centimeters) of the bo as a function of. (b) State the domain of V Page 3
Solution: (a) The volume of a rectangular bo is length times width times height. In this case, the height is, the length and width are both 88 and do the volume is V = (88 ) cubic centimeters. (b) For volume to make sense, the length, width and height all have to be positive. Thus > 0, and 88 > 0 (which means < 88 and so < ). Therefore, the domain is { 0 < < } 8. Find the equation of the line through the points (, 5) and (3, 3). Write the equation in slope-intercept form. Solution: The slope of the line is given b m = 1 1 = 3 5 3 = 1 = Plugging the point (, 5) into the equation = + b, we find 5 = ()() + b and so b = 5 ()() = 3. Thus the equation of the line in slope intercept form is = 3. 9. Find the equation of the line through the point (, 3) that is perpendicular to the line + 11 = 5. Write our answer in slope-intercept form. Solution: The solving for in the original line, we see = 11 + 5 11 and so the slope of the original line is m 1 = 11. The slope of the perpendicular line is thus m = 1/m 1 = 11/. The perpendicular line passes through (, 3) and so 3 = 11 11 11 ( + 8) = + + 3 = + 5 and therefore, = 11 + 5 is the requested equation of the perpendicular line. 10. Julie opened a lemonade stand and found that dail her profit is a linear function of the number of cups of lemonade sold. When she sells 0 cups of lemonade, she makes $3 and when she sells 300 cups of lemonade, she makes $5. (a) Find the profit function. (b) How man cups of lemonade does Julie need to sell to break even on a given da? (c) How man cups of lemonade does Julie need to sell to make $108 in a da? (d) How much would Julie make on a da when she sells 000 cups of lemonade? Page
Solution: (a) Let be the number of cups sold, then P () = m + b where m = 5 3 300 0 = 18 0 = 0.30 Thus P () =.0.30 + b, and so 3 = (0)(0.30) + b which means b = 3 7 = 3. Hence P () = 0.30 3. (b) To break even, she must have.0.30 3 = 0, and so = 3/0.30 = 10. That is, she must sell 10 cups of lemonade. (c) To make $108, we solve 0.30 3 = 108, and so = (108 + 3)/(0.30) = 80 cups. (d) She will make P (000) = 0.30(000) 3 = 5 dollars. 11. For this problem, f is the quadratic function f() = + 1 (a) Find the verte of the graph of f. (b) Write the quadratic function f() in standard form. (c) Does the graph of f open upward or downward? (d) Does f have a maimum value? If so, what is it? (e) Does f have a minimum value? If so, what is it? (f) Find the range of f. Write our answer in interval notation. Solution: (a) The verte of a quadratic function f() = a + b + c is (h, k) where h = b and k = f(h). In our case, a =, and b = so a h = ( ) = 1 Then k = f( 1) = ( 1) ( 1) + 1 = 3. So the verte is ( 1, 3). (b) The standard form of a quadratic function f() = a + b + c can be found b completing the square, or if ou alread know the verte (h, k), the standard form is f() = a( h) + k. In our case, a =, h = 1, and k = 3; so the standard form of f is f() = ( ( 1)) + (3) = ( + 1) + 3 (c) The graph opens downward because a = < 0. (d) The function f has a maimum value because a = < 0. The maimum value is f( 1) = 3. (e) The function f does not have a minimum value because a = < 0. (f) The range is (, 3]. This is because the graph opens downward and so the function takes all values less than or equal to its maimum value. Page 5
