College Level Math
1
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Evaluation of algebraic expressions by substitution
Algebraic expressions contain letters which are referred to as variables. If we replace a variable
with a number, it is referred to as substitution. Then, when we carry out any calculations, it is
called evaluating the expression.
Example 1: Find the value of ac when a=15 and c=5.
Substitute 15 for a and 5 for c and carry out the multiplication.
ac=(15)(5) =75
Example 2: m=20 and n=4. Find the value of ½(m-n).
In the expression ½(m-n), replace m with 20 and n with 4. Then carry out the
subtraction and multiplication.
1
1
1
(m-n) = 2 (20-4) = 2(16) = 8
2
Example 3: Evaluate the expression
𝑃−2𝐿
2
when P = 90 and L = 26.
Replace P with 90 and L with 26. Simplify the numerator and divide by 2.
𝑃−2𝐿
2
=
90−2(26)
2
=
90−52
2
=
38
2
= 19
Here are some problems for you to try.
Evaluate the following:
1. 2m2 when m = 8
𝑢
2. uv+𝑣 when u=12 and v=4
3. k(k+h) when k=6 and h=7
4. 3p-2s when p=9 and s=5
5. a-bc when a=12, b=6, and c=2
6.
𝑟+𝑡
2
when r=15 and t=9
7. (𝑒 + 𝑓)2 when e=8 and f=4
8. 𝑒 + 𝑓 2 when e=8 and f=4
9. 𝑒2+f when e=8 and f=4
10. (𝑤 + 𝑦)(𝑤 − 𝑦)when w=5 and y=4
𝑝+𝑞
11. 𝑝−𝑞 when p=10 and q=5
12. (𝑏 − 𝑐)2when b=7 and c=4
13. x2 when x= -3
14. –x2 when x = -3
Answers
Multiplication symbols: *
● or x
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1. 128
2. 51
3. 78
4. 17
5. 0
6. 12
7. 144
8. 24
9. 68
10. 9
11. 3
12. 9
13. 9
14. -9
Addition, subtraction, and multiplication of polynomials
When two or more terms have the same variables raised to the same powers, they are considered
like terms. Like terms can be added or subtracted. Likewise, polynomials can be added or
subtracted by “collecting” or “combining” like terms.
Example 1: Collect like terms
3a2 – 4b + 2a2 =5a2 – 4b
Example 2: Add
(-6v3 + 2v – 4) + (4v3 + 3v2 +2) = -2v3 + 3v2 +2v -2
Example 3: Subtract
(4p4q – 5p3q2 + p2q3 +2q4) – (-5p4q + 5p3q2 – 3p2q3 – 7q4) = 9p4q – 10p3q2+ 4p2q3 + 9q4
Here are some problems for you to try. The answers follow the exercise.
Collect like terms
1. 3y – 4x + 6xy2 – 2xy2
2. 3ab3 + 2a3b + 5ab3 – 8a +15 – 3a2b – 6a2b + 11a – 8
Add.
3.
4.
5.
6.
(7x5 – 5) + (3x5 – 4x2 +10)
(5a2b4 – 2a2b2 – 3b) + (-6a2b2 +3b +5)
(4ax2 + 4 bx – 5) + (-6ax2 + 8)
(13g3h +3g2h – 5h) + (g3h + 4g2h – 3gh)
Subtract
Multiplication symbols: *
● or x
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7. (-5a2 +4) – (2a2 + 3a – 1)
8. (9x3 – 2x – 4) – (-5x3 – 8)
9. (2x5 – x4 +3x3 –x2 – x – 7) – (-x5 +2x4 – 2x3 +x2 – x – 4)
10. (4x4y – 5x3y2 + x2y3 +2y4) – (–5x4y + 5x3y2 – 3x2y3 – 7y4)
Answers
1. 3y – 4x + 4xy2
2. 8ab3 + 2a3b – 9a2b + 3a + 7
3. 10x5 – 4x2 + 5
4. 5a2b4 – 8a2b2 + 5
5. –2ax2 + 4bx + 3
6. 14g3h + 7g2h – 3gh –5h
7. –7a2 – 3a +5
8. 14x3 – 2x + 4
9. 3x5 – 3x4 + 5x3 –2x2 – 3
10. 9x4y – 10x3y2 + 4x2y3 + 9y4
Solution of linear equations in 1 variable
Linear equations in 1 variable are solved using the addition and multiplication principles. The
solution determines the replacement for x which will make the equation true.
Example 1:
x + 6 = –15
check:
x + 6 = -15
x + 6 + (-6) = –15 + (-6)
-21 + 6 = -15
x = –21
Example 2:
4
5
–15 = –15
4
𝑥 = 22
5
4
5
∗ x=4∗
4 5
x=
check: 5 𝑥 = 22
22
1
55
2
4
∗
5
55
2
= 22
22=22
Here are some problems for you to try.
Solve.
1.
2.
3.
4.
5.
6.
7.
8x = 10
−3
𝑥 = 21
7
−𝑥 = −5
−𝑥
= 17
8
−4𝑥 + 2 + 5𝑥 = 3𝑥 − 15
30 + 7(𝑥 −1) =3(2x + 7)
3
2
𝑥
+
5
=
𝑥−7
4
3
Answers
Multiplication symbols: *
● or x
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1.
2.
3.
4.
5.
6.
7.
5
4
–49
5
–136
17
2
–2
–144
Solution of linear inequalities in 1 variable
Inequalities are solved using the same principles that are used for solving equalities. There is
one difference: When solving an inequality, if you multiply or divide both sides of the inequality
by a negative number, the inequality sign changes direction.
Example 1: 3 – x < 2
-x< -1
x>1
Example 2: 2m – 22 > 16 (m – 4)
2m – 22 > 16m – 64
-14m > -42
m<3
Here are some practice problems for you to try
1. x – 2 >6
2. 3x + 5 < -10
3. x – 6 > 2x – 5
4. 5t – 3 < 9 – t
5. 0.6z < -18
6. 2c – 7 < 5c – 9
7. 4(4x – 3) > 9(2x + 7)
8. 3(2 – 5x) + 2x < 2(4 + 2x)
9. 5(t + 3) + 9 < 3(t-2) +6
10. 13 – (2c +2) > 2(c + 2) + 3c
Answers
1. x > 8
2. x < –5
3. x < –1
4. t < 2
5. z < –30
6. c > 2/3
7. x < –75/2
8. x > –2/17
9. t < –12
10. c < 1
Factorization of polynomials
Multiplication symbols: *
● or x
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Factoring is the process of “unmultiplying,” that is, it is the opposite of multiplying. When you
factor a term or an expression, you find an equivalent term or expression which is a product.
Sometimes you will only be able to factor out a common term or the “greatest common factor.”
Other times you will be looking for two or more binomials. Sometimes you will need to do both.
When you multiply a monomial and polynomial, you multiply the monomial by the polynomial
using the distributive property of multiplication. When you factor, you do the opposite.
Example 1: Multiply
Factor
Example 2:
3x(6x2 + 2x – 1)
18x3 + 6x2 -3x
= 18x3 + 6x2 -3x
= 3x(6x2 + 2x -1)
Factor
10x6y2 – 4x5y3 + 2x4y4 – 2x4y2 = 2x4y2(5x2 - 2xy + y2 -1)
When you multiply two binomials, you use the FOIL method.
FOIL
(x + 6)(x – 4) = x2 – 4x + 6x – 24 = x2 + 2x – 24
and when you factor a trinomial you are looking for the two binomials whose product is the
trinomial.
Example 3:
x2 + 2x – 24 = (x + 6)(x – 4)
Example 4:
3x2 + 19x + 20 = (3x + 4)(x + 5)
Sometimes you must first take out the greatest common factor and then find the two binomials:
Example 5:
20x5 – 46x4 + 24x3 = 2x3 (10x2 – 23x + 12) = 2x3 (2x – 3)(5x – 4)
Here are some practice problems.
Factor
1.
2.
3.
4.
5.
