Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
4. Modeling with Quadratic Functions
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Exercise 37 Page 82

Check the second difference sequence.

155

Practice makes perfect

Since the data has equally-spaced inputs, we can check whether the data is linear by checking the first difference sequence.

The first difference sequence is not constant, so this data is not linear. Let's check the second difference sequence. This tells whether the data is quadratic or not.


Since the second difference sequence is constant, this data is quadratic. We do not have enough data to use either vertex form or the intercept form of a quadratic equation, so we will solve a system of three equations. We write the equations by substituting values from the table for x and y. (I): 1=a(1)^2+b(1)+c (II): 5=a(2)^2+b(2)+c (III): 11=a(3)^2+b(3)+c Since c has a coefficient equal to 1 in all of the equations, let's start by isolating it in one of the equations and substituting it into the other ones.
1=a(1)^2+b(1)+c & (I) 5=a(2)^2+b(2)+c & (II) 11=a(3)^2+b(3)+c & (III)

(I), (II), (III):Calculate power and product

1=a+b+c & (I) 5=4a+2b+c & (II) 11=9a+3b+c & (III)
c=1-a-b & (I) 5=4a+2b+c & (II) 11=9a+3b+c & (III)

(II), (III):c= 1-a-b

c=1-a-b & (I) 5=4a+2b+ 1-a-b & (II) 11=9a+3b+ 1-a-b & (III)

(II), (III):LHS-1=RHS-1

c=1-a-b & (I) 4=4a+2b-a-b & (II) 10=9a+3b-a-b & (III)

(II), (III):Subtract terms

c=1-a-b & (I) 4=3a+b & (II) 10=8a+2b & (III)
c=1-a-b & (I) b=4-3a & (II) 10=8a+2b & (III)
c=1-a-b & (I) b=4-3a & (II) 10=8a+2( 4-3a) & (III)
c=1-a-b & (I) b=4-3a & (II) 10=8a+8-6a & (III)
c=1-a-b & (I) b=4-3a & (II) 2=8a-6a & (III)
c=1-a-b & (I) b=4-3a & (II) 2=2a & (III)
c=1-a-b & (I) b=4-3a & (II) a=1 & (III)
We got the first coefficient, a=1. Let's substitute it into the second equation and find b.
b=4-3a
b=4-3( 1)

b=1

Now the only unknown value is c. We can find it by substituting a=1 and b=1 into the first equation.
c=1-a-b
c=1- 1- 1
c=- 1
Since we have all of the coefficients, we can complete the equation of the function. y=x^2+x-1 To find the number of tiles in the 12th figure, we have to substitute 12 for x in the function rule.
y=x^2+x-1
y= 12^2+ 12-1
y=144+12-1
y=155
In the context of the word problem this means that there are 155 tiles in the 12th figure.