Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Grima MAT 151 Final exam review #1-2: find an equation for the line with the given properties. Express your answer in slope intercept form. 2 1) Slope = 3; containing the point (6,5) let m = 2/3 x1 = 6 y1 = 5 use y β y1 = m(x β x1) 2 π¦ β 5 = 3 (π₯ β 6) 2 2 Hint 3 β β6 = 3 β 2 y β 5 = π₯- 4 3 +5 +5 π π Answer #1: π = π + π 2) Passing through the points (3,1) and (4,5) First find the slope 5β1 4 π = 4β3 = 1 = 4 Next let m = 4 x1 = 3 y1 = 1 use y β y1 = m(x β x1) y β 1 = 4(x β 3) y β 1 = 4x β 12 +1 +1 Answer #2: y = 4x - 11 β6 1 = β12 3 = β4 #3-4: Find the x and y-intercepts. 3) π¦ = π₯ 2 β5π₯β6 π₯+1 Find x-intercept (You can just set the numerator equal to 0, or you may cross multiply. Both methods give the same x-intercepts.) 0 1 = π₯ 2 β5π₯β6 π₯+1 Cross multiply 1(x2-5x-6) = 0(x +1) x2 β 5x β 6 = 0 (x β 6)(x + 1) = 0 xβ6=0 x+1=0 x=6 x = -1 y-intercept, let x = 0 π¦= 02 β5(0)β6 0+1 = β6 1 = β6 Answer #3: x-intercepts (-1,0) and (6,0) y-intercept (0,-6) 4) x2 + y2 = 121 let y = 0 to find x-intercept x2 + 02 = 121 x2 = 121 x2 β 121 = 0 (x + 11)(x-11) = 0 x + 11 = 0 x β 11 = 0 x = -11 x = 11 let x = 0 to find y-intercept 02 + y2 = 121 y2 = 121 y2 β 121 = 0 (y + 11)(y-11) = 0 y + 11 = 0 y β 11 = 0 y = -11 y = 11 Answer# 4: x-intercepts (11,0) and (-11,0) y-intercepts (0,11) and (0, -11) #5-8: find the domain and range. 5) 6) Domain will be the x-coordinates of each point, the Range will be the y-coordinates. Answer Domain = { 3,4,5,7} Range = {-1, 2, 6} Domain [x-coordinate far left point, x-coord far right point] Range [y-coordinate of bottom point, ycoordinate of top point] Answer Domain [0,5] Range [-5, 4] 7) Graph needs to be extended to the left. The graph goes to the far left edge of the x-axis which has an x-coordinate of ββ The domain starts at the far left edge of the xaxis or x = ββ The domain ends at the point (0,2) The domain ends at x = 0 Domain (ββ, 0] Graph extends to the bottom of the y-axis. The range starts at the bottom of the graph which is at y = ββ The range ends at the point (0,2) or more specifically at y = 0 Range (ββ, 2] Answer: Domain (ββ, π] Range (ββ, π] 8) The graph needs to be extended in both directions. It will be extended to the far left of the and far right of the x-axis and the bottom and top of the y-axis. Answer: do main (ββ, β) range (ββ, β) Final exam review #9-11: Use algebra to find the domain of each function. Write your answer in interval notation π₯β4 9) π(π₯) = π₯ 2 +6π₯β7 To find the domain, ignore the numerator Solve denominator = 0 x2 + 6x - 7 = 0 (x + 7)(x β 1) = 0 x+7=0 x = -7 xβ1=0 x=1 plot x = ββ, β7,1, β on a number line and create intervals with all round parenthesis. (ββ, β7) ββ β7 (β7,1) (1, β) 1 β Answer #9: domain (ββ, βπ) βͺ (βπ, π) βͺ (π, β) 10) π(π₯) = βπ₯ + 5 π₯+5β₯0 -5 - 5 π₯ β₯ β5 This is the answer, but it is not