6.
x2 – 14x +45
3y2 – 20y +32
9y2 + 15y + 4
14x4 – 19x3 –3x2
5x + x2 –14
56x + x2 – x3
Answers
1. (x – 9)(x – 5)
2. (3y – 8)(y – 4)
3. (3y + 1)( 3y + 4)
4. x2(7x+1)(2x – 3)
5. (x + 7)(x – 2)
6. x(8 – x)(7 + x)
There are also several “special” types of factorization with which you should be familiar.
Multiplication symbols: *
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A trinomial square factors into two binomials which are identical:
A2 + 2AB + B2 = (A + B)2
A2 – 2AB + B2 = (A – B)2
Example 1:
x2 – 4x + 4 = (x – 2)(x – 2) = (x – 2)2
Example 2:
9y2 + 12y + 4 = (3y + 2)(3y + 2) = (3y + 2)2
These problems are trinomial squares. Try them using the pattern above.
Factor.
1.
2.
3.
4.
x2 + 14x + 49
9x2 – 30x + 25
16x2 + 72xy +81y2
-8a2 + 24ab – 18b2
Answers
1.
2.
3.
4.
(x + 7)(x + 7) = (x + 7)2
(3x – 5)(3x – 5) = (3x – 5)2
(4x + 9y)(4x + 9y) = (4x + 9y)2
–2(2a – 3b)(2a – 3b) = –2(2a – 3b)2
These are examples of differences of squares:
y2 – 9
x4 – 4y4
9a2 – 36b2
And this is the pattern for factoring the difference of two squares:
A2 – B2 = (A + B)(A – B)
Example 1:
y2 – 16 = (y + 4)(y – 4)
Example 2:
4x3 – 49x = x(2x + 7)(2x – 7)
Try these problems using the pattern.
Factor.
1. x2 – 9
2. 9y2 – 36
3. 2x4 – 8y4
4. 20x2 – 5y2
5. a4 – 16b4
Answers
1. (x + 3)(x – 3)
2. 9(y + 2)(y – 2)
3. 2(x2 – 2y2)(x2 + 2y2)
4. 5(2x – y)(2x + y)
5. (a – 2b)(a + 2b)(a2 +4b2)
Multiplication symbols: *
● or x
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Another “special” form of factorization involves factoring sums or differences of cubes. Here
are the patterns that must be used:
A3 + B3 = (A + B)(A2 – AB + B2)
A3 – B3 = (A – B)(A2 + AB + B2)
Example 1:
x3 – 8 = (x3 – 23) = (x – 2)(x2 + 2x + 22) = (x – 2)(x2 + 2x + 4)
Example 2:
24a3 + 3 = 3(8a3 +1) = 3[(2a)3 + 13]
= 3(2a + 1)[(2a)2 – (2a)(1) + 12] = 3(2a + 1)(4a2 – 2a +1)
Here’s some practice for you.
Factor
1.
2.
3.
4.
a3 + 27
8 – 27b3
64x3 + 1
ab3 + 125a
Answers
1.
2.
3.
4.
(a + 3)(a2 – 3a +9)
(2 – 3b)(4 + 6b +9b2)
(4x +1)(16x2 – 4x +1)
a(b + 5)(b2 – 5b + 25)
Solution of polynomial equations by factoring
Now that you have reviewed factoring polynomials, you should be ready to solve quadratic
equations. Follow these steps to solve such an equation:
1)
2)
3)
4)
Move all the terms to one side of the equal sign so the equation is equal to zero.
Factor completely
Set each factor equal to zero
Solve the equations
Example 1: 7x + 3x2 = -2
3x2 +7x +2 =0
(3x +1)(x + 2) = 0
3x +1 =0
1
x=-3
x+2=0
x = -2
Example 2: 3x3 – 9x2 = 30x
3x3 – 9x2 – 30x = 0
3x(x2 – 3x – 10) = 0
Multiplication symbols: *
● or x
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3x(x - 5)(x + 2) = 0
3x = 0
x–5=0
x+2=0
x=0
x=5
x = -2
Here are some practice problems for you.
1. x2 + 8 = 6x
2. 5y + 2y2 = 3
3. 8b2 = 16b
4. 25 + x2 = –10x
5. x3 + x2 = 6x
6. 9y2 + 15y + 4 = 0
7. 27 + 12t + t2 = 0
8. 8y – y2 = 0
9. 2x3 – 2x2 =12x
10. 2x3 = 128x
Answers
1. 2, 4
2. -3, ½
3. 0, 2
4. -5
5. -3, 0, 2
6. -4/3, -1/3
7. -9, -3
8. 0, 8
9. -2, 0, 3
10. -8, 0, 8
Properties of square roots
The processes of “squaring” and “finding square roots” are opposite, or inverse, operations.
Example 1:
Square
Square root
32 = 9
√9 = 3
72 = 49
√49 = 7
You can also find square roots of expressions which contain variables. (Note: Throughout this
packet assume all variables under radical signs represent positive numbers.)
Example 2:
Multiplication symbols: *
Square
Square root
(3x)2 = 9x2
√9𝑥2 = √(3𝑥)2 = 3x
(ab)2 = a2b2
√𝑎2 𝑏 2 = √(𝑎𝑏)2 =ab
● or x
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Try these problems.
Simplify
1.
2.
3.
4.
5.
6.
7.
8.
9.
√225
√441
√(7𝑤)2
√(𝑥𝑦)2
√𝑥 2 𝑦 2
√(𝑥 − 11)2
√25𝑦 2
√𝑥 2 + 8𝑥 + 16
-√81
1
10. √ 𝑎2
4
Answers
1. 15
2. 21
3. 7w
4. xy
5. xy
6. x-11
7. 5y
8. x + 4
9. -9
10. 1/2a
Rational expressions
1 −5 16
50
2
1
,
8
, −3 , 𝑎𝑛𝑑
x2 + 6x + 5
x2 – 3x +2
are examples of rational numbers
and y2 – 9 are rational expressions or fractional expressions.
y+3
Because rational expression indicate division, you need to remember that a denominator (the
bottom of the fraction) of zero makes the expression undefined.
Look at the examples below:
x2 + 6x + 5 = (x+5)(x+1)
x2 – 3x + 2
(x-2)(x-1)
In this expression, replacing x with either 2 or 1 would create
a denominator a zero. Therefore, we say it is undefined for x=1 or 2.
y2 – 9
y+3
This expression is undefined for y = -3, because replacing y with
-3 would create a zero in the denominator.
Multiplication symbols: *
● or x
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Recall that a numerator (the top of the fraction) of zero does not make the expression undefined.
That is because if you divide a number into zero, the answer is zero – and zero is perfectly
acceptable as an answer!
Here are some practice problems. Remember, you may need to factor the denominator before
you can determine your answer.
Find all numbers for which the rational expressions are undefined.
1.
2.
6
2𝑥
𝑥−6
𝑥+8
3. x2 – 4y + 9
2y + 5
4.
𝑥2 – 9
𝑥 2 − 7𝑥 + 10
Answers
1. 0
2. -8
−5
3. 2
4. 2, 5
Multiplying rational expressions is just like multiplying fractions. Multiply the numerators to get
the numerator of the product and multiply the denominators to get the denominator of the
product. When multiplying rational expression, always factor both the top and bottom to reduce
the fractions, if possible, before you multiply.
Example 1:
(3𝑥+2𝑦)𝑥
3𝑥+2𝑦
𝑥
𝑥+2
𝑥 2 −4
3𝑥 2 +2𝑥𝑦
● = (5𝑥+4𝑦)𝑥 = 5𝑥 2 +4𝑥𝑦
5𝑥+4𝑦 𝑥
1
Example 2:
𝑥+2
(𝑥+2)(𝑥−2)
(𝑥+2)(𝑥−2)
● 𝑥 2 +𝑥−2 = 𝑥−3 ● (𝑥+2)(𝑥−1) = (𝑥−3)(𝑥−1)
𝑥−3
1
Dividing rational expressions is just like multiplying except you must first invert (flip over) the
second expression. The problem then becomes a multiplication.
