Download MTTC Professional Readiness Review Algebra I Solutions p. MTTC

MTTC Professional Readiness Examination
Algebra I Review
Solutions to Practice Problems
Section I. Algebraic Expressions and Functions
4
4
 3(4) 2 
 3(16)  2  28  46
2
2
1. a)
2
b) 3
1
4

2
2
3
 3  3 2 2
1
3
4
 8  4  12  4
3 3
3
2 2
3
2

2
2
4
4
  3  3 4  3  34  3  4  2  1  4  2  4  4
 1
3 1 3
1 3 3 4 3
1
9
9

4
2. a) f ( x)  3(5 2 )  4  5  1  3  25  4  5  1  3  25  75  20  1  56
b) f  g x = (3x2 – 4x + 1)(x – 3) = 3x2x – 4xx + 1x + 3x2( –3) – 4x( –3) + 1( –3)
= 3x3 – 4x2 + x – 9x2 + 12x – 3 = 3x3 – 13x2 + 13x – 3
c) f  g x = f (g(x)) = f(x – 3) = 3(x – 3)2 – 4(x – 3) + 1 = 3(x2 – 6x + 9) – 4(x – 3) + 1
= 3x2 – 18x + 27 – 4x + 12 + 1 = 3x2 – 22x + 40
Note: You compute (x – 3)2 by (x – 3)2 = (x – 3)(x – 3) = x2 – 3x – 3x + (– 3)( – 3) =
x2 – 6x + 9
d) g1(x) = x + 3 (adding 3 is the opposite of subtracting 3)
3. Substitute 10 for every instance of y in the equation.
102 + 2(10) + 5
100 + 20 + 5 125 25(5) 25
1
=
=
=
=
=8
10 + 5
15
15
3(5)
3
3
4. a) Distribute the negative sign through the parentheses, and then combine like terms.
4𝑎 + 𝑎 − (5𝑎 − 2𝑎) = 4𝑎 + 𝑎 − 5𝑎 + 2𝑎 = 5𝑎 − 5𝑎 + 2𝑎 = 𝟐𝒂
b) Distribute 2𝑥 3 and 5𝑥𝑦 through the parentheses, and then combine like terms
2𝑥 3 (3𝑦 2 − 4𝑥𝑦) + 5𝑥𝑦(𝑥 3 − 2𝑥 2 𝑦 + 4) = 6𝑥 3 𝑦 2 − 8𝑥 4 𝑦 + 5𝑥 4 𝑦 − 10𝑥 3 𝑦 2 + 20𝑥𝑦
= −3𝑥 4 𝑦 − 4𝑥 3 𝑦 2 + 20𝑥𝑦
5 a) Combine like terms. Remember, the order of the variables does not matter, but the value of
the exponents does matter. For example, 3𝑏𝑎2 = 3𝑎2 𝑏 so 3𝑏𝑎2 and 2𝑎2 𝑏 are like terms, where
3𝑎𝑏 2 and 2𝑎2 𝑏 would not be.
(2𝑎2 + 3𝑎2 𝑏 − 2𝑏) + (−𝑎2 + 2𝑎2 𝑏 + 3𝑏) =
remove parentheses 2𝑎2 + 3𝑎2 𝑏 − 2𝑏 − 𝑎2 + 2𝑎2 𝑏 + 3𝑏 =
combine like terms 2𝑎2 − 𝑎2 + 3𝑎2 𝑏 + 2𝑎2 𝑏 − 2𝑏 + 3𝑏 = 𝑎2 + 5𝑎2 𝑏 + 𝑏
b) Distribute each of the terms in the trinomial through each of the terms in the binomial, and
then combine like terms.
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Algebra I Solutions p. 2
2
(3𝑥 + 7𝑥 − 2)(𝑥 + 6) = 3𝑥3 + 18𝑥2 + 7𝑥2 + 42𝑥 − 2𝑥 − 12 = 3𝑥 3 + 25𝑥 2 + 40𝑥 − 12
c) Write the square as a product of two identical terms, then distribute (FOIL), and finally
combine like terms.
(𝑎 + 3𝑏)2 = (𝑎 + 3𝑏)(𝑎 + 3𝑏) = 𝑎2 + 3𝑎𝑏 + 3𝑎𝑏 + 9𝑏2 = 𝑎2 + 6𝑎𝑏 + 9𝑏2
d) Factor out greatest common factor first, then cancel the common factor with the denominator
10 x 2 y  6 xy 2  2 xy 2 xy(5 x  3 y  1)

