McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
6. Analyzing Functions with Successive Differences
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Exercise 33 Page 144

The differences between consecutive y-values are called first differences.

Example Solution: y=4x

Practice makes perfect

Given a set of data with x- and y-values, the differences between consecutive y-values are called first differences. Remember that the differences between x-values need to be constant! Analyzing first differences can help us to determine what type of function best models the data. Set of Data ⇓ Analyzing the Differences ⇓ Type of Function We will write a linear function that has a constant first difference of 4.

Let's choose 0 as the first value in y-values. Note that this choice is arbitrary.

Next, we find the consecutive y-values.

Recall that the differences between consecutive x-values need to be constant, too. Let's choose 1 to be the value of the differences between the x-values and 0 to be the first x-value. These choices are also arbitrary.

We will calculate the next few x-values to complete the data set in the table.

Now, let's find a linear function that models the data using slope-intercept form. y= mx+ b From the table we know that the points ( 0, 0) and ( 1, 4) belong to the linear function y= mx+ b for our function. Let's substitute these values and create a system of equations. 0=m( 0)+b & (I) 4=m( 1)+b & (II) Finally, let's solve this system of equations. We will use the Substitution Method. It is usually the best choice when one of the variables is already isolated, or has a coefficient of 1 or -1.
0=m(0)+b & (I) 4=m(1)+b & (II)
â–Ľ
Solve by substitution

(I), (II):Multiply

0=0+b & (I) 4=m+b & (II)
0=b & (I) 4=m+b & (II)
b=0 & (I) 4=m+b & (II)
b=0 & (I) 4=m+ 0 & (II)
b=0 & (I) 4=m & (II)
b=0 & (I) m=4 & (II)
Therefore, the data can be modeled by the linear function y=4x. This is only an example answer.

Alternative Solution

Another example

Here we will present the case that almost any linear function can be an example answer. Let's go back to the table with the empty row of x-values.

If we choose some complicated difference between x-values, and a complicated first x-value, we can get almost any linear function that models the data with a first difference of 4. Let's choose 3 to be the value of the differences between x-values and 2 to be the first x-value.

We will calculate the next x-values to complete the data set in the table.

Now, let's find a linear function that models the data. y= mx+ b From the table we know that the points ( 2, 0) and ( 5, 4) belong to the linear function y= mx+ b. Let's substitute these values. 0=m( 2)+b & (I) 4=m( 5)+b & (II) Finally, let's solve this system of equations.
0=m(2)+b & (I) 4=m(5)+b & (II)
â–Ľ
Solve by substitution

(I), (II):Multiply

0=2m+b & (I) 4=5m+b & (II)
- 2m=b & (I) 4=5m+b & (II)
b=- 2m & (I) 4=5m+b & (II)
b=- 2m & (I) 4=5m -2m & (II)
b=- 2m & (I) 4=3m & (II)
b=- 2m & (I) 43=m & (II)
b=- 2m & (I) m= 43 & (II)
b=- 2( 43) & (I) m= 43 & (II)
b=- 2* 43 & (I) m= 43 & (II)
b=- 83 & (I) m= 43 & (II)
Therefore, the data can be modeled by the linear function y= 43x- 83.