Example 3:
Example 4:
𝑥−2
𝑥+5
𝑥−2
𝑥−3
3𝑥 5
2
(𝑥−2)(𝑥−3)
÷
=
● 𝑥+5 = (𝑥+1)(𝑥+5)
𝑥+1 𝑥−3 𝑥+1
12𝑥 8
3𝑦 4
÷
16𝑥 3
6𝑦
=
12𝑥 8
3𝑦 4
6𝑦
3𝑥 5
● 16𝑥 3 = 2𝑦 3
y3
2
1
𝑦 2 −36
3𝑦−18
1
(𝑦+6)(𝑦−6)
Example 5: 𝑦 2 −8𝑦+16 ÷ 𝑦 2 −𝑦−12 = (𝑦−4)(𝑦−4) ●
1
Multiplication symbols: *
(𝑦−4)(𝑦+3)
3(𝑦−6)
=
(𝑦+6)(𝑦+3)
3(𝑦−4)
1
● or x
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Try these practice problems before you go on to adding and subtracting rational expressions.
Multiply or divide as indicated. Simplify if possible.
1.
2.
3.
4.
5.
6.
2𝑥 2 + 20x + 50
𝑥 2 −4
𝑥 2 −16
2𝑥+6
𝑥−4
÷ 𝑥+3
𝑥 3 −8
●
𝑥 2 −25
9𝑐 2 −1
𝑐 2 −9
𝑦 2 −64
2𝑦+10
𝑥 3 −64
𝑥 2 −16
𝑥+2
● 𝑥+5
÷
𝑥 2 +10𝑥+25
𝑥 2 +2𝑥+4
3𝑐+1
𝑐+3
𝑦+5
● 𝑦+8
𝑥 2 +5𝑥+6
÷ 𝑥 2 −3𝑥−18
Answers
1. 2(x + 5)
x−2
2. x + 4
2
3. (x− 2)(x + 5)
x-5
4. 3c – 1
c–3
5. y – 8
2
6. (x – 6)(x2 + 4x + 16)
(x + 4)(x + 2)
Adding and subtracting rational expressions follows the same rules as adding and subtracting
fractions.
If the expressions have the same denominator, add or subtract the numerators and keep the same
denominator.
3+𝑥
Example 1:
𝑥
4𝑥+5
Example 2:
𝑥+3
4
+𝑥=
3+𝑥+4
𝑥
𝑥−2
− 𝑥+3 =
=
7+𝑥
𝑥
4𝑥+5−(𝑥−2)
𝑥+3
=
4𝑥+5−𝑥+2
𝑥+3
=
3𝑥+7
𝑥+3
As with fractions, you should reduce your answer to lowest terms whenever possible.
1
4𝑥 2 −5𝑥𝑦
Example 3:
𝑥 2 −𝑦2
+
2𝑥𝑦− 𝑦2
𝑥 2 −𝑦 2
=
4𝑥 2 −3𝑥𝑦−𝑦 2
𝑥 2 −𝑦 2
=
(4𝑥+𝑦)(𝑥−𝑦)
(𝑥+𝑦)(𝑥−𝑦)
=
4𝑥+𝑦
𝑥+𝑦
1
Multiplication symbols: *
● or x
November 2012 Document1
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If the expressions have different denominators, first find the least common denominator (LCD)
and then add or subtract the numerators. This sometimes requires factoring.
Example 4:
2𝑎
3𝑏
+ 2𝑎 =
5
2𝑎
5
×
2𝑎
+
2𝑎
3𝑏
5
× 5 [LCD = (5)(2a) = 10a]
2𝑎
4𝑎2 15𝑏 4𝑎2 + 15𝑏
+
=
10𝑎 10𝑎
10𝑎
2𝑦+1
Example 5:𝑦 2 −7𝑦+6 −
(2𝑦+1)
(𝑦−6)(𝑦−1)
2𝑦 2 +3𝑦+1−𝑦 2 −2𝑦+3
=
(𝑦−6)(𝑦−1)(𝑦−1)
𝑦+3
𝑦 2 −5𝑦−6
●
(𝑦+1)
(𝑦+1)
=
−
2𝑦+1
(𝑦−6)(𝑦−1)
(𝑦+3)
(𝑦−6)(𝑦+1 )
●
−
𝑦+3
(𝑦−6)(𝑦+1)
(𝑦−1)
(𝑦−1)
=
[LCD = (y – 6)(y – 1)(y + 1)]
(2𝑦+1)(𝑦+1)−(𝑦+3)(𝑦−1)
(𝑦−6)(𝑦+1)(𝑦−1)
=
𝑦 2 +𝑦+4
(𝑦−6)(𝑦+1)(𝑦−1)
Try these problems on your own.
Add or subtract as indicated. Check each answer to see if it can be reduced.
1.
2.
3.
4.
5.
6.
7.
8.
9.
𝑥−2𝑦
𝑥+𝑦
+
4𝑎−2
𝑦−2
2𝑦−3
𝑥+𝑦
5+3𝑎
𝑎2 −49
4𝑦+3
𝑥+9𝑦
+ 49−𝑎2
𝑦−2
− 𝑦−2
4−𝑦
𝑦 2 −1
− 1−𝑦 2
𝑦−2
𝑦+3
+ 𝑦−5
𝑦+4
𝑥−2
𝑥+3
𝑥+2
+
𝑥−4
9𝑥+2
3𝑥 2 −2𝑥−8
𝑥−1
3𝑥+15
−
3𝑦+2
𝑦 2 +5𝑦−24
+
7
3𝑥 2 +𝑥−4
𝑥+3
5𝑥+25
+
Multiplication symbols: *
7
𝑦 2 +4𝑦−32
● or x
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3𝑥−1
𝑥+4
10. 𝑥 2 +2𝑥−3 − 𝑥 2 −9
Answers:
2𝑥+7𝑦
1. 𝑥+𝑦
2.
3.
4.
5.
6.
7.
8.
9.
1
𝑎+7
3𝑦+5
𝑦−2
1
𝑦−1
2𝑦 2 +22
(𝑦+4)(𝑦−5)
−11𝑥+2
(𝑥+3)(𝑥−4)
3𝑥−4
(𝑥−2)(𝑥−1)
2𝑥−14
15(𝑥+5)
3𝑦 2 −3𝑦−29
(𝑦+8)(𝑦−3)(𝑦−4)
2𝑥 2 −13𝑥+7
10. (𝑥+3)(𝑥−1)(𝑥−3)
Radical Expressions
Earlier in this packet you reviewed square roots. Hopefully, you understand that there are many
roots other than square roots – cube roots, 4th roots, 5th roots – an infinite number of possibilities.
First we will review how to simplify radicals.
To simplify a radical expression, you first must look for factors of the radicand (the number
3
3
under the radical sign) that are perfect powers. [For instance, √16 = √8 ∗ 2 where the number 8
is a perfect cube (23 = 8).} Then take the appropriate root of the resulting factors. The radical
expression is simplified when its radicand has no factors that are perfect powers. This is more
easily seen through examples.
Example 1:
Example 2:
Example 3:
Multiplication symbols: *
√50 = √25 ∗ 2 = √25 *√2 = 5√2
4
4
4
4
4
√243 = √81 ∗ 3 = √81 *√3 = 3√3
3
√6𝑥 5 = 3√6𝑥 3 𝑥 2 = 3√𝑥 3 * 3√6𝑥 2 = x * 3√6𝑥 2
● or x
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Try these practice problems.
Simplify the following radical expressions by factoring.
1. √24
2. √𝑥 4
3
3. √54
√54𝑥 6
4.
3
5.
4
6.
4
√(𝑥 + 𝑦)4
√(𝑥 + 𝑦)6
7. 3√−24𝑥 4 𝑦 5
Answers:
1. 2√6
2. x2
3
3. 3 √2
3
4. 3x2 ● √2
5. x + y
6. (x + y) ● 4√(𝑥 + 𝑦)2
7. – 2xy ● 3√3𝑥𝑦 2
Multiplying radical expression is not difficult: simply multiply the radicands and simplify, if
possible.
