PRACTICE TEST 1.1.pdf - PRACTICE TEST 1.1 MATH111 K001 Fall...

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Unformatted text preview: 2/18/2020 PRACTICE TEST 1.1 MATH111 K001 Fall 19 Assessments 4.2 Logarithmic Functions (Homework) Results PRACTICE TEST 1.1 - Grade Report Score: 94% (17 of 18 pts) Submitted: Feb 18 at 9:01pm 1/22 2/18/2020 PRACTICE TEST 1.1 Question: 1 Grade: 1.0 / 1.0 Use symmetry and intercepts to sketch the graph of y = 4 ∣ x ∣ +4. Your response Solution Test the equation for each type of symmetry. Respect to the x-Axis −y y = = Respect to the y-Axis 4 ∣ x ∣ +4 y −4 ∣ x ∣ −4 Not equivalent to y So, the graph of y = = y 4 ∣ x ∣ +4. 4 ∣ x ∣ +4 = = Respect to the Origin 4 ∣ (−x) ∣ +4 −y 4 ∣ (x) ∣ +4 y Equivalent to y = 4 ∣ x ∣ +4. = = 4 ∣ (−x) ∣ +4 −4 ∣ (x) ∣ −4 Not equivalent to y = 4 ∣ x ∣ +4. is symmetric with respect to the y-axis. Find the intercepts. -intercept: 0 x -intercept: y y = 4 ∣ x ∣ +4 = 4 ∣ 1(0) ∣ +4 ⇒ = no solution 4 Substitute 0 for y and solve for x. Substitute 0 for x and simplify to find y. So, the graph does not cross the x-axis, and it crosses the y-axis at 4. Find some additional points on the graph, on one side of the line of symmetry. Then, use the fact that for each (x, y) on a graph that is symmetric with respect to the y-axis, (x, y) is also on the graph. Plot the points and sketch the graph. 2/22 2/18/2020 PRACTICE TEST 1.1 3/22 2/18/2020 PRACTICE TEST 1.1 Question: 2 Grade: 1.0 / 1.0 Identify the type of symmetry, if any, for the graph of y = 3x 4 . Your response Symmetric with respect to the y -axis Solution Test for symmetry with respect to the x -axis. −y y = = 3x −3x 4 Substitute −y for y . 4 Divide both sides by −1 to solve the equation for y . The equations are not equivalent, so the graph is not symmetric with respect to the x -axis. Test for symmetry with respect to the y -axis. y = y 3(−x) = 3x Substitute −x for x . 4 Simplify. 4 The equations are equivalent, so the graph is symmetric with respect to the y -axis. Test for symmetry with respect to the origin. −y = −y y = 3(−x) = 3x −3x 4 Substitute −x for x and −y for y . 4 Simplify. 4 Divide both sides by −1 to solve the equation for y . The equations are not equivalent, so the graph of y = 3x 4 is not symmetric with respect to the origin. 4/22 2/18/2020 PRACTICE TEST 1.1 Question: 3 Grade: 1.0 / 1.0 Write the equation of a circle where A (1, 0) and B (3, 8) are the endpoints of a diameter. Your response (x−2) 2 + (y−4) 2 = 17 Solution Find the radius r and the coordinates of the center (h, k) , then substitute those values into the standard form equation of a circle, (x − h) 2 2 2 + (y − k) = r . Use the Distance Formula to find the length of AB, the circle‘ s diameter, and divide the diameter by 2 to find the radius. Then, use the Midpoint Formula to find the midpoint of AB, the circle‘s center. For both formulas, let A (1, 0) be (x1 , y1 ) and B (3, 8) be (x2 , y2 ) . x1 = 1, x2 = 3, y1 = 0, y2 = 8 AB = ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 + ((0) − 8) √(1 − (3)) = ‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 + (−8) √(−2) = 4‾‾‾‾ + 64 ‾ √‾ = ‾‾ ‾ √68 So the radius is center= ( 1+(3) 2 Substitute r (x − (2)) 2 = ‾‾ ‾ √68 = √17 ‾‾ ‾ . 