Admission in india

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Determine whether each relation is a function.

1. {(2, 4), (1, 3), (–3, –1), (4, 6)} 2. {(2, 6), (–3, 1), (–2, 2)}

3. {(x, y)| x = 3} 4. {(x, y)| y = 8}

5. {(x, y)| x = y2} 6. {(x, y)| x2 + y2 = 36}

7. {(a, b)| a = b3} 8. {(w, z)| w = z – 36}


1. {(2, 4), (1, 3), (–3, –1), (4, 6)}; yes, this is a function because each element of the domain is paired with exactly one element in the range.

2. {(2, 6), (–3, 1), (–2, 2)}; yes, this is a function because each element of the domain is paired with exactly one element in the range.

3. {(x, y)| x = 3}; no, this is not a function because it is a vertical line and fails the vertical line test.

4. {(x, y)| y = 8}; yes, this is a function because it is a horizontal line and passes the vertical line test.

Solutions


5. {(x, y)| x = y2}; no, this is not a function because an element of the domain is paired with more than one element in the range.Example: 4 = 22 and 4 = (–2)2

6. {(x, y)| x2 + y2 = 36}; no, this is not a function because it is a circle and fails the vertical line test.

7. {(a, b)| a = b3}; yes, this is a function because each element of the domain is paired with exactly one element in the range.

8. {(w, z)| w = z – 36}; yes, this is a function because each element of the domain is paired with exactly one element in the range.

Solutions (continued)


Periodic Function: a repeating pattern of y-values (outputs) at regular intervals. Cycle: One complete pattern of the function. A cycle can occur at any point on the graph of the functionPeriod: the horizontal length of one cycle of the function. Amplitude: half of the distance between the minimum and maximum values of the function.

Amplitude

Period Periodadmission.edhole.com

Analyze this periodic function. Identify one cycle in two different ways.

Then determine the period of the function.

Each cycle is 7 units long. The period of the function is 7.

Begin at any point on the graph.Trace one complete pattern.

The beginning and ending x-values of each cycle determine the period of the function.


Determine whether each function is or is not periodic. If it is, find the period.

The pattern of y-values in one section repeats exactly in other sections. The function is periodic.

a.

Find points at the beginning and end of one cycle.

Subtract the x-values of the points: 2 – 0 = 2.

The pattern of the graph repeats every 2 units, so the period is 2.


(continued)

b.

The pattern of y-values in one section repeats exactly in other sections. The function is periodic.

Find points at the beginning and end of one cycle.

Subtract the x-values of the points: 3 – 0 = 3.

The pattern of the graph repeats every 3 units, so the period is 3.


Find the amplitudes of the two functions in

Additional Example 2.

The amplitude of the function is 2.

a. amplitude = (maximum value – minimum value) Use definition of amplitude.

12

= [2 – (–2)] Substitute.12

= (4) = 2 Subtract within parentheses and simplify.12


(continued)

The amplitude of the function is 3.

b. amplitude = (maximum value – minimum value) Use definition of amplitude.

12

= [6 – 0] Substitute.12

= (6) = 3 Subtract within parentheses and simplify.12


The oscilloscope screen below shows the graph of the alternating

current electricity supplied to homes in the United States. Find the

period and amplitude.

1 unit on the t-axis = s1

360


The maximum value of the function is 120, and the minimum is –120.

One cycle of the electric current occurs from 0 s to s.160

period = – 0 Use the definitions.

= Simplify.

160

160

amplitude = [120 – (–120)]

= (240) = 120

12

12

The amplitude is 120 volts.The period of the electric current is s.160


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