Example 1:
√3 * √6 = √18 = √9 ∗ 2 = 3√2
Example 2:
3√25 ● 2√5 = 6 √125 = 6 ● 5 = 30
Example 3:
3
3
3
3
√2 ∗ (𝑥 + 5) * 3√4 ∗ (𝑥 + 5)4 = 3√8 ∗ (𝑥 + 5)5 = 2 * (x + 5) * 3√(𝑥 + 5)2
Dividing follows the same principles. Divide the radicands and simplify if you can.
Example 4:
Example 5:
Multiplication symbols: *
√80
80
√5
5
=√
3∗ √2
5∗ √3
=
= √16 = 4
3∗ √2
5∗ √3
∗
(√3)
(√3)
=
3 √6
15
=
√6
5
(This is called rationalizing the denominator)
● or x
November 2012 Document1
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Practice
Multiply or divide as indicated. Simplify if possible.
1. √2 ● √32
3
3
2. √3 ● √18
3
3.
4.
5.
6.
7.
8.
5 √32
3
√2
√72𝑥𝑦
2∗ √2
√52 ∗ 𝑡 4 * 3√54 ∗ 𝑡 6
3
√𝑥 3 − 𝑦3
(hint: factor x3 – y3)
√𝑥−𝑦
5
√5
3√6
2√12
Answers
1. 8
3
2. 3 ● √2
3
3. 10 ● √2
4. 3√𝑥𝑦
3
5. 25t3 ● √𝑡
6. √𝑥 2 + 𝑥𝑦 + 𝑦 2
7. √5
3√2
8. 4
When you add or subtract radical expression, sometimes it is not possible to simplify the answer.
For instance, 3 + √5 cannot be simplified. However, if two radicals have the same radicand and
the same index (the small raised number that indicates the root), you can simplify by collecting
the terms.
Example 1:
7 ● √3 + √3 = 7 ● √3 + 1●√3 = 8 ●√3
Example 2:
3 ● √8 – 5√2 = 3 ● √4 ∗ 2 – 5 √2 = 3 ● 2 * √2 – 5 ● √2 = 6 ● √2 – 5 ● √2 = √2
Example 3:
5 ● 3√16𝑦 4 + 7 ● 3√2𝑦 = 5 ● 3√8 ∗ 2𝑦 3 + 𝑦 + 7 3√2𝑦 = 5 ● 2 ● y● 3√2𝑦 + 7 3√2𝑦
= 10y ● 3√2𝑦 + 7 ● 3√2𝑦 = (10y + 7) ● 3√2𝑦
Multiplication symbols: *
● or x
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Practice problems.
Add or subtract. Simplify if possible.
1. 5 ●√2 + 8 ● √2
4
4
2. 7 ● √5𝑥 + 3 ● √5𝑥 - 7
3. 7 ● √45 – 2 ●√5
4. 3 ● 3√𝑦 5 + 4 ● 3√𝑦 2 + 3√8𝑦 6
5. √25𝑥 − 25 − √9𝑥 − 9
6. 2 ●√128 − √18 + 4 ● √32
3
3
3
7. 5 ● √32 − √108 + 2 ● √256
8. √𝑥 3 − 𝑥 2 + √9𝑥 − 9
Answers
1. 13 ● √2
4
2. 10 ● √5𝑥 – 7
3. 19 ● √5
4. (3y + 4) ● 3√𝑦 2 + 2y2
5. 2 ● √𝑥 − 1
6. 29 ● √2
3
7. 15 ● √4
8. (x + 3) ● √𝑥 − 1
Quadratic formula
Earlier you reviewed solving quadratic equations by factoring and applying the principal of zero
products. Quadratic equations can also be solved using the quadratic formula:
𝑥=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
To use the formula you must first put the equation into standard form:
ax2 + bx + c = 0
Multiplication symbols: *
● or x
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For example, the equation 5x2 + 8x = -3 becomes 5x2 + 8x + 3 = 0 and a = 5, b = 8, and c = 3.
Now substitute these values into the quadratic formula and solve for x.
𝑥=
x=
−8±√82 −4(5)(3)
2(5)
−8+2
10
−3
=
=
or x =
5
−8+ √4
10
−8−2
10
= -1
Here’s another example. Solve 3x2 = 18x − 6 for x
3x2 = 18x – 6
3x2 – 18x + 6 = 0 so a = 3, b = -18 , c = 6
𝑥=
−(−18)±√(−18)2 −4(3)(6)
2(3)
=
18±√324−72
6
=
18±√252
6
=
18±√36∗7
6
=
18±6√7
6
= 3 ± √7
So
x = 3 + √7 or x = 3 − √7
Practice problems
Solve for x using the quadratic equation.
1. x2 + 6x + 4 = 0
2. x2 – 6x – 4 = 0
3. 3x2 = -8x −1
4. h2 + 4 = 6h
5. 15x2 = 17x – 2
6. 4x + x(x−3) = 0
Answers
1. – 3 ± √5
2. 3 ± √13
3.
−4
3
±
√13
3
4. 3 ± √5
2
5. 15 , 1
6. −1, 0
Numbers to fractional powers
Multiplication symbols: *
● or x
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Fractional exponents are another way of writing radical expressions. In a fractional exponent,
the denominator of the fraction indicates the root or index. The numerator is the exponent of the
base.
Example 1:
x1/2 = √𝑥
Example 2:
(xy)1/4 = 4√𝑥𝑦
Example 3:
5
Example 4:
8 ● 3√(𝑥𝑦)2 = 8 (xy)2/3
√7𝑥𝑦 = (7xy)1/5
These kinds of expressions can often be simplified.
Example 5:
3
(27)2/3 = 3√272 = (√27)2 = 32 = 9
Rewrite the following without rational exponents. That is, rewrite using radicals.
Simplify if possible.
1. y1/4
2. x2/3
3. 45/2
4. (125)1/3
5. (a3b2c)1/5
6. 43/2
Rewrite the following with rational exponents
3
7. √19𝑎𝑏
3
8. 19 ● √𝑎𝑏
5
𝑥2𝑦
9. √ 16
10. 5√𝑎5 𝑏10
Answers
1. 4√𝑦
3
2. √𝑥 2
Multiplication symbols: *
● or x
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3. 32
4. 5
5.
√𝑎3 𝑏 2 𝑐
5
6. 8
7. (19ab)1/3
8. 19(ab)1/3
𝑥2𝑦
9. ( 16 )1/5 =
𝑥 2/5 𝑦 1/5
161/5
10. a5/5b10/5 = ab2
Systems of linear equations in 2 variables
The solution of a system of two linear equations in two variables is the ordered pair of numbers
which makes both statements true.
Example 1:
x+y=3
The order pair (-4, 7), when substituted into these two
equations, makes both statements true.
5x – y = – 27
x+y=3
-4 + 7 = 3 (true)
5x – y = -27
5(-4) – 7 = -27
-20 – 7 = -27 (true)
The ordered pair (-4, 7) can also be described by saying that x = -4 and y = 7.
Systems of equations may be solved a number of ways. One way is to graph both equations.
The point of intersection, if one exists, is the solution to the system.
Example 2:
1) y – x = 1 is the same as y = x + 1
2) y + x = 3 is the same as y = -x + 3
1
1) slope = 1
y intercept = 1
−1
2) slope = 1
y intercept = 3
Solution is (1, 2)
Note: graphing is not always accurate.
Note: If the lines are parallel, there will be no point of intersection and no solution to the system
of equations. If both equations graph as the same line, there are an infinite number of solutions
to the system of equations.
Multiplication symbols: *
● or x
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There are also other ways to solve a system of equations. One of those is the substitution
method.
Example 3:
2) x + y = 4
3) x = y +1
In this example you can substitute y + 1 for the x in the first equation.
