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Core Connections Algebra 2, 2013 View details

2. Section 12.2

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Exercise 97 Page 642

Practice makes perfect

Using the function f(x), we want to evaluate for the given value, f( 2). To do this, we need to substitute 2 for x in each instance of the x-variable and simplify the result.

f(x)=x^2+7x

Substitute

x= 2

f( 2)= 2^2+7( 2)

CalcPowProd

Calculate power and product

f(2)=4+14

AddTerms

Add terms

f(2)=18

Using the function f(x), we want to evaluate for the given value, f( -3). To do this, we need to substitute -3 for x in each instance of the x-variable and simplify the result.

f(x)=x^2+7x

Substitute

x= -3

f( -3)=( -3)^2+7( -3)

CalcPowProd

Calculate power and product

f(-3)=9-21

AddTerms

Add terms

f(-3)=-12

Using the function f(x), we want to evaluate for the given value, f( -3). To do this, we need to substitute i for x in each instance of the x-variable and simplify the result.

f(x)=x^2+7x

Substitute

x= i

f( i)=( i)^2+7 i

IDefPow

i^2=- 1

f(i)=-1+7i

Using the function f(x), we want to evaluate for the given value, f( -3.5+1.5i). To do this, we need to substitute -3.5+1.5i for x in each instance of the x-variable and simplify the result.

f(x)=x^2+7x

Substitute

x= -3.5+1.5i

f( -3.5+1.5i)=( -3.5+1.5i)^2+7( -3.5+1.5i)

▼

Simplify right-hand side

Distr

Distribute 6

f( -3.5+1.5i)=( -3.5+1.5i)^2-24.5+10.5i

ExpandPosPerfectSquare

(a+b)^2=a^2+2ab+b^2

f(-3.5+1.5i)=(-3.5)^2+2(-3.5)(1.5i)+(1.5i)^2-24.5+10.5i