 5x  3 y  1
2 xy
2 xy
 2 y  6  2( y  3) ( y  3)
e)


 y 3
2
2
1
x 2  2 x  15 ( x  5)( x  3) ( x  5)( x  3) ( x  5)
f)



 x5
3 x
3 x
x3
1
x  2 12 x 2 x  212 x 2 x  2  4  3  x  x 4 x
g)




 4x
3x x  2
3x( x  2)
3  x  ( x  2)
1
h)
1 y y 1 1 y 6
 (1  y ) 6
 ( y  1) 6
1








 1
6
6
6 y 1
6
y 1
6
y 1 1
x 2  4x  4
6
( x  2)( x  2)
6
( x  2)( x  2)2  3
3
 2




2
4 x( x  2)
( x  2)( x  2) 2  2 x( x  2)( x  2)( x  2) 2 x( x  2)
4 x  8x x  4
5
5
( x  1)
5( x  1)
5( x  1)
5
j) 2
 ( x  1)  2

 2


1
x  6x  7
x  6x  7
x  6 x  7 ( x  7)( x  1) ( x  7)
3
3
24 y x
20 xy
24 y x 3( y  2) 6  4  y  y  y  x 3( y  2)
y ( y  2)






k)
2
2
63 x  x  y 5 4  x  y
18 x y 3( y  2) 18 x y 20 xy
5x 2
i)
4
1
x6
4( x  2)
x 1
x6
4( x  2)  x  ( x  6)






x x  2 x( x  2) x( x  2) x( x  2) x( x  2)
x( x  2)
4 x  8  x  x  6)
4 x  14


x( x  2)
x( x  2)
x
x2
x
x2
x
(1)( x  2)
x
( x  2)
m)







x  5 15  3x x  5  3(5  x) x  5 3(5  x)
x  5 3(5  x)
x
( x  2)
3x
( x  2) 3x  x  2
4x  2






x  5 3( x  5) 3( x  5) 3( x  5)
3( x  5)
3( x  5)
l)
n)
7
7
1
7
1  ( x  3) 7  x  3 x  10
1 
 