2 , 0+8 2 ) ‾‾ ‾ , h √17 + (y − (4)) 2 = = = (2, 4) 2, and k 2 (√17 ‾‾ ‾ ) = ⇒ 4 into the equation of a circle with center at (h, k) , (x−2) 2 + (y−4) 2 (x − h) 2 2 + (y − k) = r 2 , and simplify. = 17 5/22 2/18/2020 PRACTICE TEST 1.1 Question: 4 Write x 2 Grade: 1.0 / 1.0 2 + y − 4x + 8y = 5 in the standard form of a circle and then identify the circle's center and radius. The standard form equation of the circle is Your Answer: Correct Answer: (x−2) (x−2) 2 2 + (y+4) + (y+4) 2 2 = 25 = 25 Your response Correct response Enter the center as an ordered pair, including the comma. Enter the center as an ordered pair, including the comma. The center is at (2,-4) (100%) The center is at (2,-4) Your response Correct response Enter the radius. Enter the radius. The radius is 5 (100%) The radius is 5 Solution Group the x-terms and the y-terms, and move the constant term to the right side. (x 2 2 − 4x) + (y + 8y) = 5 Complete the square twice to write the x -terms as a perfect square trinomial and the y -terms as a perfect square trinomial. (x (x 2 2 2 −4x) + (y +8y) 2 −4x + 4) + (y +8y + 16) (x−2) 2 = = 5 5 + 4 + 16 + (y+4) The standard form equation is (x−2)2 2 = 25 + (y+4) 2 Complete the square twice. Factor each perfect square trinomial and simplify the right side. = 25 , so the center is (2, −4) and the radius is 5 . 6/22 2/18/2020 PRACTICE TEST 1.1 Question: 5 Grade: 1.0 / 1.0 Identify the center, radius, and graph of (x − 1)2 Your response + (y + 3) 2 = 25. Correct response Enter the center as an ordered pair, including the comma. Enter the center as an ordered pair, including the comma. The center is at (1,-3) (100%). The center is at (1,-3). Your response Correct response Enter the radius exactly. Do not use a decimal approximation. Enter the radius exactly. Do not use a decimal approximation. The radius is 5 (100%). Select the graph of (x − 1) The radius is 5. 2 + (y + 3) 2 = 25. Your Answer: Correct Answer: Solution From the equation, h = 1, k = −3, and r 2 = 25, so r = 5. Therefore, the circle’s center is at (1, −3) and the radius is 5. Plot the points 5 units above, below, to the left and to the right of (1, −3) , and then sketch the circle through those points. 5 5 5 units above (1, −3) units below (1, −3) units left of (1, −3) ⇒ (1, 2) ⇒ (1, −8) ⇒ (−4, −3) 7/22 2/18/2020 5 PRACTICE TEST 1.1 units right of (1, −3) Question: 6 ⇒ (6, −3) Grade: 1.0 / 1.0 Write the equation of a circle with center at (−3, 4) and radius 3 . Your response (x+3) 2 + (y−4) 2 = 9 Solution Substitute r = 3, h = −3, 2 2 (x − (−3)) + (y − (4)) Question: 7 = and k = 2 (3) ⇒ 4 into the equation of a circle with center at(h, k) , (x+3) Grade: 1.0 / 1.0 2 + (y−4) 2 = (x − h) 2 2 2 + (y − k) = r , and simplify. 9 Write the equation of a circle with center at the origin and radius 10 . Your response x 2 + y 2 = 100 Solution Substitute r x 2 2 + y = = 10 2 (10) into the equation of a circle with center at the origin, x 2 ⇒ x 2 2 + y = 2 + y = r 2 , and simplify. 100 8/22 2/18/2020 PRACTICE TEST 1.1 Question: 8 Grade: 1.0 / 1.0 Find both the x- and y- intercepts and the graph of y = −3x 2 −3x+6. Select the correct set of intercepts. and (1, 0), (−2, 0) (0, 6) (50%) Select the correct graph. (50%) Solution Find the x -and y -intercepts. x -intercept: y 0 = x − 1 x y -intercept: y = = −3x 2 Substitute 0 for y and solve for x. −3x+6 −3(x − 1) (x + 2) = = 0 1 or x + 2 or x = = The equation is quadratic, so solve by factoring. 