1. (a) Find the coordinates of the verte of the graph of the quadratic function f defined b f() = 3 + 5 + 3 (b) Find the range of the quadratic function f from (a). notation. Epress our answer in interval Solution: (a) The verte for a function f() = a + b + c is (h, k) where h = b and a ac b k = f(h) =. In this case f() = 3 + 5 + 3 and so a = 3, b = 5 and c = 3. a Therefore, h = 5 (3) = 5. Then plugging this value into f or using the verte formula for k we find k = f ( 5 ) ( = 3 5 ( + 5 ) 5 ) + 3 or k = 11 1 In either case, k = 11 1 ; therefore, the verte is ( 5, 11 1 ) (b) Because a = 3 > 0, the graph of f opens upward, so the range of f is [ 11 1, ). 13. The height above the ground, in feet, of a projectile launched with an initial velocit of 1 feet per second from an initial height of 11 feet above the ground is a function of time t in seconds, given b h(t) = 1t + 1t + 11. (a) Find the time t when the projectile reaches its maimum height. (b) Find the maimum height attained b the projectile. (c) Find the time t when the projectile hits the ground (has a height of 0 feet). Epress answer in seconds to the nearest decimal place. Solution: (a) The maimum height occurs at t = its flight. 1 ( 1) =.5; that is.5 seconds into (b) The maimum height attained is h(.5) = 1(.5) + (1)(.5) + 11 = 335 feet above the ground. (c) h(t) = 0 implies 1t + 1t + 11 = 0, and so t = 1 ± 1 + (1)(11), t > 0 t = 1 1 + (1)(11) 9.07575 3 3 Therefore, the projectile lands after approimatel 9.1 seconds. 1. A farmer has 81 feet of fencing with which to make a rectangular enclosure that will be subdivided into two separate enclosures (see the figure below). Page
(a) Write the length l as a function of width w. (b) Write the total area as a quadratic function of w. (c) Find the dimensions of the enclosure that will produce the greatest enclosed area. Solution: (a) With length l and width w as in the figure, we have l + 3w = 81. Then l = 08 3 w. (b) Thus the area is A(w) = lw = 08w 3 w = 3 w + 08w. (c) The maimum area will occur at the verte of this parabola that opens downward. Thus, w = b a = 08 3 = 13. The dimensions of the enclosure of maimum area are then w = 13 feet and l = 08 3 (13) = 0 feet. 15. The range, in interval notation, of the quadratic function f() = 3( + ) 9 is (a) [, ) (b) (, 9] (c) [9, ) (d) (, ] (e) None of these Solution: (e) 1. The coordinates of the verte of the quadratic function f() = 3( + 3) 8 are: (a) (3, 8) (b) (3, ) (c) ( 3, 8) (d) ( 3, ) (e) None of these Solution: (d) 17. Which response best describes how the graph of g() = ( + 5) 3 relates to the graph of f() = 3? (a) The graph of g is obtained b shifting the graph of f horizontall units to the left and verticall 5 units down. (b) The graph of g is obtained b shifting the graph of f horizontall units to the right and verticall 5 units up. (c) The graph of g is obtained b shifting the graph of f horizontall 5 units to the left and verticall units down. Page 7
(d) The graph of g is obtained b shifting the graph of f horizontall 5 units to the right and verticall units up. (e) None of these Solution: (c) 18. Circle letter that best describes the function f() = 3 5 3 + 8. (a) f is even because f( ) = f(). (c) f is odd because f( ) = f() (e) None of these. (b) f is even because f( ) = f(). (d) f is odd because f( ) = f() Solution: (e) 19. Circle letter that best describes the smmetr of the graph of 3 = 3 +. (a) The graph of 3 = 3+ is smmetric about the -ais because replacing with leaves the equation unchanged. (b) The graph of 3 = 3+ is smmetric about the -ais because replacing with leaves the equation unchanged. (c) The graph of 3 = 3+ is smmetric about the -ais because replacing with leaves the equation unchanged. (d) The graph of 3 = 3+ is smmetric about the -ais because replacing with leaves the equation unchanged. (e) The graph of 3 = 3 + is smmetric about the origin because replacing (, ) with (, ) leaves the equation unchanged. Solution: (b) 0. The graph of f() = 3 is given below. Find the function g() whose graph is the graph of f shifted horizontall units to the left and verticall 1 units up, and then graph g on the graph below. f() Page 8
