MAT222 Ashford University Week 5 Relations and Functions Discussion

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MAT222

ashford university

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Week 5 - Discussion


Relations and Functions [CLOs: 4, 5]

In this discussion, you will be assigned two equations where you will complete a variety of math work related to mathematical functions. Read the following instructions in order and view the example to complete this discussion. Please complete the following problems according to your assigned number. (Instructors will assign each student their number.)

f your assigned number is

Solve the following Problems

18

22 on pages 775 – 776

36 on pages 719–723

Unformatted Attachment Preview

INSTRUCTOR GUIDANCE EXAMPLE: Week Five Discussion Relations and Functions pg 709 #31 The relation is f(x) = √(x – 1) where the -1 is included with x under the radical. f(1) = √(1 – 1) = √0 = 0 f(5) = √(5 – 1) = √4 = 2 f(5/4) = √(5/4 – 1) = √(¼) = ½ f(10) = √(10 – 1) = √9 = 3 f(2) = √(2 – 1) = √1 = 1 f(x) = √(x – 1) x 1 0 5/4 ½ 2 1 5 2 10 3 This being the graph of a square root function, it is shaped like half of a parabola laid over on its side. It has a starting point at (1, 0) and extends infinitely in the positive infinity direction (to the right). This graph falls entirely in the first quadrant and doesn’t exist in any of the other three quadrants. This function has no y intercepts and the only x intercept is the same as its starting point at (1, 0). I chose my x values carefully so that the y values were rational numbers coming out of the radical. The domain is all real numbers greater than or equal to 1, while the range is all non-negative real numbers. In interval notation it is written as follows: D = [1, ) and R = [0, ) This relation is a function because every element of the domain has one and only one value associated with it in the range, and it passes the vertical line test. pg711 #57 The relation is f(x) = 1 - |x| where the 1 and the subtraction are outside of the absolute value bars. f(-4) = 1 - |-4| = 1 – 4 = -3 f(1) = 1 - |1| = 1 – 1 = 0 f(-1) = 1 - |-1| = 1 – 1 = 0 f(4) = 1 - |4| = 1 – 4 = -3 f(0) = 1 - |0| = 1 – 0 = 1 f(x) = 1 - |x| x -4 -3 -1 0 0 1 1 0 4 -3 This is the graph of an absolute value function so it is shaped like a straight sided V with a sharp turn at the vertex. The a value is negative and so the V opens downward. The majority of the graph occurs in the third and fourth quadrants with only a tiny bit crossing into the first and second quadrants. The highest point on the graph occurs at the vertex, (0, 1) and the graph is symmetrical across the y-axis. The y-intercept is at the vertex (0, 1) as well, and the x intercepts are at (-1, 0) and (1, 0). The domain is all real numbers while the range is restricted to real numbers less than or equal to 1. This relation is also a function because it passes the vertical line test, and each domain value has one and only one range value associated with it. Let us take this same function and shift it three units upwards and 4 units to the left. How will this affect the equation? Three units upwards means we add a positive 3 outside of the absolute value bars. Four units to the left means we add a positive 4 inside of the absolute value bars. The function will now look like this: f(x) = 1 + 3 – |x + 4| so f(x) = 4 – |x + 4| and this reflects a transformation of the function three units up and four units to the left. Just for the sake of a thorough example I am also going to show the other function transformed in the same manner. Three units upwards means we add a positive 3 outside of the radical. Four units to the left means we add a positive 4 inside of the radical. The function will now look like this: f(x) = 3 + √(x – 1 + 4) so f(x) = 3 + √(x + 3) This does incorporate the two shifts into the transformed function. Week 5 - Discussion 1111 unread replies.1111 replies. Your initial discussion thread is due on Day 3 (Thursday) and you have until Day 7 (Monday) to respond to your classmates. Your grade will reflect both the quality of your initial post and the depth of your responses. Carefully review the Grading Rubric your discussion will be evaluated. for guidance on how Relations and Functions [CLOs: 4, 5] In this discussion, you will be assigned two equations where you will complete a variety of math work related to mathematical functions. Read the following instructions in order and view the example to complete this discussion. Please complete the following problems according to your assigned number. (Instructors will assign each student their number.) f your assigned number is 18 Solve the following Problems 22 on pages 775 – 776 36 on pages 719–723 For each of the above problems, do the following: • • • • • • • For each of your equation, write down five (5) points on the graph of each equation (you do not have to submit your graph. Draw the graph using paper/pencil or use online graphing calculators like Desmos Graphing Calculator (Links to an external site.)Links to an external site. to answer the question). For each of your equation, does the graph of each equation have any intercepts? If yes, what are they? If not, why not? State the domain for each of your equations. Write them in interval notation. State the range for each of your equations. Write them in interval notation. State whether each of the equations is a function or not giving your reasons for the answer. Select one of your graphs and assume it has been shifted three units upward and four units to the left. Discuss how this transformation affects the equation by rewriting the equation to incorporate those numbers. Incorporate the following four math vocabulary words into your discussion. Use bold font to emphasize the words in your writing. Do not write definitions for the words; use them appropriately in sentences describing the thought behind your math work. o Function o Relation o o Vertical Line test Transformation
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Explanation & Answer

Attached.

#22
The given relation is,
Y=0.3x
Using 5 points,
x

y

-2

-0.6

-1

-0.3

0

0

1

0.3

2

0.6

Sample calculation.
When x = -2
Y = 0.3(-2) = -0.6
This is a graph of linear relation. It has a positive slope and when x increases, Y also increases. It goes
through the origin and symmetry around the origin. This graph falls in the 1st and 3rd quadrant.
This...


Anonymous
I was struggling with this subject, and this helped me a ton!

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