1) x + y = 4
1) y + 1 + y = 4
2y + 1 = 4
2y = 3
3
y=2
3
Next, replace the y in either equation with 2; then solve for x.
x+y=4
x=y+1
3
3
x+2 =4
x =2 + 1
5
x=2
5
x=2
You may also use the elimination method which involves the addition principle. In this method,
you create—by multiplication—coefficients which are equal but which have opposite signs.
You can then add the two equations together to eliminate one of the variables and solve for the
other.
Example 4:
2x – 3y = 0
-4x + 3y = -1
In this problem the coefficients of y are already equal but
opposite.
Add the two equations to eliminate the y.
-2x + 0y = -1
-2x = -1
1
x=2
1
Now replace the x in either equation with 2; then solve for y.
2x – 3y = 0
1
2(2) – 3y = 0
1 – 3y = 0
-3y = -1
1
y=3
-4x + 3y = -1
1
-4(2) + 3y = -1
-2 + 3y = -1
3y = 1
1
y= 3
Example 5:
Multiplication symbols: *
● or x
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3x + 3y = 15
2x + 6y = 22
Multiply the first equation by -2; then add it to the second.
-6x + -6y = -30
2x + 6y = 22
-4x + 0y = -8
x=2
Now replace x with 2 in either equation and solve for y.
3x + 3y = 15
3(2) + 3y =15
6 + 3y = 15
3y = 9
y=3
2x + 6y = 22
2(2) + 6y = 22
4 + 6y = 22
6y = 18
y=3
Now you can try some yourself.
Solve the following systems of equation using any of the three methods discussed above.
1. x + y = 6
y=x+2
2. 8x + 5y = 184
x – y = -3
3. 5x + 3y = 17
-5x + 2y = 3
4. 4x + y = -1
x – 2y = 11
5. 5m + n = 8
3m – 4n = 14
6. 5r – 3s = 24
3r + 5s = 28
Answers:
1. x = 2, y = 4 2. x = 13, y = 16
5. m = 2, n = -2 6. r = 6, s = 2
3. x = 1, y = 4
4. x = 1, y = -5
Systems of Linear Equations in 3 variables
The procedure for solving three equations is as follows:
Multiplication symbols: *
● or x
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Step 1: Select two equations and eliminate a variable of your choice.
Step 2: Select two other equations and eliminate the same variable you chose in step 1.
Step 3: Solve the two equations that are left for the variable of your choice.
Step 4: Using substitution, solve for the remaining variables.
Example:
4x – 2y – 3z = 5 (1)
-8x – y + z = -5 (2)
2x + y + 2z = 5 (3)
Step 1: -8x – y + z = -5 (2)
2x + y + 2z = 5 (3)
-6x + 3z = 0 (4)
Step 2: 4x – 2y – 3z = 5 (1)
2x + y + 2z = 5 (3)
4x – 2y – 3z = 5 (1)
4x + 2y + 4z = 10 (3) multiply by 2
8x + z = 15 (5)
Step 3: -6x + 3z = 0 (4)
8x + z = 15 (5)
-6x + 3z = 0 (4)
-24x - 3z = -45 (5) multiply by 3
-30x = -45
-30 -30
3
x=2
Step 4: -6x + 3z = 0 (4)
3
-6(2) + 3z = 0
-9 + 3z = 0
3𝑧
3
=
9
3
z=3
4x – 2y – 3z = 5 (1)
3
4 (2) − 2𝑦 − 3(3) = 5
6 – 2y – 9 = 5
-2y - 3 = 5
Multiplication symbols: *
● or x
November 2012 Document1
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-2y = 8
y = -4
3
The solution is (2 , −4, 3)
Practice:
1) 2x + y – 4z = 0
x – y + 2z = 5
3x + 2y + 2z = 3
2) a + 2b + c = 1
7a + 3b – c = -2
a + 5b + 3c = 2
3) 3a – 2b +7c = 13
a + 8b – 6c = -47
7a – 9b – 9c = -3
4) x + y + z = 105
10y – z = 11
2x – 3y = 7
5) Joe, Jane and Sara can process 740 telephone orders per day. Joe and Jane together can
process 470 orders, while Jane and Sara can process 520 orders per day. How many
orders can each person process alone?
Answers:
1) (2, -2, ½)
2) (3, -5, 8)
3) (-3, -4, 2)
4) (17, 9, 79)
5) Joe 220, Jane 250, Sara 270
Solving systems that are not both linear.
Example Solve y = 𝑥 2
(1) (not a linear function)
y = x + 6 (2)
Step 1: Use substitution to create one equation in one variable
x2 = x + 6 (3)
Step 2: Solve by factoring or the quadratic formula since the new equation is a quadratic
x2 – x – 6 = 0
(x – 3)(x + 2) = 0
x = 3 or – 2
Step 3: Substitute into (1) or (2) and solve for y
y = x + 6 (2)
or
y=x+6
y=3+6
y = -2 + 6
Multiplication symbols: *
● or x
November 2012 Document1
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y=9
The solutions are (3, 9) or (-2, 4)
y=4
Practice:
1) x2 + y2 = 9 (1)
y = x + 1 (2)
2) y = √𝑥 + 4 (1)
x+y=2
(2)
3) (x – 3)2 + (y – 4)2 = 5 (1)
x+y=4
(2)
Answers
−1+ √17
1) (
2
,
1+ √17
−1− √17
2
2
) or (
,
1− √17
2
)
2) (0, 2) or (5, -3) but the (5, -3) has to be eliminated since it does not check in equation (1)
3) (2, 2) or (1, 3)
Absolute value equations
To solve equations involving absolute value, you must recall the absolute value principle:
a) For any positive number p, if |𝑥| = p, then x = -p or x = p.
b) For the equation |𝑥| = 0, the solution is x = 0.
c) |𝑥| = n
has no solution when n is a negative number
Here are examples of the preceding principle.
Example 1:
|𝑥| = 4
x = 4 or x = -4
This follows from the fact that |4| = 4 and |−4| = 4.
Example 2:
|2𝑥 + 5| = 13
2x + 5 = 13
2x = 8
x=4
or
2x + 5 = -13
2x = -18
x = -9
|𝑥| = -15 has no solution
This is because absolute value is by definition positive.
Practice on the following problems
Example 3:
1. |𝑥| = 6
Multiplication symbols: *
● or x
November 2012 Document1
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2. |𝑥| = 0
3. |𝑥| = – 8
4. |𝑥 − 4| = 1
5. |3𝑥| = 6
6. |3𝑥 − 4| = 17
7. |2𝑥 + 6| = – 3
8. |2𝑥 − 3| = 0
Answers
1. x = 6, or -6
5. x = -2 or 2
2. x = 0
3. No solution
4. x = 3 or 5
6. x = 7 or -13/3
7. No solution
8. x = 3/2
Absolute value inequalities
You can now combine your knowledge of inequalities with your understanding of absolute
value. Study these examples to understand how the two concepts work when used in the same
equation.
Example 1:
|𝑥| < 4
-4 < x < 4
This inequality indicates that x is a number which lies less than 4 units from zero
on the number line. That means that x can lie anywhere between -4 and 4.
Example 2:
|𝑥| > 4
x < -4
or x > 4
This inequality states that x lies more than 4 units from zero on the number line.
Therefore, x must be to the left of -4 or to the right of 4.
Example 3:
|6𝑥 + 7| < 5
-5 < 6x + 7 < 5
-12 < 6x < -2
-2 < x < -1/3
[The < (less than) symbol indicates that 6x + 7 is greater than -5 and less than 5.]
Example 4:
Multiplication symbols: *
|2𝑥 − 9| > 4
2x – 9 < – 4
2x < 5
5
x<2
or
2x – 9 > 4
2x > 13
13
x> 2
● or x
November 2012 Document1
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[The > (greater than) symbol indicates a distance of more than 4 units, therefore it can be
interpreted as meaning or.]
Now it’s your turn. Here are some practice problems for you.