x3
x  3 1 x  3 1  ( x  3)
x3
x3
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Algebra I Solutions p. 3
6. a) There is no common factor other than 1 among all the terms . Since the coefficient of 𝑥2 is
1, we can factor the polynomial by looking at the factors of 6. 6 factors as 16 or 23. The pair
that adds to 5 (the coefficient of the middle term) is 2 and 3. So:
𝑥2 + 5𝑥 + 6 = (𝑥 + 2)(𝑥 + 3)
b) Again, there is no common factor among all of the terms. Since the leading coefficient is 1,
it will be easier if we factor out a 1 before we start.
−𝑥2 + 3𝑥 + 18 = −1(𝑥2 − 3𝑥 − 18)
The two factors of 18 that add to 3 are 6 and 3. So
−1(𝑥 2 − 3𝑥 − 18) = −(𝑥 − 6)(𝑥 + 3)
c) In contrast to the two previous problems, the coefficient of the 𝑥 2 term is not 1. The
coefficients of 𝑥 must be factors of 6 (1 and 6) or (2 and 3), the last terms must be factors of 2 (1
and 2). We use the method of guess and check. The different possible combinations that result in
a first term of 6𝑥 2 and a last term of 2 are:
(6𝑥 + 1)(𝑥 + 2)
(6𝑥 + 2)(𝑥 + 1)
(3𝑥 + 2)(2𝑥 + 1)
(3𝑥 + 1)(2𝑥 + 2)
Checking each by multiplication shows that (3𝑥 + 2)(2𝑥 + 1) is the answer.
7. a) Does define a function. The domain is {1, 2, 3, 4, 5, 6}, the range is {2, 1, 5, 6, 8}, f(2) = 6
b) Does not define a function because f(2) cannot be 3.5 and 3. So this does not define a unique value to
each value in the x row.
c) Does define a function. The domain is {1, 3, 4, 5, 6, 8}, the range is {.5, 1.5, 2, 2.5, 3, 4}. It is not
possible to find f(2) because there is no assignment for 2. (2 is not in the domain.)
8. f(x) – g(x) = (5x4 – 6x3 + x – 7) – (–2x4 + 3x3 + 4x2 + 7x – 3)
= 5x4 – 6x3 + x – 7 + 2x4 – 3x3 – 4x2 – 7x + 3
= 5x4 + 2x4 – 6x3 – 3x3 – 4x2 + x – 7x – 7 + 3
= 7x4– 9x3– 4x2 – 6x – 4
9. x =  3 is not in the domain of f(x) because if x = 3, the denominator , x + 3, will be 0. Dividing by 0
is not possible.
10. a) Since f(2) = 4, f 1(4) = 2
b) g(8) = 4, and f(4) = 16, so f  g 8 = f(g(8)) = f(4) = 16
c) For g  f x  , we put numbers into f first, and then put what we get out into g. You can make
the table by lining up the tables for f and g.
x
1
0
1 2 3 4
f(x)
0.5 1
2 4 8 16
g(f(x)) 0.25 0.5 1 2 4 8
So the table for g  f x  looks like:
x
1
0
1 2 3 4
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Algebra I Solutions p. 4
g  f x  0.25 0.5 1 2 4 8
Notice that we can only do this because the domain (the inputs) for g is the same as the range
(the outputs) of f.
d) f 1(x) will take all the outputs from f, and send them back to the number they came from.
x
0.5 1 2 4
f 1(x) 1 0 1 2
8 16
3 4
Section II. Equations and Inequalities
1.
b) 5x – 7 = 4x + 8
a) 3x + 4 = 46
3x + 4 – 4 = 46 – 4
5x – 7 + 7 = 4x + 8 + 7
3x
5x = 4x + 15
= 42
3 x 42

3
3
5x – 4x = 4x – 4x + 15
x = 14
x = 15
c) 6x + 8 = 3x + 2
6x + 8 – 8 = 3x + 2 – 8
6x = 3x – 6
6x – 3x = 3x – 3x – 6
3x = – 6
d)
3x  6

3
3
x=–2
x
1 = 3
2
x
 1– 1 = 3 – 1
2
x
=2
2
x
 2  2 2
2
x=4
2.
a) A = bh
A bh

b
b
A
h
b
c) I = Prt
I
Prt

Pr
Pr
b)
bh
2
 bh 
2 A  2 
 2 
A
2A = bh
2 A bh

b
b
2A
h
b
I
t
Pr
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Algebra I Solutions p. 5
P  2 w 2l

2
2
P  2w
l
2
d) P = 2l + 2w
P – 2w = 2l + 2w – 2w
P – 2w = 2l
3. a) (Method of substitution) Put y = x + 8 into the first equation for y :
x + y = 10  x + x + 8 = 10
Simplify and solve for x: 2x + 8 = 10
2x + 8 – 8 = 10 – 8
2x = 2  x = 1
Put x = 1 into y = x + 8 to find y: y = 1 + 8 = 9
or:
(Method of elimination) rewrite y = x + 8 as –x + y = 8 and add equations to eliminate x
–x + y = 8
x + y = 10
2y = 18 
2 y 18
 y=9
2
2
Put y = 9 into either original equation and solve for x:
x + 9 = 10  x = 10 – 9 = 1
answer: x = 1, y = 9
b) (method of elimination) Multiply the second equation by –1 and add to eliminate x
2x + 3y = – 2
–2x + y = –10
4y = –12 
4 y  12

