Pre-Calculus Answer Completion and Graphing

This article provides answers to completion questions and graphing problems in Pre-Calculus. It covers topics such as domain, range, and inverse functions, as well as factoring and solving polynomial equations. The article includes examples and step-by-step solutions.

Pre-Calculus
Name:
Institution:
25th July 2018

Answer completion
1. K represents the y-intercept
2. Y values that are negative
3. All the non-negative real values of x
4. X=-11
Chapter 2 Multiple Choice Qs
1. D
2. A
3. C
4. A
5. D
6. C
7. B
8. C
9. C
10. D
11. C
12. B
13. B
14. B
15. C
16. D
17. D

18. D
19. B
20. A
Part 2-Graphing
1. Sketch both f (x) and g(x ) on the graphs below if g ( x )=f ( x−1 ) +2
a. f ( x )= √x
Solution
g ( x ) = √ x−1+2
b. f ( x )=− √x
Solution
g ( x )=− √ x−1+2

2. Determine the domain and range of each function
Equation Domain Range
a. f ( x )=2 √−x +3 { x ∈ R : x ≤ 0 } { y ∈ R : y ≥ 3 }
b.
g ( x )=2 √ 2
3 ( x−2 )+2
{ x ∈ R : x ≥ 2 } { y ∈ R : y ≥ 2 }
c. y−2=− √ 3(x +1) { x ∈ R : x ≥− 5
3 } { y ∈ R : y ≤ 0 }
3. Sketch the graph of each function f ( x ) and then sketch the graph g ( x )= √f ( x ) on the axes
below;
i) f ( x ) =x +2

Equation f ( x) g( x )
Domain Set of all real
numbers
{ x ∈ R : x ≥−2 }
Range Set of all real
numbers
{ y ∈ R : y ≥ 0 }
ii) f ( x ) =x2−4
Solution

Equation f (x) g( x )
Domain Set of all real numbers { x ∈ R : x ≤−2∨x ≥ 2 }
Range { y ∈ R : y ≥−4 } { y ∈ R : y ≥ 0 }
iii) f ( x ) = ( x −2 ) 2+ 4
Solution
Equation f (x) g( x )
Domain Set of all real
numbers
Set of all non-negative real numbers
Range { y ∈ R : y ≥ 4 } { y ∈ R : y ≥ 2 }
4. Solve the following equation algebraically
3 √2 x +4 +9=12

Solution
3 √2 x +4=12−9
3 √2 x +4=3
( 3 √ 2 x + 4 ) 2=32
9 ( 2 x+ 4 )=9
18 x+ 36=9
18 x=−27
x=−27
18 =−3
2 =−1.5
5. Determine the equation of the following radical function in the form y=a √ b( x−h)+k
Solution
From the graph we have;
h=3
k =3
a=3.5
b=1
y= 7
2 √ ( x−3)+3
6. Area of the tennis ball
Solution
r =1
2 √ A
π
r2= 1
4 ( A
π )
A=4 π r2= 4∗22
7 ∗3.32=136.9029 cm2

Section One: Multiple Choice
1. D
2. C
3. A
4. C
5. A
6. C
Section 2:
7. Quiz
Solution
The hshows the shift either upwards or downwards while the k shows the y-intercept
of the curve
8. Quiz
Solution
The value of x must be multiplied with 3 and then the function has to be multiplied by
2
x f(x)
0 0
1 3.46410
2
4 6.92820

3
9 10.3923
9. Quiz
Solution
We have the values as
(8,16) and (4,8)
Gradient is;
Gradient= 16−8
8−4 = 8
4 =2
Thus the equation is;
y−8
x−4 =2
y−8=2(x −4)
y=2 x−8+ 8=2 x
y=2 x
10. Graph of the reflection through x-axis

Solution
11. Filling the table
Equation Words Mapping
y=4 f (x) 4 times the f(x) function 4
y=f ( 3 ( x ) )+ 2 Multiply the f(x) function
by 3 and then add 2
3
y=f (−(x +3)) f(x) function added 3 then
the negative
-1
12. image point
a) (-4,18)
b) (6,25)
c) (13,24)
d) (3,21)
13. Inverse
a) E
b) C
c) B
d) A
e) D

