College Algebra

Chapter 3 Review

Created by: Lauren AtkinsonMath Coordinator, Mary Stangler Center for Academic Success

Note:

This review is composed of questions from the chapter review at the end of chapter 3. This review is meant to highlight basic concepts from chapter 3. It does not cover all concepts presented by your instructor. Refer back to your notes, unit objectives, handouts, etc. to further prepare for your exam.

Use the graph to find the following: a) Sign of the leading coefficient

b) Vertex

c) Axis of symmetry

d) Intervals where f is increasing and were f is decreasing

a) Positive 𝑎 > 0b) −3,2c) 𝑥 = −3d) Increasing: [−3,∞)

Decreasing: (−∞,−3]

Write 𝑓(𝑥) in the form 𝑓 𝑥 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐and identify the leading coefficient.

𝑓 𝑥 =1

3(𝑥 + 1)2−2

𝑓 𝑥 =1

3𝑥 + 1 𝑥 + 1 − 2

𝑓 𝑥 =1

3𝑥2 + 𝑥 + 𝑥 + 1 − 2

𝑓 𝑥 =1

3𝑥2 +

2

3𝑥 +

1

3− 2

𝑓 𝑥 =1

3𝑥2 +

2

3𝑥 −

5

3

Use the graph of the quadratic function f to write it as 𝑓 𝑥 = 𝑎(𝑥 − ℎ)2+𝑘

Vertex= −2,−4 meaning our equation will become 𝑓 𝑥 = 𝑎(𝑥 + 2)2−4 and we can use the point (0,4) to find 𝑎. 4 = 𝑎 0 + 2 2 − 44 = 4𝑎 − 48 = 4𝑎2 = 𝑎So 𝑓 𝑥 = 2(𝑥 + 2)2−4

Write in the form 𝑓 𝑥 = 𝑎(𝑥 − ℎ)2+𝑘 and identify the vertex.

𝑓 𝑥 = 2𝑥2 + 4𝑥 − 5

It is easier to first identify the vertex using 𝑥 =−𝑏

2𝑎.

𝑥 =−4

2(2)=

−4

4= −1

𝑓 −1 = 2(−1)2+4 −1 − 5 = −7 so our vertex is −1,−7 and we can plug that back into our equation as h and k. The leading coefficient 𝑎 stays the same as the leading coefficient of our original equation, 𝑎 = 2. Our equation then becomes 𝑓 𝑥 = 2(𝑥 + 1)2−7

Use the vertex formula to determine the vertex on the graph of f.

𝑓 𝑥 = 𝑥2 + 8𝑥 + 5

Vertex formula is (−𝑏

2𝑎, 𝑓

−𝑏

2𝑎)

In our case, −𝑏

2𝑎=

−8

2(1)=

−8

2= −4

And 𝑓 −4 = (−4)2+8 −4 + 5 = −21So our vertex is (−4,−21)

Solve the quadratic equation.

3𝑥2 + 4𝑥 = −13𝑥2 + 4𝑥 + 1 = 0You can either factor, or use the quadratic formula. Factoring: 3𝑥2 + 4𝑥 + 1 = 3𝑥 + 1 𝑥 + 1 = 03𝑥 + 1 = 0 OR 𝑥 + 1 = 0

𝑥 =−1

3OR 𝑥 = −1

Quadratic: 𝑥 =−𝑏± 𝑏2−4𝑎𝑐

2𝑎=

−4± 42−4(3)(1)

2(3)=

−4± 4

6=

−4±2

6=

−4+2

6OR

−4−2

6= −

1

3𝑂𝑅 − 1

−5𝑥2 − 3𝑥 = 0You again can either factor or use the quadratic formula.Factoring: −5𝑥2 − 3𝑥 = 𝑥 −5𝑥 − 3 = 0𝑥 = 0 OR −5𝑥 − 3 = 0

𝑥 = −3

5

Quadratic: − −3 ± −3 2−4(−5)(0)

2(−5)=

3± 9

−10=

3±3

−10=

3+3

−10𝑂𝑅

3−3

−10= −

3

5𝑂𝑅 0

Solve the quadratic equation.