1. |𝑥| < 3
2. |𝑦| > 12
3. |𝑥 + 4| < 9
4. |𝑥 − 2| > 6
5. |5𝑥 + 2| < 13
6. |9 − 4𝑥| > 14
7. |−5 − 7𝑥| < 30
8. |2 − 9𝑥| > 17
9. |𝑚 + 5| + 9 < 16
10. 7 – |3 − 2𝑥| > 5
Answers
1. -3<x<3
2. y < -12 or y > 12
5. -3 < x < 11/5 6. x < -5/4 or x > 23/4
9. -12 < m < 2
10. ½ < x < 5/2
3. -13 < x < 5
7. -5 < x < 25/7
4. x < -4 or x > 8
8. x < -5/3 or x > 19/9
Location and identification of points on the plane
On a number line every number can be represented by a point on the line. When you graph
equations in two variables, it is necessary to use two numbers to designate each point. Points are
graphed on a coordinate plane which is made up of two perpendicular lines. The pairs of points
are referred to as ordered pairs because it is important what order the pair of numbers is in. The
first number refers to the distance and direction from zero along the horizontal axis; the second
number refers to the vertical distance and direction from zero.
Multiplication symbols: *
● or x
November 2012 Document1
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Notice that the points (3,2) and (2,3) are not the same point; similarly, the points (-4,5) and (5,-4)
are not the same point. This is because the first coordinate indicates horizontal distance and the
second indicates vertical distance.
Distance formula
The formula 𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 is used to find the distance between any two points.
Example 1:
Find the distance between (4, -3) and (-5, 4).
First, decide which is point 1 and which is point 2.
x1 = 4, y1 = -3 and x2 = -5, y2 = 4
(note: it does not matter which point you call point 1 and which you call
point 2. The answer will come out the same.)
𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
𝑑 = √(−5 − 4)2 + (4 − (−3))2
𝑑 = √(−9)2 + (7)2
𝑑 = √81 + 49
𝑑 = √130 ≈ 11.402
Try these problems.
Find the distance between the two points.
1. (2, 6) and (-4, -2)
2. (-2, 1) and (4, 2)
Multiplication symbols: *
● or x
November 2012 Document1
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3. (-3, 1) and (6, -7)
4. (10, -7) and (8, -3)
Answers
1. 10
2. √37 ≈ 6.083
3. √145 ≈ 12.042
4. √20 ≈ 4.472
Exponential and logarithmic equations
Equations which have variables in exponents are called exponential equations. An example of
such an equation is 23x = 64. Often these types of equations can be solved by writing both sides
using the same base. If the bases are equal, then the exponents must also be equal.
Example 1:
23x = 64
23x = 26
Since the bases are the same, the exponents are equal.
3x = 6
x=2
Example 2:
54x+7 = 125
54x+7 = 53
4x + 7 = 3
4x = – 4
x = -1
Sometimes it’s impractical – or seemingly impossible – to write both sides of an equation using
the same base. In that case you can solve the equation by taking either the common logarithm or
the natural logarithm of both sides.
Example 3:
5x = 12
𝑙𝑜𝑔 5x = log 12
x log 5 = log 12
𝑙𝑜𝑔12
1.0792
x=
≈
≈ 1.544
𝑙𝑜𝑔5
0.6990
In a similar way logarithmic equations can sometimes be changed to exponential equations to
make it easier to solve them.
Example 4:
log2 x = 3
x = 23
x=8
Solve for x using any method that seems appropriate.
Multiplication symbols: *
● or x
November 2012 Document1
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1. 2x = 8
2. 32x = 9
3. 42x-3 = 64
4. 7x = 20
5. 2x = 55
6. log5x = 2
7. log3(5x + 7) = 2
8. log2 x = -5
Answers
1. 3
2. 1
3. 3
4. Approx. 1.54
5. Approx. 5.78
6. 25
7. 2/5 8. 1/32
Factorials
Products such as 8●7●6●5●4●3●2●1 are often used in higher mathematics and especially in
statistics. Because it is seen so frequently, this type of product has its own symbol, “!”. The
product 8●7●6●5●4●3●2●1 can be written 8!, which is read “8” factorial.” 3! = 3●2●1 = 6.
You may also need to perform some arithmetic operations involving factorials. To make it
possible for certain statistical formulas to work properly, 0! is defined as equal to 1.
Example 1:
Example 2:
5! = 5●4●3●2●1 = 120
10!
5!
=
10∗9∗8∗7∗6∗5∗4∗3∗2∗1
5∗4∗3∗2∗1
=
3,628,800
120
= 30,240
Evaluate the following factorials.
1. 7!
2.
3.
9!
5!
8!
4!
4. 1!
5. 0!
6.
10!
7!∗3!
Multiplication symbols: *
● or x
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7. (8-5)!
8.
7!
(7−2)!
Answers
1. 5,040
2. 3,024
3. 1, 680
4. 1
5. 1
6. 120
7. 6
8. 42
Evaluation of functions
A function is a specific, carefully defined relationship between two sets of items. For the
relationship to be considered a function, it must be a one-to-one relationship, that is each item in
the first set must correspond to exactly one item in the second set. Let’s look at some examples
that do not involve numbers.
Example 1: Each state in the United States has one and only one capital, therefore it is a oneto-one relationship. It could be considered a function.
Example 2: If you want to buy a bicycle, each bike has exactly one price. This is a one-to-one
correspondence and could be considered a function.
Example 3: If you drive your car at 50 mph for 3 hours, you will travel 150 miles. If you
drive at a rate of 60 mph for 4 hours, you will travel 240 miles. Each rates and time correlates to
a distance. Distance is said to be a function of rate and time.
The same principle applies to many algebraic relationships. Consider this set of ordered pairs:
{(-2, 5), (5, 7), (0, 1), (4, -2)}. This relations is a function because none of the pairs have the
same x coordinate. Looking at the pairs in a different form may help clarify this idea.
x
-2
5
0
4
→
→
→
→
→
y
5
7
1
-2
Now consider this set of ordered pairs: {(9, -5), (9, 5), (2, 4)}. This is a not a function because
two of the ordered pairs have the same first coordinate and different second coordinates.
x
9
→
→
2
→
y
-5
5
4
Functions used in mathematics are often given as equations. They are evaluated in the same way
that other equations are; only the notation differs.
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In a previous section we evaluated polynomials by substituting numbers for variables as follows:
For x = 3, evaluate y = x2 – 5.
y = 32 – 5
y=9–5
y=4
Since this relationship is a function, you could also write f(x) = x2 – 5. This is read “f of x equals
x squared minus five.” (Note: The use of the letter f is arbitrary. You could also use other
letters such as g(x), h(x), d(f). The meaning is the same regardless of the letters used.) If you
were being asked to evaluate the expression for x = 3, it would appear something like this:
Find f(3) when f(x) = x2 – 5.
f(x) = x2 – 5
f(3) = 32 – 5
f(3) = 9 – 5
f(3) = 4
Functions can also be evaluated by replacing the x with another letter:
Find h(t) when h(x) = 7x + 4
h(x) = 7x + 4
h(t) = 7t + 4
Evaluate these functions for the values of x given.
1. g(x) = 3x2 – 2x +1
a) g(0)
b) g(-1)
c) g(3)
2. f(x) = 5x2 + 4x
a) f(0)
b) f(3)
c) f(-1)
𝑥−4
3. f(x) = 𝑥+3
a) f(5)
b) f(-3)
c) f(4)
Answers
1. a) 1
b) 6
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c) 22
2. a)0
b) 57
c) 1
3. a) 1/8
b) undefined
c) 0
Functions can also be combined mathematically, this is, they can be added, subtracted,
multiplied or divided.
Example 1:
f(x) = x + 2 and g(x) = x2 + 1
(f+g)(x) = (x + 2) + (x2 + 1)
= x2 + x + 3
(f – g)(x) = (x + 2) – (x2 + 1)
= -x2 + x + 1
(g – f)(x) = (x2 + 1) – (x + 2)
= x2 – x – 1
f(x) ● g(x) = (x + 2) ● (x2 + 1)
= x3 + 2x2 + x + 2
𝑓(𝑥)
𝑥+2
𝑔(𝑥)
= 𝑥 2 +1
𝑔(𝑥)
𝑥 2 +1
=
𝑓(𝑥)
𝑥+2
for x ≠ -2 (division by 0 is undefined)
Okay, it’s time for you to try it. Find (f+g)(x), (f – g)(x), (g – f)(x), fg(x), ff(x), f/g(x), and g/f(x)
for each pair of functions.