 y = –3
4
4
Put y = –3 into either of the original equations to solve for x. 2x + 3(–3) = – 2 
2x – 9 = –2  2x = –2 + 9 = 7 
answer: x 
2x 7
7
  x
2
2
2
7
, y = –3
2
c) (method of substitution) Put y = 2x – 5 into the second equation and simplify:
3y – x = 5  3(2x – 5) – x = 5  6x – 15 – x = 5  5x – 15 = 5 Solve for x:
5x = 5 + 15 = 20 
5 x 20

x=4
5
5
put x = 4 into y = 2x – 5: y = 2(4) – 5 = 8 – 5 = 3
or:
(method of elimination)
y – 2x = – 5
3y – x = 5
You can multiply the first equation by –3 to cancel y or the second equation by –2 to cancel x. It doesn’t
matter which. We’ll multiply the second equation by –2.
y – 2x = – 5
– 6y + 2x = – 10
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= –15 
–5y
Algebra I Solutions p. 6
 5 y  15

y=3
5
5
put y = 3 into either of the original equations and solve for x:
3 = 2x – 5  3 + 5 = 2x – 5 + 5
8 = 2x  x = 4
answer: x = 4, y = 3
d) Using method of elimination: Multiply the first equation by 3 and the second equation by – 2 and add
to cancel x:
6x – 9y = 15
– 6x – 10y = 4
– 19y = 19 
 19 y
19

y=–1
 19
 19
Put y = – 1 into either of the original equations and solve for x:
2x – 3(–1) = 5  2x – (–3) = 5
2x + 3 = 5  2x = 5 – 3 = 2  x = 1
answer: x = 1, y = – 1
(Note: You could multiply the first equation by 5 and the second equation by 3 to cancel y instead.)
2(𝑥 + 5) + 3𝑥 = 20 − 5𝑥
distribute the 2: 2𝑥 + 10 + 3𝑥 = 20 − 5𝑥
Combine like terms: 5𝑥 + 10 = 20 − 5𝑥
Add 5x to and subtract 10 from both sides: 10𝑥 = 10
10𝑥
10
Divide by 10 and simplify: 10 = 10  𝑥 = 𝟏
4. a)
b)
1
2
5
2𝑥 + 3 𝑥 − 5 = 𝑥 + 2
Multiply both sides by the least common denominator (30)
1
2
3
5
5