Solution: To obtain a shift of units to the left, we replace 3 with ( + ) 3, and then to shift it verticall 1 units up, we add 1 to it. Therefore, g() = ( + ) 3 + 1. The graph is as follows: g() f() 1. The graph of = f() is given below in (a) and (b). (a) 8 8 (b) 8 On the graph of (a) above, sketch = f( 5), and on the graph (b) above, sketch = f( 5) +. 8 Solution: In (a), the graph of f is shifted horizontall 5 units to the right and in (b), the graph of f is shifted horizontall 5 units to the right and verticall units up. Page 9
(a) 8 8 (b) 8 8. The graph of a function f() is given below. Describe how the graph of the function g() = f( + ) relates to the graph of f, and then sketch the graph of g(). 8 Solution: The negative sign in front of f( + ) reflects the graph over the -ais, and replacing with + shifts the graph horizontal units to the left. Therefore the graph of g is: 8 3. The graph of = f() is given below in (a) and (b). Page 10
(a) 8 8 (b) 8 8 (i) On the graph of (a) above, sketch the graph of = 3f(). (ii) On the graph of (b) above, sketch the graph of = 3f(). Solution: (i) The graph of = 3f() is a vertical stretching of the graph of f() b a factor of 3. (ii) The graph of = 3f() is the graph in (i) verticall shifted down units.. The graphs are as follows (a) 8 8 (b) 8 8. Let f() = 8 + 7 and g() =. Find the domains of f + g, f g, fg and f g. Write our answers in interval notation. Solution: First, is in the domain of f if 8+7 0 7 8, and so domf = [ 7 8, ). Similarl, is in the domain of g if 0, and so domg = (, ]. The domain of f + g is intersection of the domains of f and g, thus The domain of f + g is [ 78 ), (, ] = [ 78 ],. Page 11
Similarl, the domain of fg and of f g is [ 78 ],. The domain of f is all in the domain of f and g so that g() 0. Noting that g() = 0 g when =, we conclude that the domain of f [ g is 7 ). 8, 5. Let f() = 3 + and g() = 3 9. (a) Find (f g)() (b) Find (g f)() Solution: (a) (f g)() = f(g()) = (g()) 3g() + = (3 9) 3(3 9) + = (9 5 + 81) 9 + 7 + = 3 1 + 3 9 + 33 = 3 5 + 357 (b) This composition is quicker than (a): (g f)() = g(f()) = 3f() 9 = 3( 3 + ) 9 = 1 9 + 18 9 = 1 9 + 9. Let f() = + 3. Find and simplif the difference quotient f( + h) f(). h Solution: For f() = + 3, the difference quotient is f( + h) f() h = ( + h) ( + h) + 3 ( + 3) h = ( + h + h ) h + 3 + 3 h = 8h + h h h = h(8 + h ) h = 8 + h 7. Use long division to find (3 3 + + 7) ( + ). Page 1
Solution: The long division is as follows 3 1 + 3 3 0 + + 7 (3 + 1 ) 3 1 ( 3 8) 1 + 1 ( 1 8) 1 + 55 Therefore, (3 3 + + 7) ( + ) = 3 1 + Note: to check our answer ou can check to see if That is ou should check to see if quotient divisor + remainder = dividend. 1 + 55 + (3 1)( + ) + 1 + 55 is 3 3 + + 7 8. Use snthetic division to find (3 5 10 3 + + 5) ( 3) Solution: The snthetic division with c = 3 is as follows 3 3 0 10 5 9 7 51 159 71 3 9 17 53 157 7 Therefore, (3 5 10 3 + + 5) ( 3) = 3 + 9 3 + 17 + 53 + 157 + 7 3 9. A polnomial P with integer coefficients satisfies P ( ) = 0 and P (3) =. Which of the following must be true concerning P ()? (a) ( ) is a factor of P (), and the remainder of P () ( 3) is. (b) ( + ) is a factor of P (), and the remainder of P () ( 3) is. (c) ( ) is a factor of P (), and the remainder of P () ( + 3) is. (d) ( + ) is a factor of P (), and the remainder of P () ( + 3) is. (e) None of the above. Page 13
