Pre-Calculus

This article covers various topics in Pre-Calculus including average cost, line equation, graph transformation, quadratic formula, absolute value, function zero and more. It provides step-by-step solutions to questions and includes graphs and tables for better understanding.

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