0 −2 2 −3(0) −3(0)+6 = 6 Substitute 0 for x and solve for y. There are two x − intercepts, (1, 0) and (−2, 0) , and one y − intercept, (0, 6) . Plot the x − and y − intercepts and then sketch a line through the intercepts. 9/22 2/18/2020 PRACTICE TEST 1.1 10/22 2/18/2020 PRACTICE TEST 1.1 Question: 9 Grade: 1.0 / 1.0 Find both the x- and y-intercepts and the graph of y = −3+2x. Select the correct intercepts. (0, −3) and ( 3 2 , 0) (50%) Select the correct graph. (50%) Solution Find the x-intercept and the y-intercept. x -intercept. 0 = −3+2x y -intercept. y = −3+2(0) ⇒ ⇒ x y = = 3 Substitute 0 for y and solve for x. 2 −3 The x- and y-intercepts are (0, −3) and ( 3 2 Substitute 0 for x and solve for y. , 0) . The equation is linear, so the graph forms a line. Plot the x- and y-intercepts and then sketch a line through the intercepts. 11/22 2/18/2020 PRACTICE TEST 1.1 Question: 10 Grade: 1.0 / 1.0 Choose the table and graph of y = x 2 +3x+2. Select one table and one graph. Choice Selected x y −1 2 0 4 1 6 2 13 x y −1 0 0 2 1 6 2 12 x y −1 0 0 −2 1 −6 2 −12 Points No Yes +1 No No 12/22 2/18/2020 PRACTICE TEST 1.1 Yes +1 No Solution Choose several values for x. Here, −1, 2 0, 1 and 2 are chosen. Substitute each x -value into the equation to find the corresponding y -value. x y y = x −1 0 y = (−1) 0 2 y = 2 (0) +3 (0) +2 = 2 1 6 y = 2 (1) +3 (1) +2 = 6 +3x+2 2 +3 (−1) +2 = 0 13/22 2/18/2020 2 12 PRACTICE TEST 1.1 y = 2 (2) +3 (2) +2 = 12 Plot each ordered pair from the table on a coordinate plane and then draw a line through the points. 14/22 2/18/2020 PRACTICE TEST 1.1 Question: 11 Grade: 1.0 / 1.0 Choose the table and graph of y Select one table and one graph. Choice = 2+2x. Selected Points No Yes x y −1 4 0 2 1 0 2 −2 x y −1 0 0 2 1 4 +1 No Yes +1 15/22 2/18/2020 2 PRACTICE TEST 1.1 6 x y −1 −4 0 −2 1 4 2 2 No No Solution Choose several values for x. Here, −1, x y y = 2+2x −1 0 y = 2+2 (−1) 0 2 y = 2+2 (0) = 2 1 4 y = 2+2 (1) = 4 2 6 y = 2+2 (2) = 6 = 0, 1 and 2 are chosen. Substitute each x -value into the equation to find the corresponding y -value. 0 Plot each ordered pair from the table on a coordinate plane and then draw a line through the points. 16/22 2/18/2020 Question: 12 PRACTICE TEST 1.1 Grade: 0.0 / 1.0 Classify the figure formed by A ( 0, −4 ), B ( 4, −4 ), and C ( 2, −4 − 2 √3 ‾ ). Your response right isosceles triangle Solution Use the Distance Formula to find the distance between each pair of points, AB, ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 AB = √((0) − (4)) + ((−4) − (−4)) = BC, and AC. ‾‾ ‾ √16 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ 2 ‾ 2 BC = √((4) − (2)) + ((−4) − (−4 − 2√( 3)) = ‾‾ ‾ √16 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ 2 ‾ 2 AC = √((0) − (2)) + ((−4) − (−4 − 2√( 3)) = ‾‾ ‾ √16 Since AB + BC ≠ AC, the points are not collinear and therefore they form a triangle. The side lengths are equal, so the triangle is equilateral. 