Solution: (b) 30. Determine the far right and far left behavior of the polnomial P () = 5 1 + 300 11 7 3 + 7 7. (i) What is the leading coefficient of P? (ii) What is the degree of P? Is it even or odd? (iii) Because the degree of P is (even or odd) and because the leading coefficient of P is (positive or negative) we know the long-term behavior of P is (choose best response) (a) Up to far left and up to far right (b) Up to far left and down to far right (c) Down to far left and down to far right (d) Down to far left and up to far right (e) None of the above. Solution: (i) The leading coefficient of P is 5 which is negative. (ii) The degree of P is its highest power is 1 which is even. (iii) The leading coefficient of P is 5 which is negative, and the highest power of P is 1 which is even and so the graph of P goes down to the far left and down to the far right, i.e. the answer is (c). 31. Use the Intermediate Value Theorem to determine whether P has a zero between a and b. P () = 3 + 7; a =, b = 0 First, find P ( ) = and P (0) =. Then choose the best response from the following: (a) Because P ( ) and P (0) have opposite signs, we know that P has at least one real zero between and 0. (b) Because P ( ) and P (0) have opposite signs, we do not know if P has at least one real zero between and 0. (c) Because P ( ) and P (0) have the same sign, we know that P has at least one real zero between and 0. (d) Because P ( ) and P (0) have the same sign, we do not know if P has at least one real zero between and 0. Page 1
Solution: First and P ( ) = ( ) 3 ( ) ( ) + 7 = 13 P (0) = (0) 3 (0) (0) + 7 = 7. Because P ( ) < 0 and P (0) > 0, i.e. the have opposite signs, the Intermediate Value Theorem ensures that P has at least one zero between and 0, so (a) is the appropriate choice. 3. Let P () = ( ) 8 ( 5) 3. Which of the following is true? (a) The graph of P () crosses the -ais at (, 0) and intersects but does not cross the -ais at ( 5, 0). (b) The graph of P () crosses the -ais at ( 5, 0) and intersects but does not cross the -ais at (, 0). (c) The graph of P () crosses the -ais at (5, 0) and intersects but does not cross the -ais at (, 0). (d) The graph of P () crosses the -ais at (, 0) and intersects but does not cross the -ais at (5, 0). (e) None of the above Solution: (c) 33. Determine the -intercepts for the graph of P. For each -intercept, use the Even and Odd Powers of ( c) Theorem to determine whether the graph of P crosses the -ais or intersects but does not cross the -ais at that intercept. P () = ( 1) ( + 3) 5 ( ) 7 Solution: We rewrite P as follows P () = ( + 1) ( 1) ( + 3) 5 ( ) 7 So the zeros of P from smallest to biggest are = 3, 1, 1,. At the -intercept ( 3, 0) the graph of P crosses the -ais because the power of ( + 3) is 5 which is odd. At the -intercept ( 1, 0) the graph of P intersects but does not cross the -ais because the power of ( + 1) is which is even. At the -intercept (1, 0) the graph of P intersects but does not cross the -ais because the power of ( 1) is which is even. At the -intercept (, 0) the graph of P crosses the -ais because the power of ( ) is 7 which is odd. Page 15
3. Use the polnomial P given below in both standard and factored form to answer the following questions. P () = 3 8 + 3 = ( 3) ( + ) (a) Determine the far right and far left behavior of P. (b) List the -intercepts, and at each intercept determine whether the graph of P crosses or intersects but does not cross the -ais. (c) Find the -intercepts. (d) Use the above information to sketch a rough graph of P. Solution: (a) The degree of P is odd (i.e. it is 3), and its leading coefficient is which is positive, and so the graph goes down to the far left and up to the far right. (b) One -intercept is (, 0) and the graph of P crosses at this intercept because the