in interval notation. Answer #10: [βπ, β) 11) f(x) = 2x β 6 This is a polynomial and there is no algebra required to find the domain. Answer #11: π πππππ (ββ, β) 12) 12) Find the following: a) the interval(s) where the function graphed is increasing The graph increases from the beginning which is x = -β To the first turning point x = -3 It also increases from the second turning point x = -1 to x = β Answer 12a) increasing (ββ, βπ) βͺ (βπ, β) b) the interval(s) where the function graphed is decreasing The graph is decreasing between the turning points x = -3 to x = -1 Answer 12b) decreasing (-3,-1) c) The values of x (if any) where the function has a local maximum The x-coord of the high point 12c) answer x = -3 d) The local maximum value (if any) Asking for y coord of max 12d) answer y = -8 e) The values of x (if any) where the function has a local minimum asing for x-coord of low point. Answer 12e) x = -1 f) The local minimum values (if any) Wants y-coord of low point. Answer 12f) y = -12 13) Find the average rate of change of f(x) = x3 + 6x2 from 0 to 2 Create two points. We are given the x-coordinates and need to find y-coordinate. f(0) = (0)3 + 6(0)2 f(0) = 0 + 0 f(0) = 0 First point (0, 0) f(2) = (2)3 + 6(2)2 f(2) = 8 + 24 f(2) = 32 Second point (2, 32) Average rate of change is the slope of the line that connects the points. 32β0 2β0 Average rate of change = = 32 2 Answer #13: Average rate of change = 16 14) Let π(π₯) = π₯ 2 14a) Find f(x+2) - 4 The x + 2 needs to go inside the parenthesis as it is inside the parenthesis. This will move the graph left 2. The minus 4 will go after the parenthesis and move the graph down. Answer #14a: f(x + 2) β 4 = (x + 2)2 - 4 14b) Describe how the graph of the given function relates to the graph of π(π₯) = π₯ 2 Answer #14b: left 2 and down 4 15) Let π(π₯) = βπ₯ 15a) Find -g(x-3) - 5 The negative in front of the g will go in front of the square root. It will reflect the graph over the x-axis. The x β 3 will go under the radical, and it is okay to put it in a parenthesis, but the parenthesis is not needed. It will move the graph right. The minus 5 goes after the square root. It will move the graph down. Answer #15a: -g(x β 3) β 5 = ββπ β π β π 15b) Describe how the graph of the given function relates to the graph of π(π₯) = βπ₯ Answer #15b: reflect over x-axis, right 3 and down 5 16) Let β(π₯) = |π₯| 16a) Find h(x-3) + 4 The x β 3 needs to go inside the absolute value, it can be left in a parenthesis but the parenthesis are not needed. This will move the graph right 3. The plus 4 will go after the absolute value and move the graph up. Answer #16a: h(x β 3) + 4 = |π β π| + π 16b) Describe how the graph of the given function relates to the graph of β(π₯) = |π₯| Answer #16b: right 3 and up 4 17) f(x) = x2 17a) Find f(x-3) + 4 and describe the transformation as compared to the function f(x) = x2. Answer #17a: f(x β 3) + 4 = (x β 3)2 + 4. shifts graph right 3 and up 4. 