30 (2𝑥 + 𝑥 − ) = 30 (𝑥 + )
Simplify
30 × 2𝑥 +
30
3
𝑥−
2
30×2
5
= 30𝑥 +
30×5
2
60𝑥 + 10𝑥 − 12 = 30𝑥 + 75
70𝑥 − 12 = 30𝑥 + 75
Add 12 to and subtract 30𝑥 from both sides
40𝑥 = 87
Divide both sides by 40
𝑥=
𝟖𝟕
𝟒𝟎
c) (𝑥 − 3)2 = 25
Take the square root of both sides
𝑥 − 3 = ±√25
𝑥 − 3 = 5 or
𝑥 − 3 = −5
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Algebra I Solutions p. 7
Solve the two equations 𝑥 = 𝟖
or 𝑥 = −𝟐
4
−5
𝑥
d)
3
= 𝑥−3
Multiply by the least common denominator, 𝑥(𝑥 − 3)
4𝑥(𝑥−3)
− 5𝑥(𝑥 −
𝑥
3) =
3𝑥(𝑥−3)
𝑥−3
Cancel denominators and simplify: 4(𝑥 − 3) − 5𝑥(𝑥 − 3) = 3𝑥
4𝑥 − 12 − 5𝑥 2 + 15𝑥 = 3𝑥
−5𝑥 2 + 19𝑥 − 12 = 3𝑥
Subtract 3𝑥 from both sides, multiply by – 1: 5𝑥2 − 16𝑥 + 12 = 0
Factor: (5𝑥 − 6)(𝑥 − 2) = 0
Set each factor equal to 0, and solve the two equations.
𝟔
𝑥 = 𝟓 or 𝑥 = 𝟐
𝑥
1
1
e)
+ 𝑥+5 = 𝑥+11
𝑥+11
Multiply by the least common denominator of (𝑥 + 5)(𝑥 + 11)
𝑥(𝑥 + 5)(𝑥 + 11) (𝑥 + 5)(𝑥 + 11) (𝑥 + 5)(𝑥 + 11)
+
=
𝑥 + 11
𝑥+5
𝑥 + 11
𝑥(𝑥 + 5) + (𝑥 + 11) = 𝑥 + 5
Simplify, subtract 𝑥 and subtract 5 from both sides.
𝑥2 + 5𝑥 + 𝑥 + 11 = 𝑥 + 5
𝑥2 + 5𝑥 + 6 = 0
Factor
(𝑥 + 2)(𝑥 + 3) = 0
Set each factor equal to 0, and solve the two equations.
𝑥 = −𝟐 or 𝑥 = −𝟑
f)
√𝑥 + 5 − 3 = 4
Isolate the radical by adding 3 to both sides
√𝑥 + 5 = 7
2
Square both sides (√𝑥 + 5) = 72
𝑥 + 5 = 49
Solve the equation by subtracting 5 from both sides
𝑥 = 44
5. a)
x2 – 12 = 4x
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Algebra I Solutions p. 8
Subtract 4x from both sides: x2 – 4x – 12 = 0
Solving by factoring: (x – 6)(x + 2) = 0
x – 6 = 0 or x + 2 = 0
x = 6 or x = – 2
Alternate method: Use quadratic formula: a = 1, b = –4, c = –12
x
 b  b 2  4ac

