# Worksheet-3_Sol.pdf - Department of Mathematics University...

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Unformatted text preview: Department of Mathematics, University of Wisconsin-Madison Math 114 Worksheet Sections 1.1-1.4, 2.1 1. Find the distance between the point P1 = (−4, −3) and P2 = (6, 2). Solution: d(P1 , P2 ) p (x2 − x1 )2 + (y2 − y1 )2 p = (6 − (−4))2 + (2 − (−3))2 p = 102 + 52 √ 125 = √ = 5 5 = 2. Find the midpoint of the line segment joining P1 = (2, −3) and P2 = (4, 2). Solution:  M = = = x1 + x2 y1 + y2 , 2 2   2 + 4 −3 + 2 , 2 2 1 (3, − ) 2  3. Find all points on the y-axis that are 6 units from the point (4, −3). Solution: Let P1 = (0, y) and P2 = (4, −3). We want d(P1 , P2 ) = 6. So 1 6 = p 36 = 16 + (−3 − y)2 20 = 9 + 6y + y 2 (4 − 0)2 + (−3 − y)2 = y 2 + 6y − 11 p −6 ± 36 − 4(−11) y = √ 2 −6 ± 80 = 2 √ −6 ± 4 5 = 2 √ = −3 ± 2 5 √ √ Answer: (0, −3 + 2 5) and (0, −3 − 2 5) 0 4. Find the intercepts of the equation 5x + 2y = 10. You might want to plot the points to illustrate the intercepts. Solution: 1) the y-intercept is found when x = 0. So, 5(0) + 2y = 10 =⇒ y = 5. Answer: (0, 5) 2) the x-intercept is found when y = 0. So, 5(x) + 2(0) = 10 =⇒ x = 2. Answer: (2, 0) 6 4 2 −3 −2 −1 1 2 3 −2 −4 −6 5. For the following points, find the point that is symmetric to it with respect to (a) the x-axis; (b) the y-axis; (c) the origin. You might want to plot the points to illustrate the symmetry. (a) (4, −2) Solution: For (a) symmetry with respect to the x-axis, we replace y by −y, we get (4, 2) For (b) symmetry with respect to the y-axis, we replace x by −x, we get (−4, −2) 2 For (c) symmetry with respect to the origin, we replace x by −x and y by −y, we get (−4, 2) 6 4 (c) (-4,2) (a) (4,2) 2 −4 −2 2 4 −2 (b) (-4,-2) (4,-2) −4 −6 (b) (4, 0) Solution: For (a) symmetry with respect to the x-axis, we replace y by −y, we get (4, 0) For (b) symmetry with respect to the y-axis, we replace x by −x, we get (−4, 0) For (c) symmetry with respect to the origin, we replace x by −x and y by −y, we get (−4, 0) 6 4 2 −4 −2 2 4 −2 −4 −6 6. Test the equation x2 − y − 4 = 0 for symmetry. Solution: To test for symmetry with the x-axis, we replace y by −y. We get x2 − (−y) − 4 = 0 =⇒ x2 + y − 4 = 0. Since x2 − y − 4 = 0 is not equivalent to x2 + y − 4 = 0, the graph of the equation is not symmetric with respect to the x-axis. To test for symmetry with the y-axis, we replace x by −x. We get (−x)2 − y − 4 = 0 =⇒ x2 − y − 4 = 0. Since (−x)2 − y − 4 = 0 is equivalent to x2 − y − 4 = 0, the graph of the equation is symmetric 3 with respect to the y-axis. To test for symmetry with the origin, we replace x by −x and y by −y. We get (−x)2 − (−y) − 4 = 0 =⇒ x2 + y − 4 = 0. Since x2 − y − 4 = 0 is not equivalent to x2 + y − 4 = 0, the graph of the equation is not symmetric with respect to the origin. 10 5 −4 −2 2 4 −5 −10 7. For the following, find an equation for the line with the given properties. (a) Containing the points (−3, 4) and (2, 5). 5−4 Solution: m = 2−(−3) = 51 1 y − 5 = 5 (x − 2) so the equation is 5y − x = 23 or y = 15 x + 23 5 (b) x-intercept of (−4, 0) and y-intercept of (0, 4). Solution: m = 4−0 0−(−4) = 1 and b = 4 so the equation is y = x + 4or y − x = 4 (c) Vertical; containing the point (3, 8). Solution: The slope is undefined so the equation is x = 3 (d) Parallel to the line x − 2y = −5; containing the point (0, 0). Solution: x − 2y = −5 =⇒ y = 12 x + 52 so m = 12 Here b = 0 so the equation is y = 21 x or x − 2y = 0 4 (e) Perpendicular to the line x − 2y = −5; containing the point (0, 4). Solution: We found in 7d) that m = 12 so the slope of the perpendicular line has a slope of m = −2 Here b = 4 so the equation is y = −2x + 4 or 2x + y = 4 8. Find the standard form of the equation of each circle. (a) r = 5; (h, k) = (4, −3) Solution: The standard form is (x−h)2 +(y −k)2 = r2 so the equation is (x−4)2 +(y +3)2 = 25 (b) With endpoints of a diameter at (4, 3) and (0, 1). Solution: The center is the midpoint of the line segment joining P1 = (4, 3) and P2 = (0, 1).   x1 + x2 y1 + y2 M = , 2 2   4+0 3+1 = , 2 2 = (2, 2) The length of the radius is the distance between the center (2, 2) and one of the endpoints of the diameter, let’s take (0, 1) p d = (2 − 1)2 + (2 − 0)2 √ 5 = So the equation is (x − 2)2 + (y − 2)2 = 5 (c) x2 + y 2 − 6x + 2y + 9 = 0 Solution: x2 + y 2 − 6x + 2y + 9 2 2 x − 6x + y + 2y 2 = 0 = −9 2 x − 6x + 9 + y + 2y + 1 = −9 + 9 + 1 (x − 3)2 + (y + 1)2 = 1 (d) 2x2 + 2y 2 + 8x + 7 = 0 5 Solution: 2x2 + 2y 2 + 8x + 7 2 2x + 8x + 2y 2 2(x + 4x) + 2y 2(x + 2) + 2y = −7 = −7 2 (x + 2)2 + y 2 9. For f (x) = = −7 + 8 = 1 1 = 2 x2 − 1 , find the following. x+4 (a) f (−1) Solution: (−1)2 − 1 (−1) + 4 1−1 = (−1) + 4 0 = 3 =0 f (−1) = (b) f (−x) Solution: (−x)2 − 1 (−x) + 4 x2 − 1 = (−x) + 4 x2 − 1 = 4−x f (−x) = (c) −f (x) 6 0 2 2(x2 + 4x + 4) + 2y 2 2 = 2 Solution: x2 − 1 ) x+4 1 − x2 = x+4 −f (x) = −( (d) f (x + 1) Solution: (x + 1)2 − 1 (x + 1) + 4 (x2 + 2x + 1) − 1 = (x + 1) + 4 2 x + 2x = x+5 x(x + 2) = x+5 f (x + 1) = (e) f (2x) Solution: (2x)2 − 1 (2x) + 4 4x2 − 1 = 2x + 4 f (2x) = 10. For f (x) = 3x2 + 2, find f (x + h) − f (x) . Be sure to simplify. h Solution: f (x + h) − f (x) (3(x + h)2 + 2) − (3x2 + 2) = h h 3(x + h)2 − 3x2 = h 3(x2 + 2xh + h2 ) − 3x2 = h 3h(2x + h) = h = 3(2x + h) 7 ...
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