1. f(x) = x2 – 1
2. f(x) = x 2
3. f(x) = x2 – 4
g(x) = 2x + 5
g(x) = √𝑥
g(x) = x2 + 2
Answers
1. f + g = x2 + 2x + 4
f – g = x2 – 2x – 6
g – f = -x2 + 2x + 6
f – g = 2x3 + 5x2 – 2x – 5
f ● f = x4 – 2x2 + 1
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𝑓
𝑥 2 −1
𝑔
2𝑥+5
5
= 2𝑥+5 , x≠-2
𝑔
= 𝑥 2 −1, x≠1 or -1
𝑓
2. f + g = x2 + √𝑥
f – g = x 2 - √𝑥
g – f = √𝑥 – x2
f – g = x 2 ● √𝑥
f ● f = x4
𝑓
𝑥2
=
𝑔
√𝑥
𝑔
√𝑥
𝑓
, x>0
= 𝑥 2 , x±0
3. f + g = 2x2 – 2
f – g = -6
g–f=6
f ● g = x4 – 2x2 – 8
f ● f = x4 – 8x2 + 16
𝑓
𝑥 2 −4
𝑔
𝑥 2 +2
= 𝑥 2 +2
𝑔
= 𝑥 2 −4, x±2 or -2
𝑓
Composition of functions
If one function’s output (answer) depends on the output or answer of another function, they are
called composite functions. One real life example of functions depending on each other is the
calculation of your state income tax: the amount of state income tax you must pay depends on your
adjusted gross income reported on your federal tax return, which in turn depends on your annual
earnings. These functions are composites.
The symbol for composition of functions an open circle: f ○ g. This is read “f composed with g,”
“the composition of f and g,” and “f circle g.”
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To compose one function with another, the variable in the first function can be replaced by the
entire second function.
Example 1:
Given that f(x) = 2x – 5
and g(x) = x2 – 3x + 8
(f ○ g)(x) = 2(x2 – 3x + 8) – 5 = 2x2 – 6x + 11
And
(g ○ f)(x) = (2x – 5)2 – 3(2x – 5) + 8 = 4x2 – 26x + 48
You can also compose two functions by replacing the variable in one function with the answer to
the other function.
Example 2:
given the functions in example 1, find (f ○ g)(7) and (g ○ f)(7)
For (f ○ g)(7)
g(7) = 72 – 3(7) + 8 = 36
then (f ○ g)(7) = 2 ●36 – 5 = 67
For (g ○f)(7):
(f)(7) = 2 ● 7 – 5 = 9
Then (g ○ f)(7) = 92 – 2 ● 9 + 8 = 62
Composition of functions can be tricky. Remember, you are replacing the variable in one function
with the other function. Practice with these functions. Find both f ○ g and g ○ f.
1. f(x) = √𝑥
and
g(x) = x – 3
2. f(x) = x5
and
Answers
1. (f ○ g)(x) = √𝑥 − 3
(g ○ f)(x) = √𝑥 -3
g(x) = 2x – 3
2. (f ○ g)(x) = (2x – 3)5
(g ○ f)(x) = 2x5 – 3
Now repeat the composition of functions for x = 12.
Answers
1. (f ○ g)(12) = 3
(g ○ f)(12) = √12 – 3 or 2√3 – 3
2. (f ○ g)(12) = 215
(g ○ f)(12) = 2(125) – 3
Trigonometry
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1. Right triangle trigonometry
If
sin 𝜃 =
, then
cos 𝜃 =
𝑎𝑑𝑗.𝑙𝑒𝑔
ℎ𝑦𝑝𝑜𝑡
𝑜𝑝𝑝.𝑙𝑒𝑔
ℎ𝑦𝑝𝑜𝑡
(soh)
(cah)
𝑜𝑝𝑝 𝑙𝑒𝑔
tan 𝜃 = 𝑎𝑑𝑗 𝑙𝑒𝑔 (toa)
ℎ𝑦𝑝𝑜𝑡
csc 𝜃 = 𝑜𝑝𝑝 𝑙𝑒𝑔
ℎ𝑦𝑝𝑜𝑡
sec 𝜃 = 𝑎𝑑𝑗 𝑙𝑒𝑔
𝑎𝑑𝑗 𝑙𝑒𝑔
cot 𝜃 = 𝑜𝑝𝑝 𝑙𝑒𝑔
2. Special right triangle
a)
So
√2
2
√2
= 2
sin 45 ° =
csc 45 °
tan 45 ° = 1
And so on
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b)
So
1
sin 30° = 2
1
cos 60° = 2
And so on
The triangles in (a) and (b) are based on geometry relationships.
3) The triangles in #2 allow us to extend the concepts to other angles. We’ll use the unit
circle [radius 1, center at (0,0)]
a)
Draw a 135o angle. The reference angle is 45o (measure from the terminal side of 135o to
√2
√2
the x axis). So, the sin 135o is the same as sin 45o ( 2 ) but cos 135o is –cos 45o (- 2 ) since
in QII, the x number is negative.
b) This concept allows for exact values of any trig function whose reference angle is 30 ͦ,
45 ͦ, or 60 ͦ.
c) We can also determine the value of any trig function whose terminal side falls on an
axis.
sin 90° = 1
cos (−90 ͦ°) = 0
tan 0 ° = 0
4) Radian measure of angles
a) So far, angles have been given in degrees. A much more common measure for angles
is radian measure. A complete revolution around our circle is 360 ͦ. A radian is the
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ratio of the arc length to the radius. In our circle the circumference is 2π. (C=2πr
2πr
where r = 1). The size of the circle doesn’t matter as the ratio 𝑟 is always 2 π.
b) Thus
360 ͦ = 2𝜋R
Or
180 ͦ = πR
c) The superscript R will be dropped. Any angle measure given in terms of π or just a
number will be assumed to be a radian. The degree symbol must continue to be used.
d) Converting degrees to radians and vice versa (write a proportion and solve).
Example 1) 150 ͦ = _________R
180 ͦ
150 ͦ
Solution:
= 𝑥
𝜋
180x = 150π
x=
5𝜋
6
Example 2) 3R = ___________ͦ
Solution:
180 ͦ
𝜋
𝑥
=3
πx = 540
x=
540
𝜋
x = 171.9 ͦ
Problems:
1) Solve-
b=
c=
B=
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2) Solve
m
A=
m
B=
3) Given sin 𝜃 =
4
5
and θ is in QII: Find the value of the other five trig functions.
√3
4) Given tan 𝜃 = 5 and θ is in QIII: Find the value of the other five trig functions. (all
answers in simplest radical form).
5) Evaluate the following (all trig evaluations are exact values).
a) 120 ͦ = _____R
g) tan(−45 ͦ ) =
b) 5R = ______ ͦ
h) csc 270 ͦ =
c) 250 ͦ = ____ R
i) sec 6 =
d)
3𝜋
4
= ______ ͦ
e) 750 ͦ = _____R
f) sin 315° =
Answers
1) b = 13.74
j) csc(−𝜋) =
k) cos(
B = 70 ͦ
cot 𝜃 =
4) sin 𝜃 =
2) m
A = 53.1 ͦ
m
B = 36.9 ͦ
cos 𝜃 =
csc 𝜃 =
3) cos 𝜃 =
tan 𝜃 =
−3
sec 𝜃 =
5
−4
3
cot 𝜃 =
5
csc 𝜃= 4
Multiplication symbols: *
4𝜋
3
)=
l) cot 0 =
sec 𝜃 =
c = 14.62
M
𝜋
5) a)
−5
3
−3
4
−√21
14
−5√7
14
−2√21
3
−2√7
5
5√3
3
2𝜋
3
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b)
c)
900 °
𝜋
g) -1
= 286.5 ͦ
h) -1
25𝜋
18
i)
d) 135 ͦ
e)
f)
2√3
3
j) undefined
25𝜋
6
k)
−√2
2
−1
2
l) undefined
5) Non-right triangle trigonometry
a) Two relationships allow us to solve triangles that are not right triangles.
