Exercise 1 Page 80

When data has equally-spaced inputs, we can analyze patterns in the differences of the output to determine what type of function can be used to model the data. Let's see two examples.

Example 1

Consider a function f(x) represented by a table of values.

We see that the table has equally-spaced x-values. Moreover, the first differences are all equal to 2, and therefore they are constant. This means we can model the data with a linear function. Therefore, a quadratic model is not the most appropriate one for this set of data.

Example 2

We see that the first differences are not constant. This means that a linear model is not the most appropriate one. To determine if a quadratic model is the most appropriate one, we will calculate the second differences. To do so, we need to subtract consecutive terms of the first differences.

The second differences are all equal to 2, and therefore they are constant. This means we can model the data with a quadratic function. g(x)=ax^2+bx+c From the first table, we know that the ordered pair (0,1) is a point on the parabola. Let's substitute x= 0 and g(x)= 1 in the equation above.

g(x)=ax^2+bx+c

SubstituteII

x= 0, g(x)= 1

1=a( 0)^2+b( 0)+c

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Solve for c

CalcPow

Calculate power

1=a(0)+b(0)+c

ZeroPropMult

Zero Property of Multiplication

1=c

RearrangeEqn

Rearrange equation

c=1

We can partially write the function. g(x)=ax^2+bx+1 To find the values of a and b, we will create a system of equations by substituting any other two ordered pairs of the table for x and g(x) in the above equation. For simplicity, we will choose (- 1,2) and (1,2). c|c 2=a( - 1)^2+b( - 1)+1 & 2=a( 1)^2+b( 1)+1 ⇕ & ⇕ a-b=1 & a+b=1 Using the obtained equations, we can write the system. a-b=1 & (I) a+b=1 & (II) It looks like the Elimination Method is the easiest way of solving the system. Let's add Equation (I) to Equation (II).

a-b=1 a+b=1

SysEqnAdd

(II): Add I

a-b=1 a+b+( a-b)=1+ 1

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(II): Solve for a

RemovePar

(II): Remove parentheses

a-b=1 a+b+a-b=1+1

AddSubTerms

(II): Add and subtract terms

a-b=1 2a=2

DivEqn

(II): .LHS /2.=.RHS /2.

a-b=1 a=1

We will now substitute 1 for a in Equation (I).

a-b=1 a=1

Substitute

(I): a= 1

1-b=1 a=1

SubEqn

(I): LHS-1=RHS-1

- b=0 a=1

MultEqn

(I): LHS * (- 1)=RHS* (- 1)

b=0 a=1

Finally, we can write the quadratic model for the given table. g(x)=1x^2+0x+1 ⇔ g(x)=x^2+1