power of the ( + ) factor is 1 which is odd. The other -intercept is (3, 0) and the graph of P intersects but does not cross the -ais at this intercept because the power of ( 3) is which is even. (c) As with an polnomial, the -intercept is (0, P (0)) which is (0, 3) (d) A graph of P is as follows: 50 0 30 0 10 5 3 10 1 3 5 0 30 0 50 35. According to Descartes Rule of signs, the number of possible positive real zeros of P () = 3 3 + + 5 1 is (a) or (b) 3 or 1 or 0 (c) 3 or 1 (d) or 0 (e) None of these. Solution: (c) 3. Use the rational zero theorem to determine the possible rational zeros of P () = 1 8 7 7 + + Page 1
Solution: The possible rational zeros of P are ± all factors of all factors of 1. The factors of are 1,, 3, and the factors of 1 are 1,, 7, 1. Therefore, the possible rational zeros of P are: ±1, ±, ±3, ±, ± 1, ±, ±3, ±, ±1 7, ± 7, ±3 7, ± 7, ± 1 1, ± 1, ± 3 1, ± 1 then simplifing and eliminating repeats, we obtain that the possible rational zeros of P are as follows ±1, ±, ±3, ±, ± 1, ±3, ±1 7, ± 7, ±3 7, ± 7, ± 1 1, ± 3 1 37. Find the zeros of the polnomial P b using snthetic division to successivel divide out two rational zeros of P and then solve the remaining quadratic to find the remaining two zeros. P () = 3 + + 9 Hint. Two of the four numbers 0, 3, 1 and 11 are zeros. Solution: Clearl 0 is not a zero. According to the rational zero theorem, 3 is a possible rational zero, we will tr dividing it out 3 1 9 3 3 9 9 1 1 3 3 0 Thus P () = ( 3)( 3 + 1 3 3). According to the rational zero theorem, 1 is a possible zero of the factor 3 + 1 3 3 which we now tr to divide out with snthetic division. 1 1 1 3 3 1 0 3 1 0 3 0 Therefore, P () = ( 3)( 3 + 1 3 3) = ( 3)( + 1)( 3) the remaining zeros of P satisf 3 = 0 and so = ± 3. Thus the zeros of P are 1, 3, 3, 3. 38. Use the graph of the polnomial P given below to answer the following questions. Page 17
5 3 1 3 1 1 1 3 5 3 5 (a) Use the far right and far left behavior to determine whether the degree of P is even or odd. (b) Use the far right and far left behavior to determine whether the leading coefficient of P is positive or negative. (c) List the zeros of P and for each zero, determine whether it has even multiplicit or odd multiplicit. Solution: (a) Because the far right and far left behaviors are different, the degree of P is odd. (b) Because the graph goes down to the far right, the leading coefficient of P is negative. (c) The zeros of P are the -values where the graph crosses the -ais. Therefore, the zeros of P are 1, and. The zero = 1 has odd multiplicit because the graph of P crosses the -ais at ( 1, 0). The zero = has odd multiplicit because the graph of P crosses the -ais at (, 0). The zero = has odd multiplicit because the graph of P crosses the -ais at (, 0). 39. (a) Find a polnomial P of degree 3 that has zeros, and. You ma leave our answer in factored form. (b) Find a polnomial Q of degree 3 that has zeros, and such that Q(0) = 7. You ma leave our answer in factored form. Solution: (a) One such answer is P () = ( + )( + )( ) and in fact, an answer of the form P () = k( + )( + )( ) where k 0 is valid. (b) From part (a), we can let Q() = k( + )( + )( ) and so Q(0) = k(0 + )(0 + )(0 ) = 7k and because Q(0) = 7 was given, we know 7k = 7, and so k = 7 7 = 1. Page 18
