Grossmont Cuyamaca Community College District Algebra MathLab Task

User Generated

bcbyv31

Mathematics

Grossmont Cuyamaca Community College District

Description

-Need 2 chapters of Homework on math lab completed

-Both need to be done by 2/14/2021 or before.

-Must have understanding of college algebra.

-Easy problems but many of them

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Explanation & Answer

I have completed homework 2 earlier today. Attached are the solutions. Is it correct that you also need homework 3 done?

Homework 2 – Ch 2
1.

F(x) = 7x + 8

F(5) = 7*5 + 8 = 35 + 8 = 43
F(x+6) = 7(x+6) + 8 = 7x + 42 + 8 = 7x + 50
F(-x) = 7*-x + 8 = -7x + 8

2. F(r) = sqrt(r+2) - 6
F(-2) = -6
F(79) = 3
F(x-2) = sqrt x - 6

3. F(x) = (7x^2 – 1) /x^2
F(3) = 62/9
F(-3) = 62/9
F(-x) = (7x^2 – 1) /x^2

4. F(x) = x / abs x
F(6) = 1
F(-6) = -1
F(r^2) = 1

5. Graph.

6. Graph

7. Use vertical line test to determine if y is a function of x in the graph.
Since the graph passes the vertical line test at all points, y is a function of x

8. Graph is a ellipse, failing the vertical line test. y is not a function of x.

9. Parabola. Y is a function of x.

10. Graph of a trig function.

F(4) = 7

11. Graph of piecewise function.
G(x) = -9
X=2

12. Parabola with a vertex at (0,1) opening down
Domain = (-infinity, infinity)
Range = (-infinity, 1]
X intercepts : -1,1
Y intercept is 1
F(-2) = -3
F(2) = -3

13. Absolute value function with vertex at (-2,-3) opening down.
Domain = (-infinity, infinity)
Range = (-infinity, -3]
X intercepts : none
Y intercept is -5
F(-3) = -4
F(0) = -5

14. Half of a parabola
Domain: (-3,0]
Range: [-1,8)
X intercept = -1
Y intercept = -1
F(-2) = 3

15. A negative square root function.
Domain: [0, infinity)
Range: (-infinity, -2]
X intercept = none
Y intercept = -2
F(1) = -3

16. Line segment graph.
Domain: [-4,1]
Range: [-4,6]
X intercept: -1
Y intercept: -2
F(-4) = 6

17. A company manufactures bicycles has a fixed cost of 90000. It costs 150 to produce each
bicycle. The total cost for the company is the sum of the fixed cost and variable cost.
C(x) = 90000 + 150 x
C(110) = 106500. It costs $106500 to produce 110 bicycles

18. If f(a) > f(x) in an open interval containing a, x does not equal a, then the function value f(a) is a
relative __maximum__ of f.
If f(a) < f(x) in an open interval containing a, x does not equal a, then the function value f(a) is a relative
__minimum__ of f.

19. Graph of a parabola.
Function is increasing on (-infinity, -1)
Function is decreasing on (-1, infinity)
There is no interval where the function is constant.

20. Increasing on [4, infinity)

Nowhere is the function decreasing or constant.

21. The graph is increasing on (-4,6)
No intervals for decreasing or constant.

22. Even odd or neither for a piece wise function.
Neither

23. Cubic function.
The function is odd (f(-x) = -f(x))

24. F(x) = x^5 – 4x
F(-x) = (-x)^5 – 4(-x) = -x^5 + 4x
-f(x) = -x^5 + 4x
Since f(-x) = -f(x), function is odd
Odd functions are symmetric to the origin.

25. G(x) = x^2 – x
G(-x) = x^2 + x
Neither.
No symmetry.

26. F(x) = x^6 + x^10 + 4
F(-x) = (-x)^6 + (-x)^10 + 4 = x^6 + x^10 + 4
F(-x) = f(x), function is even.
Symmetric to the y axis.

27. Piecewise function
F(x) = 4x + 4 if x < 0
2x + 5 if x >= 0
F(-3) = 4*-3 + 4 = -8
F(0) = 2*0 + 5 = 5
F(1) = 2*1 + 5 = 7

28. Graphing of piecewise function.

29. Difference quotient for f(x) = -8x + 7
F(x+h) = -8x – 8h + 7
F(x+h) - f(x) = -8x – 8h + 7 - (-8x + 7) = -8h
F(x+h) - f(x)/h = -8h / h = -8

30. Difference quotient for f(x) = 2x^2 – x + 2
F(x+h) = 2(x+h)^2 - (x+h) + 2 = 2(x^2 + 2xh + h^2) - x – h + 2 = 2x^2 + 4xh + 2h^2 – x – h + 2
F(x+h) - f(x) = 2x^2 + 4xh + 2h^2 – x – h + 2 - (2x^2 – x + 2) = 4xh + 2h^2 – h
F(x+h) - f(x) / h = (4xh + 2h^2 – h) / h = h(4x + 2h – 1) / h = 4x + 2h - 1

31. Difference quotient for f(x) = sqrt(2x)
F(x+h) = sqrt(2(x+h)) = sqrt(2x + 2h)
F(x + h) - f(x) = sqrt(2x + 2h) - sqrt x

F(x+h) - f(x)/ h = sqrt(2x + 2h) - sqrt x/h

32. The slope, m, of a line through the distinct points (x1,y1) and (x2,y2) is given by the formula m =
(y2- y1)/(x2-x1).

33. The slope of a horizontal line is zero.

34. The slope of a vertical line is undefined.

35. The point-slope form of the equation of a nonvertical line with a slope m that passes through
the point (x1, y1) is y – y1 = m(x – x1).