2a
 (4)  (4) 2  4(1)( 12) 4  16  (48) 4  16  18



2(1)
2
2
4  64 4  8

2
2
4  8 12
48  4

 6 or x =

 2
So x =
2
2
2
2
b)
y2 – 49 = 0
factoring: (y – 7)(y + 7) = 0
y – 7 = 0 or y + 7 = 0
y = 7 or y = 7
This one we can also solve by taking square roots. First add 49 to both sides:
y2 = 49
Then take square roots of both sides, remembering to get the positive and negative root.
y 2   49  y = 7, y = 7
( you could also use the quadratic formula with a = 1, b = 0, c = –49)
c)
2x2 – 7x –15 = 0
Solving by factoring: (x – 5)(2x + 3) = 0
x – 5 = 0 or 2x + 3 = 0
x = 5 or 2x = – 3
3
So x = 5 or x = 
2
Alternate method: Use quadratic formula: a = 2, b = –7, c = –15
 (7)  (7) 2  4(2)( 15) 7  49  (120) 7  49  120
 b  b 2  4ac
x




2a
2(2)
4
4
7  169 7  13

4
4
7  13 20
7  13  6
3

 5 or x =


So x =
4
4
4
4
2
6. a)
3z + 1 > 7
Subtract 1 from both sides: 3z > 6
Divide by 3, reverse the arrow: z <  2
Answer as set: {z: z <  2} As number line:
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Algebra I Solutions p. 9
5 + 2x – (x + 3) < 3x – 4
Simplify: 5 + 2x – x – 3 < 3x – 4
2 + x < 3x – 4
Add 4 and subtract x from both sides: 6 < 2x
Divide both sides by 2: 3 < x
Answer as set: {x: x > 3} As number line:
b)
c)
5  4x  1  17
Subtract 1 from all 3 parts: 4  4x  16
Divide all 3 parts by 4: 1 < x < 4
Answer as set: {x: 1 < x < 4} As number line:
x3
0
5
Multiply all 3 parts by -5, reverse the arrows: 5 > x  3 > 0
Add 3 to all three parts: 8 > x > 3
Answer as set: {x: 3 < x < 8} As number line:
1 
d)
7. Use distance = ratetime. d Fred  rFred  t Fred and d T ed  rT ed  t T ed . And we also know that
rFred  4 , rT ed  6 , if we call t T ed  t , then t Fred = t + 15 minutes, which is t +
1
4
hour.
And we also have d Fred  d T ed  2 , because when they meet, their combined distance will cover
the whole 2 mile trail.
Putting all these together we get: 4(t + 14 ) + 6t = 2
Simplifying and solving:
4t + 1 + 6t = 2
10t + 1 = 2
10t = 1
t = 1/10 of an hour or 6 minutes
8. Let x and y be the two numbers. Write an equation for the product of the two numbers and a
second equation for the sum. Then solve the system of two equations by substitution.
𝑥𝑦 = 56
Equation for product
𝑥 + 𝑦 = 18
Equation for sum
𝑥 = 18 − 𝑦
Solve sum equation for x
(18 − 𝑦)𝑦 = 56
Substitute the expression for x into the product equation
2
18𝑦 − 𝑦 = 56
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𝑦2 − 18𝑦 + 56 = 0
(𝑦 − 14)(𝑦 − 4) = 0 Solve by factoring
𝑦 = 14 or 4 If 𝑦 is 14, 𝑥 = 18 − 14 = 4. If 𝑦 is 4, 𝑥 = 18 − 4 = 14.
Either way, the two numbers are 4 and 14
9. a) The sum of two numbers is x + y. The difference of two numbers is x – y . So the system of
equations is: x + y = 115, and x – y = 21 Add the equations to cancel y: x + y = 115
x – y = 21
2x = 136  x = 68
Put x = 68 into either of the original and solve for y: 68 + y = 115  y = 115 – 68 = 47
Answer: 68 and 47.
b) The system is: x + y = 26.4 and x = 5y.
Solve by substitution: 5y + y = 26.4  6y = 26.4  y =
26.4
= 4.4  x = 5(4.4) = 22
6
Answer: 4.4 and 22
10. a) The profit that the company makes when they make and sell 40 units is the amount you get from
putting in 40 for x. P(40) = –3600 + 400(40) – 4(40)2 = –3600 + 400(40) – 4(1600) = –3600 + 16000 –
6400 = $6000
b) Putting in 100 for x we get P(100) = –3600 + 400(100) – 4(100)2 = –3600 + 400(100) – 4(10000) =
–3600 + 40000 –40000 = –3600. Since the profit is negative, this means the company loses money if they
make and sell 100 units.
11. Distance = rate  time, so time =
distance
rate
350
= 7 hours
50
350
Time for John is:
= 5 hours
70
Time for Joe is:
Since John left 1 hour after Joe, he will get there 6 hours after Joe left, which is an hour before Joe
arrives.
12. Let x = the number of pounds of cheap coffee used by the grocer
Let y = the number of pounds of gourmet coffee used by the grocer.
Then x  y  30 , since there are 30 lb. of house blend.
Also, the cost of the cheap coffee plus the cost of the gourmet coffee should give the cost of 30 lb of
house blend: 3.5 x  8 y  4.25(30) .
Now solve the equations simultaneously. (We used elimination.)
x  y  30
  8 x  8 y  240 