17b) make a table of values and sketch a graph x 1 2 3 4 5 f(x β 3) + 4 (1 β 3)2 + 4 = 8 (2 β 3)2 + 4 = 5 (3 β 3)2 + 4 = 4 (4 β 3)2 + 4 = 5 (5 β 3)2 + 4 = 8 point (1,8) (2,5) (3,4) (4,5) (5,8) 17c) state the domain and range of the function Domain β the graph extends from the far left to the far right of the x-axis unbroken so the domain is (ββ, β) Range β the y-coordinate of the bottom of the graph is y = 4. The y-coordinate of the top is β so the range is [4, β) Answer #17c: domain (ββ, β) range [π, β) 17d) state the intervals where the function in increasing and decreasing the graph increases from the x-coordinate of the vertex (x= 3) to the end of the graph ( x = β) The graph decreases from the start of the graph ( x = β) to the x-coordinate of the vertex (x = 3) Answer #17d: increasing (π, β) decreasing (ββ, π) 17e) state if the function has a local maximum, if it does state the local maximum value Answer #17e: The function has no local minimum 17f) state if the function has a local minimum, if it does state the local minimum value Answer #17f: There is a local minimum (the vertex) at x = 3, the local minimum value is y = 4. 18a) Use synthetic division to factor the polynomial, f(x) = x3 + x2 + 2x + 2 here is a graph of f(x) I will use the number (-1) for my synthetic division as it is the only x-intercept. since x = -1 is a zero, I know (x+1) is a factor of f(x) the synthetic division will get me the remaining factors. 1 1 -1 1 0 The result of my synthetic division gives me -1 π₯ 3 +π₯ 2 +2π₯+2 π₯+1 2 -2 0 = π₯ 2 + 2 πππππππππ 0 so now I can factor f(x) f(x) = x3 + 2x2 β 5x β 6 = (x+1)(x2+2) Answer #18a: f(x) = (x+1)(x2 + 2) 18b) Solve x3 + x2 + 2x + 2= 0 (x + 1)(x2 + 2) = 0 x+1=0 x = -1 2 0 2 x2 + 2 = 0 x2 = - 2 βπ₯ 2 = ±ββ2 π₯ = ±β2π Answer #18b: x = -1, π = ±βππ (the x2 + 2 is prime, so this is completely factored) #19 β 20: For each problem find the following: a) Domain b) Vertical Asymptote (if any) c) Horizontal asymptote, or slant asymptote d) x- intercept(s) if any e) y-intercept(s) if any f) Sketch a graph of the function : label all the features found in parts b β e 19. f ( x) ο½ 2x ο« 6 xο3 19a) Domain x- 3 = 0 x=3 Answer: all real numbers except 3 19b) Vertical Asymptote (if any) Answer: x = 3 19c) Horizontal asymptote, or slant asymptote 2 π¦= =2 1 Answer: y = 2 19d) x- intercept(s) if any 2x + 6 = 0 2x = -6 x = -3 Answer (-3,0) 19e) y-intercept(s) if any π(0) = 2β0+6 0β3 6 = β3 = β2 Answer #19e: (0,-2) 19f) Sketch a graph of the function : show all the features found in parts b β e y 8 6 4 2 x -18 -16 -14 -12 -10 -8 -6 -4 -2 2 (-3,0) -2 (0,-2) -4 -6 -8 4 6 8 10 12 14 16 18 20. f ( x) ο½ x ο1 x ο« 4 x ο 21 2 20a) Domain x2 + 4x β 21 = 0 (x+7)(x-3) = 0 x +7=0 xβ3 =0 x = -7 x = 3 Answer: all real numbers except -7, 3 20b) Vertical