𝑎
𝑏
𝑐
Sin Law
= sin 𝐵 = sin 𝐶
sin 𝐴
(use in AAS or SSA situations)
a2 = b2 + c2 - 2bc cos 𝐴
b2 = a2 + c2 – 2ac cos 𝐵
c2 = a2 + b2 – 2ab cos 𝐶
(use in SSS or SAS situations)
Cos Law
b) Example:
C = 180 ͦ - (45 ͦ+ 35 ͦ) = 100 ͦ
m
Now use Sin Law since this is an AAS situation
10
sin 45 °
𝑏
= sin 35 °
b = 8.11
10
sin 45 °
𝑐
= sin 100 °
c = 13.93
c) Example
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m
m
m
A= ____________
B=_____________
C=_____________
Since this is SSS, we’ll use Cos Law
402 = 202 + 302 – 2(20)(30) cos 𝐶
1600 = 400 + 900 – 1200 cos 𝐶
1600 = 1300 -1200 cos 𝐶
300 = -1200 cos 𝐶
1
- = cos 𝐶
4
1
cos -1 (- 4) = C
104.5 ͦ = C
Now you can use Sin Law or Cos Law to get the next angle.
sin 104.5 °
40
=
sin 𝐵
20
29 ͦ = B
m
m
A = 180 – (104.5 ͦ + 29 ͦ)
A = 46.5 ͦ
Practice
1) Solve (nearest whole number)
2) Solve (nearest whole number)
3) Solve (nearest whole number)
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4) The shape below is formed by steel girders.
a) Find a (round to hundredths)
b) Find b (round to hundredths)
5) In the figure below XY is the distance across a lake. (round to nearest whole number) find
XY.
6) Two holes are drilled in a circular metal disc whose center is at point C. Find “d”, the centerto-center distance between the two holes. Round to hundredths.
Answers
1) m X = 22 ͦ
m Y = 48 ͦ
y = 40
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2) d = 27
m E = 38 ͦ
m F = 100 ͦ
3) a = 18
c=7
m C = 19 ͦ
4) a = 2.02m
b = 3.39m
5) XY = 1548 m
6) d = 4.97cm
Graphing non-linear functions
1) Equations of the form y = ax2 + bx + c (quadratics)
a) Quadratics graph into parabolas.
b) The parabola opens up if a>0
c) The parabola opens down if a<0
−𝑏
d) The x-coordinate of the vertex occurs at 2𝑎
Example: Graph y = 2x2 + 4x – 3
Step 1: opens up (2>0)
−4
−4
Step 2: x coordinate of vertex is 2(2) = 4 = - 1
Step 3: y coordinate of vertex (substitute -1 into the equation for x and solve for y)
y = 2(-1)2 + 4(-1) – 3
y=2–4–3
y=-5
vertex (-1, -5)
Step 4: find another ordered pair
x
0
y
-3
Step 5: graph
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Example: Graph f(x) = -3x2 + 4
Step 1: opens down (-3<0)
0
Step 2: x coordinate of vertex is 2(−3) = 0
Step 3: y coordinate of vertex is 4 vertex (0, 4)
Step 4: Find another ordered pair
x y
1 1
Step 5: Graph
2) Absolute value functions
These will be graphed by making a table of ordered pair. You will have to pick enough
until you “see” the V
Example: y = │x│ + 3
x
-1
-2
0
1
y
4
5
3
4
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Example: y = │5 - x│
x
3
4
5
6
y
2
1
0
-1
Example: y = 3 - │x│
x
-2
-1
0
1
2
y
1
2
3
2
1
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3) Exponential and log functions
Example: y = 3x
Step 1: make a table of ordered pairs
x y
-2 1/9
-1 1/3
0 1
1 3
2 9
Note: this graph will never cross the x-axis since 3 to any power is positive
Example: y = - (2)x
Step 1: make a table of ordered pairs
x
-2
-1
0
1
y
-1/4
-1/2
-1
-2
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2
-4
Note: this graph will never cross the x axis since y is always negative
1
Example: g (x) = (4)x
Step 1:
x y
-1 4
0 1
1 ¼
2 1/16
Step 2:
In exponential functions y = ax, a is always > 0
Example: h(x) = log4x
Recall, any log has an equivalent exponential form.
h(x) = log4x ↔ y = log4x ↔ 4y = x
Step 1: To graph this make a table of ordered pairs picking values of y first.
x
1/4
1
4
y
-1
0
1
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4) Trig Functions
a) The graphs of all of the trig functions are periodic—that is the graph repeats itself
over a certain interval. sin, cos, sec, and csc have periods of 2π. tan and cot have
periods of π.
b) Several of the basic graphs are shown below with corresponding tables of ordered
pairs.
y = sin x
x
0
π/2
π
3π/2
2π
y
0
1
0
-1
0
Note: The graph repeats itself after the 2π interval.
y = tan x
x
-π/2
-π/4
0
Multiplication symbols: *
y
Undefined
-1
0
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π/4
π/2
1
undefined
The dashed lines here indicate vertical asymptotes and two periods are graphed.
y = sec x
x
-π/2
0
π/2
π
3π/2
π/3
y
Undefined
1
Undefined
-1
Undefined
2
Again, vertical asymptotes are pictured by dashed vertical lines.
c) By placing numbers in front of the trig function or in front of the x, it is possible to
stretch or shrink these graphs both vertically and horizontally.
Example: y = 3 cos x
x
0
Multiplication symbols: *
y
3
● or x
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π/2
π
3π/2
2π
0
-3
0
3
The three caused a vertical stretch.
A coeff of ½ would cause a vertical shrink.
A coeff of -5 would cause a vertical stretch and a reflection over the x-axis.
Example: y = 2 sin 2x
We now know that the coeff of 2 will cause a vertical stretch. The 2 in front of the x will
cause a horizontal (since it’s on the x) shrink. Any coeff on the x causes the opposite effect of
what you see.
x
y
0
0
π/4 2
π/2 0
3π/4 -2
π
0
Note: the period is now π (original period ÷ by 2)
5) Inequalities
The same techniques discussed earlier will apply with the addition of a test point to
determine shading.
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a) Example: y ≤ 2x + 3 (this is a line with slope ½ and y intercept = 3)
Now a test point (any point not on the line) will determine shading. Use (0,0)
0≤ 0 + 3 is true so shade over (0,0)
Answer:
b) Example: 2x + 3y > 6 (again a linear boundary).
Use the intercept method to graph.
x
y
0
2
3
0
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The boundary is dashed because there is no equal sign.
Again a test point of (0,0).
0+0>6 is false, so shade away from (0,0)
Answer:
c) y>x2 + 2x – 4
Since this is a quadratic, we need its vertex. We know it opens up.
−2
V(2
V (-1,
y = (-1)2 + 2(-1) -4
y=1–2–4
y = -5
V(-1, -5)
x
y
0
-4
Test (0,0)
0>0+0-4
0>-4 is T so shade towards (0,0)
Answer:
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d) y < │x-2│
x
y
0
2
1
1
2
0
3
1
--until you see a V
Test (0,0)
0< │0-2│
0< 2 is true, so shade towards (0,0)
Answer:
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Practice
Graph the following
1) y = │x-6│
7) y = 5x
1
1
2
4
2) y = 3x2 + 6x – 7
8) y = sin x
3) y = cot x
9) y = -2(x+2)2+ 5
4) y = log2 x
10) y < - │x│
5) y = 5 cos x
1
6) x + 2 𝑦 ≤ 4
Answers:
1)
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2)
3)
4)
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5)
6)
7)
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8)
9)
10)
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