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
MAT222 Week 5 Discussion Example
to complete this discussion.
Solve the following Problems
1. y = 0.3x Graph each function and state the domain and range.
2. Sketch the graph of each function and state the domain and range.
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.)
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.
·
Function
·
Relation
·
Vertical Line test
·
Transformation
·
Your initial post should be at least 250 words in length.
INSTRUCTORGUIDANCE 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
x f(x) = √(x – 1)
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
x f(x) = 1 – |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.
