University:
Glendale Community College
Course:
MAT 140 | College Mathematics
Academic year:

2022
Views:

264

Pages:

12
Author:

Tiana Stone

Chapter 6 Practice Test Chapter 6: Quadratic Equations #1 - 7: Solve the equations using the square root property. 1) 36𝑧 2 = 49 2) 𝑚2 = 150 3) 𝑥 2 = −40 4) (𝑥 − 2)2 = 64 5) (2𝑥 − 7)2 = 25 6) (𝑥 − 4)2 = 54 7) (𝑥 − 5)2 = −36 #8 - 10: Solve using the quadratic formula. 8) 2𝑤 2 + 2𝑤 – 8 = 0 9) 3𝑡 2 – 11𝑡 − 4 = 0 10) 𝑦 2 − 8𝑦 – 14 = 0 #11 - 12: Rewrite in the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 with 𝑎 > 0, then solve using the quadratic formula 11) 𝑥 2 = −7 – 5𝑥 12) (𝑥 + 1)(𝑥 – 4) = 14 #13 – 14: Make a table of values and sketch the function. Identify the vertex and axis of symmetry. Identify whether the vertex is a maximum or minimum point, then state the maximum or minimum value. 13) 𝑔(𝑥) = −2𝑥 2 + 5 x -2 g(x) -3 -1 3 0 5 1 3 2 -3 computation 𝑔(−2) = −2(−2)2 + 5 = −2 ∗ 4 + 5 = −8 + 5 = −3 𝑔(−1) = −2(−1)2 + 5 = −2 ∗ 1 + 5 = −2 + 5 = 3 𝑔(0) = −2(0)2 + 5 = −2 ∗ 0 + 5 = 0 + 5 = −3 𝑔(1) = −2(1)2 + 5 = −2 ∗ 1 + 5 = −2 + 5 = 3 𝑔(2) = −2(2)2 + 5 = −2 ∗ 4 + 5 = −8 + 5 = −3 Vertex (0, 5) Axis of Symmetry x = 0 vertex is a maximum point maximum y-value is y = 5 which occurs when x = 0 #13 – 14: Make a table of values and sketch the function. Identify the vertex and axis of symmetry. Identify whether the vertex is a maximum or minimum point, then state the maximum or minimum value. 14) 𝑚(𝑥 ) = 3(𝑥 − 4)2 − 6 Change sign of −4 inside the parenthesis, place +4 in the middle of the xcolumn Vertex (4, -6) Axis of Symmetry x = 4 vertex is a minimum point minimum y-value is y = -6 which occurs when x = 4 x 2 3 4 5 6 m(x) computation 6 𝑚(2) = 3(2 − 4)2 − 6 = 3(4) − 6 = 6 -3 𝑚(3) = 3(3 − 4)2 − 6 = 3(1) − 6 = −3 -6 𝑚(4) = 3(4 − 4)2 − 6 = 3(0) − 6 = −6 -3 𝑚(5) = 3(5 − 4)2 − 6 = 3(1) − 6 = −3 6 𝑚(6) = 3(6 − 4)2 − 6 = 3(4) − 6 = 6 #15: Find the vertex using the vertex formula. Make a table of values and sketch the function. Identify the vertex and axis of symmetry. Identify whether the vertex is a maximum or minimum point, then state the maximum or minimum value. 15) ℎ(𝑥) = −2𝑥 2 + 12𝑥 – 13 Use − 𝑏 2𝑎 formula to get the number for the middle of x-column a = -2 b = 12 − x 1 2 3 4 5 𝑏 2𝑎 = −12 2(−2) h(x) -3 3 5 3 -3 = −12 −4 = 3 (place 3 in the middle of the table’s x-column) computation ℎ(1) = −2(1)2 + 12(1) − 13 = −3 ℎ(2) = −2(2)2 + 12(2) − 13 = 3 ℎ(3) = −2(3)2 + 12(3) − 13 = 5 ℎ(4) = −2(4)2 + 12(4) − 13 = 3 ℎ(5) = −2(5)2 + 12(5) − 13 = −3 Vertex (3, 5) Axis of Symmetry x = 3 vertex is a maximum point maximum y-value is y = 5 which occurs when x = 3 16) An object is launched. The equation for the object's height (ℎ(𝑡)) in meters at time 𝑡 seconds after launch is ℎ(𝑡) = – 8𝑡 2 + 64𝑡 . a) Make a table of values and sketch of graph of the function. Points for graph (2,96) (3,120) (4,128) (5,120) (6,96) b) What do the numbers on the x (horizontal axis) represent? Time in Seconds (how long since the object was launched) c) What do the numbers on the y (vertical axis) represent? The object’s height in feet d) Identify the vertex. (4,128) e) What do the coordinates of the vertex represent? The object will be 128 feet high 4 seconds after it is launched. f) When does the ball reach its maximum height? 4-seconds g) What is the maximum height of the ball? (128 feet) h) Find the 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡(𝑠) using Algebra. (0,0) and (8,0) i) Give a physical interpretation of any 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. The object starts on the ground at time equals zero seconds. It lands after 8 seconds of flight. j) Find the practical domain of the function. [0,8] k) Find the practical range of the function. [0,128]