Asymptote (if any) Answer: x = -7 and x = 3 20c) Horizontal asymptote, or slant asymptote There is no work to find the horizontal asymptote. The largest exponent is in the denominator so x-axis which has the equation y = 0 is the horizontal asymptote. Answer: y = 0 (the x-axis) 20d) x- intercept(s) if any Itβs enough to just set the numerator equal to zero. xβ1=0 x=1 Answer: (1,0) 20e) y-intercept(s) if any 0β1 β1 1 π(0) = 02 +4β0β21 = β21 = 21 Answer #20e: (0, 1/21) 20f) Sketch a graph of the function : show all the features found in parts b - e #21-22: Write each side with the same base then solve. Be sure to check your answer. 21) 3x-3 = 27 3x-3 = 27 3x-3 = 33 xβ3=3 Answer #21: x = 6 1 π₯β2 2 22) ( ) 1 π₯β2 (2) 1 π₯β2 (2) = 1 8 13 = 23 1 3 = (2) xβ 2 = 3 Answer #22: x = 5 #23-26: Solve the logarithmic equations, round to 2 decimals when needed. 23) log2(x+1) = 5 log2(x+1) = 5 25 = x + 1 32 = x + 1 Answer #23: x = 31 24) log4(x-5)=3 Log4(x-5) = 3 43 = x β 5 64 = x β 5 Answer #24: x = 69 25) log2 (x+14) β log2 (x+6)= 1 πππ2 21 = 2 1 = π₯+14 π₯+6 =1 π₯+14 π₯+6 π₯+14 π₯+6 2(x +6) = 1(x+14) 2x + 12 = x + 14 2x β x = 14 β 12 Answer #25: x = 2 26) log2 (x+2)+log2 (x+6) = 5 Log2(x+2)(x+6) = 5 Log2(x2 + 8x+ 12) = 5 25 = x2 + 8x+ 12 32 = x2 + 8x + 12 0 = x2 + 8x β 20 0 = (x + 10)(x β 2) x+ 10 = 0 x = -10 (doesnβt check) Answer #26: x = 2 xβ2=0 x=2 Final Exam Review 27) Solve using the substitution method 4 4 1 xο« y ο½5 5 4 x ο½ y ο«1 1 20 β 5 π₯ + 20 β 4 π¦ = 20 β 5 16x + 5y = 100 16(y+1) + 5y = 100 16y + 16 + 5y = 100 21y + 16 = 100 21y = 84 y = 84/21 y=4 Solve for x x=y+1 x=4+1 x=5 Answer #27: (5,4) 2 28) Solve using the elimination method. 4(2x + 3y = 13) 3(5x β 4y = 21) 8x + 12y = 52 15x β 12y = 63 23x = 115 x = 115/23 x=5 Solve for y: 2x + 3y = 13 2(5) + 3y = 13 10 + 3y = 13 3y = 3 y=1 Answer #28: (5,1) 2 x ο« 3 y ο½ 13 5 x ο 4 y ο½ 21 29) Solve each system of equations, by hand without matrices 2π₯ + 4π¦ β 5π§ = β5 βπ₯ + π¦ + 2π§ = 7 π₯ β 3π¦ + 3π§ = 4 (pair the middle equation with the other 2 and drop out the xβs) Step 1: 2(βπ₯ + π¦ + 2π§ = 7) 2π₯ + 4π¦ β 5π§ = β5 -2x + 2y + 4z = 14 2x + 4y β 5z = -5 6y - z = 9 Step 2: 6y β z = 9 3(-2y + 5z = 11) 6y βz = 9 -6y + 15z = 33 14z = 42 z=3 Solve for y 6y β 3 = 9 6y = 12 y=2 Step 3 Solve for x π₯ β 3π¦ + 3π§ = 4 x β 3(2) + 3(3) = 4 xβ6+9=4 x+3=4 x=1 Answer #29: (3,2,1) βπ₯ + π¦ + 2π§ = 7 π₯ β 3π¦ + 3π§ = 4 -2y + 5z = 11 #30-31: Solve the following systems of equations. 