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# Applications of Graphs Help (page 2)

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By McGraw-Hill Professional
Updated on Oct 4, 2011

## Application of Graphs Practice Problems

#### Practice

1. A sales representative’s pay is based on his sales. Table 3.2 shows his salary during one year.

Table 3.2

 Month Pay January (1) 2100 February (2) 2000 March (3) 2400 April (4) 2700 May (5) 2500 June (6) 3000 July (7) 3500 August (8) 3600 September (9) 2500 October (10) 2000 November (11) 2000 December (12) 2100

How much did his monthly pay change on average between January and July? Between July and December? Between October and December?

2. Find the average rate of change between the indicated points of the function whose graph is given in Figure 3.14.

Fig. 3.14 .

3. Find the average rate of change for f ( x )= 2 − x 3 between x = −2 and x = 1.

4. Find the average rate of change for f ( x ) = 6 x − 3 between x = − 5 and x = 3 and between x = 0 and x = 8.

#### Solutions

1. The average monthly increase between January and July is the slope of the line containing the points (1,2100) and (7,3500).

The average monthly decrease between July and December is the slope of the line containing the points (7,3500) and (12,2100).

The average monthly increase from October to December is the slope of the line containing the points (10,2000) and (12,2100).

2. x 1 = 0, y 1 = − 1 and x 2 = 2, y 2 = 8

3.

y 1 = f ( x 1 ) = f (−2) = 2 − (−2) 3 = 10

y 2 = f ( x 2 ) = f (1) = 2 - (1) 3 = 1

4. For x 1 = −5 and x 2 = 3—

y 1 = f ( x 1 ) = f (−5) = 6(−5) − 3 = −33

y 2 = f ( x 2 ) = f (3) = 6(3) − 3 = 15

For x 1 = 0 and x 2 = 8—

y 1 = f ( x 1 ) = f (0) = 6(0) − 3 = − 3

y 2 = f ( x 2 ) = f (8) = 6(8) − 3 = 45

Average rate of change

The average rate of change between any two points on a linear function is the slope.

Newton’s Quotient gives the average rate of change of f ( x ) between x 1 = a and x 2 = a + h .

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