Algebra

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

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.