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Core Connections Algebra 1, 2013

CC

Core Connections Algebra 1, 2013 View details

1. Section 5.1

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Exercise 8 Page 199

Practice makes perfect

To find f(1), we have to substitute x=1 into f(x)= 62x-3 and evaluate.

f(x)=6/2x-3

x= 1

f( 1)=6/2( 1)-3

a * 1=a

f(1)=6/2-3

Subtract term

f(1)=6/-1

MoveNegDenomToFrac

Put minus sign in front of fraction

f(1)=-6/1

a/1=a

f(1)=-6

Again, to find f(0), we substitute x=0 in f(x)=6/2x-3 and evaluate.

f(x)=6/2x-3

x= 0

f( 0)=6/2( 0)-3

Zero Property of Multiplication

f(0)=6/-3

MoveNegDenomToFrac

Put minus sign in front of fraction

f(0)=-6/3

Calculate quotient

f(0)=-2

Similar to above, we will substitute x=-3 into the given rule.

f(x)=6/2x-3

x= -3

f( -3)=6/2( -3)-3

a(- b)=- a * b

f(-3)=6/-6-3

Subtract term

f(-3)=6/-9

MoveNegDenomToFrac

Put minus sign in front of fraction

f(-3)=-6/9

a/b=.a /3./.b /3.

f(-3)=-2/3

For the last time in this exercise, we will substitute x=1.5 into the given rule.

f(x)=6/2x-3

x= 1.5

f( 1.5)=6/2( 1.5)-3

Multiply

f(1.5)=6/3-3

Subtract term

f(1.5)=6/0

Since we cannot divide by 0 we cannot find the value for the given number.

This time we are given the value and asked to find x when the function reaches this value. To find it, we will substitute f(x)=4 into the given rule and solve for x.

f(x)=6/2x-3

f(x)= 4

4=6/2x-3

LHS * (2x-3)=RHS* (2x-3)

4(2x-3)=6/2x-3*(2x-3)

FracMultDenomToNumber

a/2x-3* 2x-3 = a

4(2x-3)=6

Distribute 4

8x-12=6

LHS+12=RHS+12

8x=18

.LHS /8.=.RHS /8.

x=18/8

a/b=.a /2./.b /2.

x=9/4

When x= 94 the value of the function, f(x) equals 4.

Subchapter links

5.1.1 Representing Exponential Growth p.199-201

5.1.2 Rebound Ratios p.204-205

5.1.3 The Bouncing Ball and Exponential Decay p.208-209