14. Inverse
f ( x )=7 x g ( x )=−3 x+ 4 h ( x )= x+ 4
3 k ( x )= x
3 −5
f−1 ( x )= 1
7 x g−1 ( x )= 4−x
3
h−1 ( x )=3 x +4 k−1 ( x )=3 ( x+5)
15. The equation of the semi-circle is;
Solution
g ( x )=4 f ( 3 ( x−4 ) ) +4
Multiple choice
1. A
2. C
3. B
4. A
5. B
6. D
7. D
8. C
9. B
10. B
11. B

12. A
Short Answer
1. Show that x+1 is a factor of P ( x )=x3+ 2 x2+3 x +2. Explain how you know it is a
factor
Solution
P ( x )=x3+ 2 x2+3 x +2= ( x+1 ) (x2 + x +2)
This means that
P ( x ) = ( x+ 1 ) (x2 +x +2)
Thus ( x +1 ) is a factor of P(x )
2. Factor the equation y=x3−2 x2−5 x+ 6 and then sketch a graph of it
Solution
y=x3−2 x2−5 x+ 6= ( x −1 ) ( x+ 2)( x−3)
Sketch

3. Factor each of the following fully.
a) x3+ 6 x2 +11 x+ 6
Solution
( x +1 ) ( x +2 ) (x +3)
b) 4 x3 −11 x2−3 x
Solution
4 x3 −11 x2−3 x=4 x3−12 x2 + x2−3 x=4 x2 ( x−3 ) + x ( x−3 )= ( 4 x2 + x ) ( x−3 )=x (4 x+1)( x−3)
Part 2-Short Answer
1. Determine the fully factored form of x4 +10 x3+7 x2−162 x−360
Solution

The factors are;
(x +6)( x +5)( x+3)(x−4)
2. Factor fully
i) x4 −81
Solution

( x−3)(x +3)( x2+ 9)
ii) 2 x3 +5 x2−14 x−8
Solution
(2 x+ 2)( x−2)(x +4 )
3. A swimming pool has the shape of a rectangular prism and a volume of 2100 m^3.
The dimensions of the pool are x metres deep by 25x metres long by 10x + 1 metres
wide. What are the actual dimensions of the pool?
Solution
x (25 x)(10 x +1)=2100
¿>250 x ³+25 x ²=2100so, 250 x ³+25 x ²−2100=0
¿>10 x ³+ x ²−84=0
When x=2, 10 x ³+x ²−84=0....so, x−2 is a factor
so , 10 x ³+ x ²−84=( x−2)(10 x ²+bx +42)¿>1=b−20i .e . b=21
Therefore , 10 x ³+ x ²−84=( x−2)(10 x ²+21 x +42)=0 x=2 isthe only solution as ¿
so, the pool measures 2 metres by 50 metres by 21 metres

4. Fill in the chart below for this graph
Degree ad name Degree 3
Leading coefficient sign (+
or -)
-3
End Behavior As x →+∞ , f ( x ) →−∞ , x →−∞ , f ( x ) →+∞
x-intercepts and multiplicity -2, 0, 1; (x +2)( x)(x−1)
y-intercept 0
Domain Set of all real numbers
Range { y ∈ R : y ≥ 0 }
5. For the polynomial equation below, determine the information to fill in the chart
below it. Sketch it below f ( x ) =2 ( x +3 ) 2 (x−1)( x+ 1)
Solution
Degree and name Degree 4 i.e. x4
Leading coefficient sign (+
or -)
+2
End Behavior As x →+∞ , f ( x ) →+∞ , x →−∞ , f ( x ) →−∞
x-intercepts and multiplicity -3, -1, 1; ( x +3)(x+1)(x−1)
y-intercept -18
Domain Set of all real numbers
Range { y ∈ R :16 y +51 √17+107 ≥ 0 }