30) x ο 7 y ο½ ο4 x 2 ο« y 2 ο½ 10 Solve top for x x β 7y = -4 +7y + 7y x = 7y β 4 substitute into bottom (7y-4)2 + y2 = 10 (7y -4)(7y β 4) + 1y2 = 10 49y2 β 28y β 28y + 16 + 1y2 = 10 50y2 β 56y + 16 = 10 50y2 β 56y + 6 = 0 Divide by 2 25y2 β 28y + 3 = 0 Factor (25y β 3)(y β 1) = 0 25y β 3 = 0 25y = 3 y = 3/25 yβ1=0 y=1 Solve for x x = 7y β 4 x = 7(3/25) β 4 used calculator x = -79/25 x = 7(1) β 4 x=3 Answer #30: (-79/25, 3/25) (3,1) 31) x ο« 2 y ο½ 11 x 2 ο« y ο½ 13 Solve bottom equation for y x2 + y = 13 -x2 - x2 y = 13 β x2 Substitute into top equation x + 2(13 β x2) = 11 x + 26 β 2x2 = 11 -x - 26 +2x2 +2x2 - x -26 0 = 2x2 β x β 15 0 = (2x + 5)(x - 3) 2x + 5 = 0 xβ3= 0 2x = -5 x=3 x = -5/2 Solve for y y = 13 β x2 y = 12 β (-5/2)2 solved on my calculator y = 13 β 32 y=4 y = 27/4 Answer #31: (-5/2, 27/4) (3,4) 32) Find the difference quotient for the function f(x) = x2 +3x - 5: π(π₯ + β) β π(π₯) β f(x+h) = (x + h)2 + 3(x+h) β 5 f(x + h) = (x + h)(x + h) + 3(x + h) β 5 f(x + h) = x2 + 1xh + 1xh + h2 + 3x + 3h - 5 f(x + h) = x2 + 2xh + h2 + 3x + 3h -5 (π₯ 2 +2π₯β+β 2 +3π₯+3ββ5)β(π₯ 2 +3π₯β5) π(π₯+β)βπ(π₯) β = π(π₯+β)βπ(π₯) β = π₯ 2 +2π₯β+β 2 +3π₯+3ββ5βπ₯ 2 β3π₯+5 β π(π₯+β)βπ(π₯) β = 2π₯β+β 2 +3β 5 π(π₯+β)βπ(π₯) β = β(2π₯+β+3) β Answer #32: β π(π+π)βπ(π) π = ππ + π + π Answers: 2 1) π¦ = 3 π₯+1 2) y = 4x β 11 3) x-intercepts (-1,0) and (6,0) y-intercept (0,-6) 4) x-intercepts (11,0) and (-11,0) y-intercepts (0,11) and (0, -11) 5) domain {3,4,5,7} Range {-1,2,6} 7) ππππππ (ββ, 0] πππππ (ββ, 2] 6) Domain [0,5] range [-5,4] 8) ππππππ (ββ, β) πππππ (ββ, β) 9) domain (ββ, β7) βͺ (β7,1) βͺ (1, β) 10) ππππππ [β5, β) 11) ππππππ (ββ, β) 12a) (ββ, β3) βͺ (β1, β) 12b) (-3,-1) 12e) x = -1 13) average rate of change = 16 12f) y = -12 12c) x = -3 12d) y = -8 14a) f(x+2) β 4 = (x + 2)2 - 4 14b) Same shape as f(x) = x2, except moved down 4 units and left 2 units 15a) -g(x-3) β 5 = ββπ₯ β 3 β 5 15b) Same shape as g(x) = βπ₯, except moved down 5 units, right 3 units and reflected over x-axis 16a) h(x-3) + 4 = |π₯ β 3| + 4 16b) Same shape as h(x) = |π₯|, except shifted up 4 units and right 3 units 17a) f(x-3) + 4 = (x-3)2 + 4 The graph has the same shape as f(x) = x2, except it is shifted right 3 units and up 4 units. 17b) x 5 4 3 2 1 f(x) or y 8 5 4 5 8 Computation, use calculator to get y - column (5-3)2 + 4 (4-3)2 + 4 (3-3)2 + 4 (2-3)2 + 4 (1-3)2 + 4 17c) Domain (ββ, β) Range [4, β) (see me for help if you need some finding the domain and range) 17d) The graph is increasing (3, β) and decreasing from (ββ, 3) 17e) The graph does not have a high point so it has no local maximum 17f) The low point is the local minimum. We say there is a local minimum at x = 3 and the local minimum value is y = 4 18a) f(x) = (x+1)(x2 + 2) 18b) x = -1, ±πβ2 19a) all real numbers except 3 19d) (-3,0) 19b) x = 3 19c) y = 2 19e) (0, -2) 19f) y 8 6 4 2 x -18 -16 -14 -12 -10 -8 -6 -4 -2 2 (-3,0) 4 6 8 10 12 14 16 18 -2 (0,-2) -4 -6 -8 20a) all real numbers except -7,3 20b) x = -7 and x = 3 20c) y = 0 (the x-axis) 11d) (1,0) 20e) (0, 1/21) 20f) y 4 3 2 1 x -14 -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 -1 -2 -3 -4 21) x = 6 22) x = 5 23) x = 31 24) x = 69 25) x = 2 26) x = 2 27) (5,4) 28) (5,1) 29) (1,2,3) 30) (-79/25, 3/25) and (3,1) 32) 2x + h+ 3 31) (-5/2, 27/4) and (3,4)