36. The slope-intercept form the equation of a line is y = mx + b where m represents the slope and b
represents the y intercept.

37. Find the slope of the line passing through the pair of points. Then indicate whether the line
rises, falls is horizontal or vertical.
(-5,2) and (3,4)
M = (4 – 2) / (3 - -5) = 2 / 8 = ¼
Slope is positive, line rises.

The line rises from left to right.

38. Same instructions as previous problem.
(1,7) and (5,-5)
M = (-5 – 7) / (5 – 1) = -12 / 4 = -3
Negative slope, line falls

The line falls from left to right.

39. (-2,5) and (-2,3)
M = (5 – 3) / (-2 – -2) = 2 / 0 = undefined.
The slope is undefined.

The line is vertical

40. Write the point slope form of a line satisfying the given conditions. Then use point slope to
convert to slope intercept form.
Slope = m = 8, passing through (-6,1)
Y – y1 = m(x –x1)
Y – 1 = 8(x +6)
Solve equation for y to convert to slope intercept form, y = mx + b
Y – 1 = 8x + 48
Y = 8x + 49

41. Slope = -1/3, passing through (3, -9)
Point slope form:
Y + 9 = -1/3(x-3)
Solve for slope intercept.
Y + 9 = -1/3 x + 1
Y = -1/3 x - 8

42. Passing through (-5,-2) and (5,14)
Point-slope form.
Solve for the slope.
M = (14 - -2) / ( 5 - -5) = 16 / 10 = 8/5
Y –14 = 8/5(x – 5)
Solve for slope intercept.
Y – 14 = 8/5 x – 8

Y = 8/5 x + 6

43. Give the slope and y intercept of the line. Then graph.
Y = 5x + 4
Slope is 5
Y intercept is 4

44. Y = -¾ x + 3
Slope = -¾
Y intercept = 3

45. Y = 3
Horizontal line at y = 3.

46. X = -5
Vertical line at given x value.

47. Rewrite the equation 3x + y + 1 = 0 in slope intercept form.
Y = -3x – 1
Slope of the line is –3 and the y intercept is -1

48. Fill in the blank:
If (x1, f(x1) ) and (x2, f(x2) ) are distinct points on the graph of a function f, the average rate of change of
f from x1 to x2 is ____.
F(x2) - f(x1) / (x2-x1)

49. Write an equation for the line L in point slope form and slope intercept form. L is parallel to y =
4x through (-4,2)
M=4

Y – 2 = 4(x +4) [point slope]
Y – 2 = 4x + 16
Y = 4x + 18 [slope intercept]

50. Perpendicular to 3x through (-1,-3)
M = -1/3
Y + 3 = -1/3(x + 1) [point slope]
Y + 3 = -1/3 x – 1/3
Y = -1/3 x – 10/3 [slope intercept]

51. Passing through (-8,-5) and parallel to y = -5x + 4
M = -5
Y +5 = -5(x+8) [point slope]
Y + 5 = -5x – 40
Y = -5x – 45 [slope intercept]

52. Find average rate of change of the function f(x) = x^2 + 4x from x1 = 2 to x2 = 3
F(x1) = f(2) = 2^2 + 4*2 = 4 + 8 = 12
F(x2) = f(3) = 3^2 + 4*3 = 9 + 12 = 21

Average rate of change = (21 – 12) / (3-2) = 9/1 = 9

53. Graph should shift to the right 3 units and up 2 units.

54. Use f(x) = x^2 to graph the following function: g(x) = x^2 + 5
Select the transformations that are required:
Shift the graph up 5 units.
Graph

55. Use f(x) = x^2 to graph the following function: g(x) =(x-4)^2
Select the transformations that are required:
Shift the graph right 4 units.
Graph

56. G(x) = -2(x – 1)^2 –2
Transformations.
Shift graph 1 unit to the right.
Stretch the graph vertically by a factor of 2.
Reflect the graph over the x axis.
Shift the graph 2 units down.

57. Use transformations to graph g(x) = sqrt x – 2
Shift the graph 2 units down.
Graph.

58. Use transformations to graph g(x) = 3 sqrt(x-4) -1
Stretch the graph vertically by a factor of 3.
Shift the graph 4 units to the right.
Shift the graph 1 unit down.

59. Transformations to absolute value function h(x) = abs(x) + 3
Transformation needed?
Vertical translation (up 3)

60. Transformation of h(x) = abs(x-4)
Horizontal translation (to the right 4 units)

61. G(x) = -4 abs(x + 6) + 3
Transformations needed?
Vertical stretch/shrink
Reflection about the x axis
Horizontal shift
Vertical shift

62. G(x) = (x-6)^3
Move graph 6 units to the right.

63. Transformations to g(x) = cbrt(x) + 3
Shift the graph 3 units up.

64. Transformations to g(x) = cbrt(x-3)
Shift 3 units to the right.

65. H(x) = - cbrt(x+4)
Shift 4 units to the left. Reflect over the x axis.

66. Find the domain of the function: f(x) = 2(x + 5)
Domain = (-infinity, infinity)

67. Domain of f(x) = 14 / (x^2 + 23x + 132)
Denominator cannot be equal to zero.
X^2 + 23x + 132 = 0
(x+11) (x+12) = 0
X = - 11, x = -12

Domain: (-infinity, -12) U (-12, -11) U (-11, infinity)

68. Find the domain. f(x) = 2 / (x^2 + 9) + 1/(x^2 –1)
No denominator can be equal to zero.
X^2 + 9 ...


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