17/22 2/18/2020 PRACTICE TEST 1.1 Question: 13 Grade: 1.0 / 1.0 A ( 1, 12 ) , B ( 4, 15 ) , and C ( 16, 24 ) are collinear. True or false? Your response The points are not collinear Solution Use the distance formula to find the distance between each pair of points, AB, ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 AB = √(1 − 4) + (12 − 15) = BC, and AC. ‾‾ ‾ √18 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 BC = √(4 − 16) + (15 − 24) = ‾‾‾‾ √225 ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ AC = √(1 − 16)2 + (12 − 24)2 = ‾‾‾‾ √369 Since AC is greater than AB and BC, check whether the sum of AB and BC is equal to AC. Use a calculator to simplify the expression for AB + BC = √18 ‾‾ ‾ + √225 ‾‾‾‾ ≈ 19.2426 and the expression for AC AB + BC = Therefore, A, ‾‾ ‾ + √225 ‾‾‾‾ ≠√369 ‾‾‾‾ √18 B, = = ‾‾‾‾ ≈ 19.2094. √369 AC and C are not collinear. 18/22 2/18/2020 PRACTICE TEST 1.1 Question: 14 Grade: 1.0 / 1.0 is the midpoint between R(x, y) and S(10, 3). Find the coordinates of R. M (3, −2) Enter the answer as an ordered pair, including the comma. The coordinates of R are (-4,-7) (100%) Solution Substitute the x -coordinate from the midpoint and the x -coordinate from S into the equation xM x+10 3 = 6 = x + 10 Multiply both sides by 2. −4 = x Subtract 10 from both sides. 2 = y+3 2 x +x 1 2 2 . Substitute. Substitute the y -coordinate from the midpoint and the y -coordinate from S into the equation yM −2 = = y +y 1 2 2 . Substitute. −4 = y + 3 Multiply both sides by 2. −7 = y Subtract 3 from both sides. 19/22 2/18/2020 PRACTICE TEST 1.1 Question: 15 Grade: 1.0 / 1.0 Determine the midpoint of the two points (6, 4) and (12, 11) . Enter the answer as an ordered pair, including the comma. The midpoint is (9,15/2) (100%) Solution Substitute the coordinates from the ordered pairs (points) into the Midpoint Formula M (x, y) Let (x1 , y1 ) 6+12 ( 2 18 ( 2 , , = (6, 4) 4+11 2 ) 15 and (x2 , y2 ) = x +x 1 2 ( 2 , y +y 1 2 2 ) and simplify. = (12, 11) . Substitute. Simplify the numerators. 2 ) ( 9 , 15 ) Simplify the fraction. 2 Question: 16 Grade: 1.0 / 1.0 Determine the distance between the two points (3, 0) and (6, 4). The distance between the points is 5 (100%) Solution Substitute the coordinates from the ordered pairs (points) into the Distance Formula d Let (x1 , y1 ) = (3, 0) and (x2 , y2 ) = (6, 4) and simplify. . ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 + (4 − −0) √(6 − 3) Substitute. = ‾‾‾‾‾‾‾‾‾‾ ‾ 2 2 + (4) √(3) Simplify within the parentheses. = 9‾‾‾‾ + 16 ‾ √‾ Evaluate the powers. = ‾‾ ‾ √25 Add. = 5 Evaluate the square root. d = ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ 2 2 ‾ = √(x − x ) + (y − y ) 2 1 2 1 20/22 2/18/2020 Question: 17 PRACTICE TEST 1.1 Grade: 1.0 / 1.0 Choose the graph that corresponds to the table. x y 0 5 5 1 −4 0 4 −5 Your response Solution To plot the point (0, 5) , move 5 units up from the origin, to 5 on the y -axis. To plot the point (5, 1) , move 5 units to the right of the origin, to 5 on the x -axis. Then from 5 on the x -axis, move up 1 unit. To plot the point (−4, 0) , move 4 units to the left of the origin, to −4 on the x -axis. To plot the point (4, −5) , move 4 units to the right of the origin, to 4 on the x -axis. Then from 4 on the x -axis, move down 5 units. 21/22 2/18/2020 Question: 18 PRACTICE TEST 1.1 Grade: 1.0 / 1.0 In which area of the coordinate plane does the point (0, −7) lie? Your response on the y -axis Solution To plot the point (0, −7) , move down 7 units on the y -axis. The point (0, −7) is on the y -axis since the x -coordinate is 0. 22/22 ...
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