  4.5 x  112.5  x  25
3.5 x  8 y  127.5 3.5 x  8 y  127.5
Since x  y  30 , y  30  x  30  25  5 .
Answer: the grocer used 25 pounds of cheap coffee and 5 pounds of gourmet coffee.
13. This is a distance  rate  time problem. You can set up a chart to help you organize your thoughts.
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Let t = the number of hours each car is traveling. The time is the same for each car.
Rate Time Distance
Car 1 40
t
40t
Car 2 60
t
60t
Since they are going in opposite directions, the sum of their distances will be 10 when they are 10 miles
apart. So the equation is: 40t  60t  10  100t  10  t 
1
1
60minuts
hr, Or
hr 
 6 minutes.
10
10
1hr
14. Let x be the number of hours it takes the two pigs working together to build a house.
1
of a house in 1 hour.
4
1
Since his brother builds1 house in 3 hours, the brother builds of a house in 1 hour.
3
1
Since together they build 1 house in x hours, together they build
of a house in 1 hour.
x
1 1 1
12
Equation:   . Multiplying each term by 12x gives: 4 x  3x  12  7 x  12  x 
hr.
3 4 x
7
Since the little pig builds 1 house in 4 hours, he builds
(or approximately 1 hour and 43 minutes)
Section III. Graphs
1. a)
Finding the x-intercept: set y = 0: 2x + 5(0) = 10
2x = 10
Divide by 2: x = 5
Finding the y-intercept: set x = 0: 2(0) + 5y = 10
5y = 10
Divide by 5: y = 2
Intercepts are: (5, 0) and (0, 2)
Graph:
b) If x – 3 = 0, x = 3 for all values of y. So this is a special case. This is a vertical line. The xintercept is (3, 0) but there is no y-intercept because x is never 0. The slope of a vertical line is
undefined. (because there is no change in x, so the
change in x is 0, and you can’t divide by 0)
Graph:
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b
2.
a) slope =
2
2
= 1 y-intercept = (0, – 2)
b) slope = 
1
4
y-intercept = (0, – 1)
equation: y =  14 x – 1
equation: y = x – 2
c) slope = 12 = 2 y-intercept = (0, 0)
equation: y = 2x
d) slope = 0 y-intercept = (0, 3)
equation: y = 3
3. a) ( 2, 1), (1,  3)
b) Using the distance formula it is:
d
x2  x1 2   y2  y1 2

1  (2)2   3  12

32   42
 9  16  25  5
c) Using the midpoint formula, the midpoint is:
 x1  x2 y1  y 2    2  1 1  (3)    1  2 
,
,


 ,
   12 ,1
2
2
2
2
2
2
 


 
4. Slope is:
m
y2  y1
72
5
Since the slope is positive, the line is rising.


x2  x1 5  (3) 8
5. First find the slope.
m
y2  y1
3 1
2
1



x2  x1  2  4  6
3
Then y =  13 x + b. Use either point to find b. Using x = 4 when y = 1 to solve for b we get:
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1 =  13 (4) + b  1 =  43 + b
Multiply both sides by 3: 3(1) =  43 (3)+ b(3)
3 =  4+ 3b
Add 4 to both sides: 7 = 3b
Divide both sides by 3: 73 = b
 Equation: y =  13 x +
7
3
6. Since this line has an undefined slope, it must be vertical. So it has the form x = c, and the
x-coordinate is the same for all the points on this line. Because it contains the point (5,3), the
x-coordinate must be 5 for all the points on the line. So the line is x = 5.
7. If the line is parallel to the line y = 2x + 5 it will have the same slope, which is 2. If it passes through
the point (0, 4), the y-intercept is 4. So the slope-intercept form is: y = 2x + 4
8. a) The domain is all the x values used. If the graph projected a shadow on the horizontal axis, which
values would it cover: {x: 3 < x < 5}
b) The range is all the y values used. If the graph projected a shadow on the vertical axis, which values
would it cover: {y: 2 < y < 3}
Domain: if the graph projected a shadow
onto the x-axis it would extend from 3 to 5.
Range: If the graph projected a shadow onto the
y-axis it would extend from 2 to 3
c) {x: 4 < x < 4} d) {y: 4 < y < 4} e) The graph goes through the point (3, 2), so f(3) = 2.
f) The graph crosses the x-axis at 2 and 1, so the x-intercepts are (2, 0) and (1, 0).
g) (0, 1)
h) f(4) = 3, because the graph goes through the point (4, 3). And g(4) = 4.
So (f + g)(4) = f(4) + g(4) = 3 + 4 = 3  4 = 1.
i) g(1) = 3, and f(1) = 2. So (g  f)( 1) = g(1)  f(1) = 3  (2) = 3 + 2 = 5
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