25𝑧2 = 9𝑧2 =

9

25

𝑧 =± 9

± 25=

±3

±5

= ±3

5

Quadratic: −0± 02−4(25)(−9)

2(25)=

± 900

50= ±

30

50= ±

3

5

1

2𝑡2 +

3

4𝑡 +

1

4= 0

First get rid of all of the fractions by multiplying by 4: 2𝑡2 + 3𝑡 + 1 = 0Then either factor or use the quadratic formula. Factor: 2𝑡 + 1 𝑡 + 1 = 02𝑡 + 1 = 0 OR 𝑡 + 1 = 0

𝑡 = −1

2OR 𝑡 = −1

Quadratic: −3± 32−4(2)(1)

2(2)=

−3± 1

4=

−3±1

4=

−3+1

4𝑂𝑅

−3−1

4= −

1

2𝑂𝑅 − 1

Solve the quadratic equation.

𝑥 6 − 𝑥 = −16

First, you must make sure the equation equals zero, 6𝑥 − 𝑥2 + 16 = −𝑥2 + 6𝑥 + 16 = 0Factor: −𝑥 + 8 𝑥 + 2 = 0−𝑥 + 8 = 0 OR 𝑥 + 2 = 0𝑥 = 8 OR 𝑥 = −2

Quadratic: −6± 62−4(−1)(16)

2(−1)=

−6± 100

−2=

−6±10

−2=

−6+10

−2𝑂𝑅

−6−10

−2= −2 𝑂𝑅 8

(𝑘 + 2)2= 7

𝑘 + 2 2 = 7

𝑘 + 2 = ± 7

𝑘 = ± 7 − 2

a) State whether 𝑎 > 0 or 𝑎 < 0

b) Estimate the real solutions.

c) Is the discriminant positive, negative or zero?

a) Since our graph is “frowning”, 𝑎 < 0b) There are no places where our graph

crosses the x-axis, therefore there are no real solutions. NONE

c) Since we have no real solutions, our discriminant is negative (we have two imaginary solutions)

Solve by completing the square.

𝑥2 − 3𝑥 = 3

𝑥2 − 3𝑥 + __________ = 3

The number that fills in the blank is the one we use to complete the square: −𝑏

2𝑎

2=

−3

2 1

2=

9

4and that value gets added on both sides of the equation

𝑥2 − 3𝑥 +9

4= 3 +

9

4

Factor the left side of the equation

𝑥 −3

2𝑥 −

3

2=

21

4

𝑥 −3

2

2=

21

4

𝑥 −3

2

2=

21

4

𝑥 −3

2= ±

21

4

𝑥 =± 21

2+

3

2

𝑥 =3± 21

2

Solve ℎ = −1

2𝑔𝑡2 + 100 for 𝑡

ℎ = −1

2𝑔𝑡2 + 100

ℎ − 100 = −1

2𝑔𝑡2

−2 ℎ − 100 = 𝑔𝑡2

−2(ℎ−100)

𝑔= 𝑡2

±−2(ℎ−100)

𝑔= 𝑡

Write each expression in standard form.

a) 2 − 3𝑖 + −3 + 3𝑖2 − 3𝑖 − 3 + 3𝑖 = −1

b) −5 + 3𝑖 − (−3 − 5𝑖)

−5 + 3𝑖 + 3 + 5𝑖 = −2 + 8𝑖

c) (3 + 2𝑖)(−4 − 𝑖)

−12 − 3𝑖 − 8𝑖 − 2𝑖2 = −12 − 11𝑖 − 2 −1 = −10 − 11𝑖

d) 3+2𝑖

2−𝑖

3+2𝑖

2−𝑖∙2+𝑖

2+𝑖=

6+3𝑖+4𝑖+2𝑖2

4+2𝑖−2𝑖−𝑖2=

6+7𝑖+2 −1

4− −1=

4+7𝑖

5=

4

5+

7

5𝑖

Use the graph and the given 𝑓(𝑥) to complete the following

a) find any x-intercepts

b) Find the complex zeros of 𝑓 a) About 1.5

b) 𝑓 𝑥 = 𝑥2 − 3𝑥 +9

4

So we use the quadratic:

−(−3)± (−3)2−4(1)(9

4)

2(1)

3± 0

2=

3

2

Use the graph and the given 𝑓(𝑥) to complete the following

a) find any x-intercepts

b) Find the complex zeros of 𝑓a) Our graph never touches

the x-axis so NONEb) 𝑓 𝑥 = −1 + 2𝑥 − 2𝑥2

So we can use the quadratic:−2± 22−4(−2)(−1)

2(−2)=

−2± −4

−4−2±2𝑖

−4=

−1±𝑖

−2=

1

2±

1

2𝑖

Find all complex solutions.

2𝑥2 + 3 = 2𝑥

Since it asks for the complex solutions, we know that we have to use the quadratic formula because the equation can not be factored. 2𝑥2 + 3 = 2𝑥2𝑥2 − 2𝑥 + 3 = 0− −2 ± −2 2−4(2)(3)

2(2)=

2± −20

4=

2± −1 4 5

4=

2±2𝑖 5

4=

1±𝑖 5

2=

1

2±

5

2𝑖

The table contains test values for a quadratic function. Solve each inequality.

a) 𝑓(𝑥) < 0

b) 𝑓(𝑥) ≥ 0

x -6 -5 0 3 4

f(x) 9 0 -15 0 9

a) We want to look at values in our table that are less than zero, those values include -5, 0 and 3 where -5 and 3 are not included: −5 < 𝑥 < 3

b) We want to look at vlues in our table that are greater than or equal to zero, those values include ⋯− 6,−5 𝑎𝑛𝑑 3, 4,⋯ which gives us the following: 𝑥 ≤ −5 or 𝑥 ≥ 3

Solve the inequality. Write the solution in interval notation.

𝑥2 − 2𝑥 ≥ 0

𝑥2 − 2𝑥 ≥ 0 can be solved by either factoring or the quadratic formula. Factoring: 𝑥2 − 2𝑥 = 𝑥(𝑥 − 2) ≥ 0 meaning𝑥 ≥ 0 OR 𝑥 − 2 ≥ 0

𝑥 ≥ 2But those intervals overlap so you need to look at the graph and determine where our graph is above the x-axis. You have only found the critical points where our graph crosses the x-axis.

Our graph sits above the x-axis on the following intervals: −∞, 0 ∪ [2,∞)

Solve the inequality. Write the solution in interval notation.

9𝑥2 − 4 > 0

This equation is best solved algebraically: 9𝑥2 > 4

𝑥2 >4

9

𝑥 >± 4

± 9= ±

2

3and again our intervals overlap so we have to look at the graph to

determine where our graph is above the x-axis

Our graph falls above the x-axis from

(−∞,−2

3) ∪ (

2

3, ∞)

Solve the inequality. Write the solution in interval notation.

𝑛2 + 4 ≤ 6𝑛

This equation cannot be solved with factoring so we must use the quadratic formula:𝑛2 − 6𝑛 + 4 ≤ 0−(−6)± (−6)2−4(1)(4)

2(1)=

6± 20

2=

6± 4 5

2=

6±2 5

2= 3 ± 5

Again, our intervals overlap so we must look at the graph.

We want to know where our graph falls below the x-axis and we can see that this occurs in the following interval:

[3 − 5, 3 + 5]

Use the transformations to sketch a graph of f.

a) 𝑓 𝑥 = 3 𝑥 − 2 + 1

b) 𝑓 𝑥 = −4 −𝑥

c) 𝑓 𝑥 = − 𝑥 − 3

The rest of the chapter review covers the application problems. There will be application problems on the exam so study accordingly.