Therefore the requested polnomial is Q() = ( + )( + )( ) 0. Find all zeros of the polnomial P defined b given that = i is a zero of P. P () = 5 3 + 5 0 + Solution: Because comple (non-real) zeros of polnomials with real coefficients come in conjugate pairs, we know that both i and i are zeros of P (). We now use snthetic divi- i 1 5 5 0 sion divide the factors with these zeros out of P (). i 10i 0 + i 1 5 + i 1 10i i 0 Now use snthetic division again to divide out the factor ( + i). i 1 5 + i 1 10i i i 10i i 1 5 1 0 This means P () = ( i)( + i)( + 5)( 5 + 1). To find the remaining two zeros of P, we solve solve 5 + 1 = 0. According to the quadratic formula, the solutions are = 5 ± (5) (1)(1) = 5 ± 1 Thus, the zeros of P are i, i, 5 1 and 5 + 1. 1. Find a polnomial P with real coefficients of smallest degree that has zeros i, and. Solution: Because comple zeros come in conjugate pairs, the zeros of the polnomial of smallest degree would be i, i, and. Therefore the polnomial must have factors ( i), ( + i) and ( ). So, one possible polnomial of smallest degree is P () = ( i)( + i)( ) = ( + 1)( ) = 3 + Page 19
. Which is true concerning the horizontal asmptote(s) of f() = 3 + 5 + 3 1. (a) The horizontal asmptote is = 5. (b) The horizontal asmptote is = 3. (c) The horizontal asmptote is = 0. (d) The horizontal asmptote is = 3/5. (e) None of the above. 3. The slant asmptote for the rational function F () = 3 + 3 + + 1 (a) = + 3 (b) = + 1 (c) = 3 + (d) = 0 (e) None of these. Consder the rational function F defined b F () = 13 + 0 1 + 5 (a) Find the domain of F ; write our answer in interval notation. (b) Find all -intercepts of F, if there are an. (c) Find the -intercept of F, if there is one. (d) Write equations of all vertical asmptotes of F, if there are an. Solution: (a) The domain of F is all numbers for which the denominator of F is not 0. In this case, we factor the denominator as 1 + 5 = ( 5)( 9) and so the domain of F is { : 5, 9} which in interval form we have domf = (, 5) (5, 9) (9, ) (b) The -intercepts will occur at the zeros of the numerator of F that are in the domain of F. In this case, we factor the numerator as 13 + 0 = ( 5)( 8) Thus the possible intercepts occur when = 5 and = 8, but because 5 is not in the domain of F, the graph cannot have an intercept there, and so the onl -intercept is when = 8, that is (8, 0) is the -intercept. (c) The -intercept is alwas (0, F (0)) as long as 0 is in the domain of F. In this case it is, and we find F (0) = 0 13(0) + 0 0 1(0) + 5 = 0 5 = 8 9 so the -intercept is (0, 8 9 ). (d) For asmptotes, we simplif F, that is F () = 13 + 0 1 + 5 = ( 5)( 8) ( 5)( 9) = 8 9, 5, 9 Once F is simplified, the vertical asmptotes will occur where the denominator is 0, in this case where 9 = 0. Thus = 9 is the equation of the vertical asmptote. is: Page 0
5. Find all horizontal asmptotes, or eplain wh there is no horizontal asmptote, for the rational functions F, G and H defined below. (a) F () = 179 + 37 5 1 3 5 + 17 8 (b) G() = 179 + 37 5 1 17 11 3 5 + (c) H() = 1 + 375 + 17 9 3 5 + 17 9 Solution: (a) There is no horizontal asmptote because the degree of the numerator of F is 9 and this is greater than the degree of the denominator of F which is 8. (b) The horizontal asmptote is = 0 because the degree of the numerator of G is 9 and this is less than the degree of the denominator of G which is 11. (c) The degrees of the numerator and denominator of H are both 9, and so the horizontal asmptote is determined b the ratio of the leading coefficients of H, that is the horizontal asmptote has equation = 17 17 = 1. Find all slant and horizontal asmptotes for the rational function F () = 83 3 + 3 + 5 Solution: There are no horizontal asmptotes because the degree of the numerator is larger than the degree of the denominator. There is a slant asmptote because the degree of the numerator is the degree of the denominator plus 1. The slant asmptote is found b long division as follows. 3 + 5 8 3 3 + 3 (8 3 + 0) 3 ( 15) 3 + 18 Therefore, F () = 3 + and the slant asmptote is the line = 3. 18 3 + 5 7. Use the rational function F defined below to answer the following questions. F () = ( + )( ) 1 (a) Find the domain of F. Epress answer in interval notation. Page 1
