Pre-Calculus Quiz Solution for MATH 115 Summer 2018

This document presents a comprehensive solution to a pre-calculus quiz, likely from a MATH 115 course during the Summer 2018 semester. The solutions cover a range of pre-calculus concepts, including calculating average cost per part, determining the equation of a line given two points, understanding transformations of functions (specifically square root and cubic functions), solving quadratic equations using various methods (including the quadratic formula), graphing quadratic functions and identifying key features like the vertex, axis of symmetry, and intercepts, solving radical equations, working with absolute value equations and inequalities, and finding the zeros and multiplicities of polynomial functions. The solutions are detailed, providing step-by-step explanations to facilitate understanding and learning. The quiz encompasses several key areas of precalculus to test the students on their knowledge and problem-solving skills.

Running head: PRE-CALCULUS 1
Pre-Calculus
Name
Institution

PRE-CALCULUS 2
Question 1
Average cost per part¿ Total cost of parts
Total parts produced
¿ 300+3100
100+500 =3400
600 =$ 5.67 per part
Question 2
Line ( 4,7 )∧(6,2)
Gradient= y2− y1
x2−x1
= 2−7
6−4 =−5
2
Let the line pass through an arbitrary point with coordinates ( x , y ). Using the point ( 4,7 ) and the
arbitrary point, the gradient equals −5
2 . That is,
y−7
x−4 =−5
2
2 ( y−7 )=−5(x−4)
2 y−14=−5 x+20
2 y=−5 x +20+14=−5 x+ 34
y=−5 x +34
2 =−5
2 x +17
Therefore, the equation of the line is,
y=−5
2 x+17
Question 3

PRE-CALCULUS 3
The graph of f ( x )=− √x +4−5 is obtained by first shifting the graph of y= √ x four units to the
left followed by a reflection along the x-axis. Then, a shift of five units downwards.
Question 4
The graph of f ( x )=x3 +3 is obtained from the graph of y=x3 by shifting the graph of y=x3
three units up.
Question 5
x2+ 6 x+9=4
We rewrite the equation in the form, y=a x2 +b x +c then apply the quadratic formula as follows.
x2+ 6 x+9−4=4−4
x2+ 6 x+5=0
So that, a=1 , b=6 ,∧c=5
x=−6 ± √62−4 ×1 ×5
2 ×1 =−6 ± √16
2 =−6 ± 4
2 =−3 ± 2
x=−3+ 2=−1∨x=−3−2=−5
Therefore , x=−1∨x=−5
Question 6
x2+ 35=5 x
We rewrite the equation in the form, y=a x2 +b x +c then apply the quadratic formula as follows.
x2+ 35−5 x=5 x−5 x, x2−5 x+ 35=0

PRE-CALCULUS 4
So that, a=1 , b=−5 ,∧c=35
x= 5 ± √52−4 ×1 ×35
2× 1 = 5 ± √−115
2 = 5 ± j √115
2 =2.5 ± j √115
2
x=2.5+ j √115
2 ∨x=2.5− j √ 115
2
Question 7
The graph of f ( x )=x2 +2 x−4 is shown below.
From the graph the vertex is point (−1 ,−5),the axis of symmetry is x=−1,the x-intercepts are
x=−3.236 and x=1.236, and y-intercept is y=−5.
Question 8

PRE-CALCULUS 5
√2 x +3− √x +1=1
√ 2 x +3= √ x +1+1
Then, square both sides to obtain,
( √ 2 x +3 ) 2= ( √ x+1+1 ) 2
2 x+3=2 ( √ x +1 ) +( x+ 1)+1
2 x+3− ( x +1 )−1=¿2√ x+1
x +1=2 √ x +1
x +1
√ x +1 =2
x +1
√x +1 = √ x+1=2
Square the equation again to obtain,
( √ x+ 1 ) 2=22
x +1=4
x=4−1=3
Therefore, x=3
Question 9
| 1
5 x−8 |=−3
There is no solution since absolutes cannot be negative

PRE-CALCULUS 6
Question 10
|5 x +8|<3
We can break the function down into,
5 x+ 8<3 and 5 x+ 8>−3
5 x+ 8<3 , 5 x <3−8
5 x ←5 , x ←1
5 x+ 8>−3
5 x>−3−8
5 x>−11, x > −11
5 , x>−2.2
The two inequalities become, x ←1 and x >−2.2
Combining the two ranges we get, −2.2< x ←1
In interval notation, [−2.2 ,−1]
Question 11
−2<|2 x +2|≤6
First, we test each absolute of its positive and negative ranges. That is,
2 x+2 ≥ 0 for x ≥−1 ,therefore for x ≥−1 ,|2 x +2|=2 x +2
2 x+2 ≥ 0
2 x ≥−2

PRE-CALCULUS 7
x ≥−1
2 x+2< 0 for x ←1,therefore for x ←1, |2 x+2|=−(2 x+2)
2 x+2< 0
2 x←2
x ←1
So we have, x ≥−1 , x ←1
Then, we evaluate the expression in the ranges x ≥−1 , x ←1 as follows.
For x ←1:
−2<−(2 x +2)≤ 6
− ( 2 x+2 ) >−2
( 2 x+2 )<2
2 x<2−2 , x <0
−(2 x +2)≤ 6
( 2 x+2 ) ≥−6
2 x ≥−2−6
x ≥−4
Therefore, for x ←1:−4 ≤ x <0
For x ≥−1:

PRE-CALCULUS 8
−2<2 x+ 2≤ 6
2 x+2>−2
2 x>−2−2, x >−2
2 x+2 ≤ 6
2 x ≤ 6−2 , x ≤ 2
Therefore, for x ≥−1:−2< x ≤2
Combining the ranges −4 ≤ x <0 and −2< x ≤ 2 we obtain −4 ≤ x ≤2
In interval notation, [−4 ,2]
Question 12
4 x2 <16
x2< 16
4
x2< 4
x < √ 4
x <2 , x >−2
Therefore ,−2< x <2
Question 13
f ( x )=5 ( x+ 8)2(x −8)3
Given the function f ( x ) ,

PRE-CALCULUS 9
For zeros of f ( x ) ,
f ( x ) =5 (x+ 8)2(x −8)3=0
( x +8)2=0 ,∨(x−8)3=0
(x +8)2=0 , x+ 8=0 , x=−8
x=−8 is a zero of multiplicity 2
(x−8)3=0 , x−8=0 , x=8
x=8 is a zero of multiplicity 3
Question 14
f ( x )=−(−x−4)4
x 2 2.5 3 4 5 5.5 6
f ( x ) -16 -5.0625 -1 0 -1 -
5.0625
-16
The graph of the function is shown in the figure below.

PRE-CALCULUS 10