(b) Find the -intercept(s) of F, if there are an. (c) Find the -intercept(s) of F, if there are an. (d) Find the horizontal asmptote(s) of F, if there are an. (e) Find the slant asmptote(s) of F, if there are an. (f) Find all vertical asmptotes of F, if there are an. For each vertical asmptote, determine the behavior of F just to the right, and just to the left of the asmptote. (g) Use the information from (a) through (f), along with plotting some additional points as necessar to sketch the graph of F along with all of its asmptotes. Solution: (a) The domain of f is (, 1) (1, ). (b) The intercepts are all -values in the domain where the numerator of f is zero. So ( + )( ) = 0, that means the intercepts are (, 0) and (, 0). (c) There is a -intercept since 0 is in the domain of F, and it is (0, F (0)) where F (0) = (0 + )(0 ) 0 1 = That is, the -intercept is (0, ). (d) There is no horizontal asmptote because the degree of the numerator of F is larger than degree of the denominator of F (i.e. > 1). (e) There is a slant asmptote, and it is found b performing the division on F (either long, or snthetic in this case), and ou can check F () = ( + )( ) 1 = + 1 + 3 1 so the slant asmptote is = + 1. (f) The vertical asmptote is = 1 (since the denominator of F is 0 when = 1 but the numerator is not). To check the behavior the graph of F just to the left of 1, notice that when is slightl smaller than 1 we get F () 3 small negative number which is a large positive number and so the graph of F increases without bound to the left of = 1. To check the behavior the graph of F just to the right of 1, notice that when is slightl larger than 1 we get F () 3 small positive number which is a large negative number and so the graph of F decreases without bound to the right of = 1. (g) A graph of f is as follows, the asmptotes are the dashed lines. Page
0 1 1 8 8 1 1 0 8. Let f() = 1 + 3 11 + 30. (a) Simplif f and find its domain. (b) Find equations for the vertical asmptote(s) for the graph of f. (c) Find the - and -intercepts of the graph of f. (d) For each vertical asmptote found in part (b), determine the behavior of f just to the right and just to the left of the vertical asmptote. (e) Find all values of c for which there is a hole in the graph of f above = c. (f) Find all horizontal asmptote of f. (g) Use the information above and plot additional points a necessar to graph f. Solution: (a) The domain of f is { : 5, } (see the simplified form of f below): f() = ( )( ) ( 5)( ) = 5 5, (b) The zeros in the denominator of the simplified rational function provide the locations of the vertical asmptotes. Therefore, the graph of f has vertical asmptote = 5, there will be a hole in the graph at =. (c) The -intercept is (, 0) and the -intercept is (0, 5 ). (d) When is close to, but larger than 5 we have f() = ( ) ( 5) 1 small positive number When is close to, but smaller than 5 we have f() = ( ) ( 5) = 1 small negative number = large negative number. = large positive number. Page 3
Thus the graph decreases without bound as approaches 5 from the right and increases without bound as approaches 5 from the left. (e) There would be a hole in the graph above = (since is not in the domain of f, but there is no vertical asmptote at = ). (f) The degrees of the numerator and denominator are the same, so the horizontal asmptote is found b setting equal to the ratio of the leading coefficients, that is = 1 (g) A graph of f along with its asmptotes is as follows 10 8 10 8 10 10 Page