MATH 115 Algebra: Complete Solution to Midterm Exam, Summer 2018

This document presents a detailed solution to a MATH 115 midterm exam from Summer 2018. The solution covers a range of algebra topics, including determining the slope of a line, writing slope-intercept equations, finding the midpoint of a segment, identifying symmetry in graphs, determining the domain of functions, solving quadratic equations, finding composite functions, identifying the vertex of a function, solving inequalities with absolute values, solving equations with square roots, finding vertical asymptotes, finding zeros of polynomials, solving rational equations, determining the inverse of a function, and analyzing the behavior of polynomial functions. Each question is worked out step-by-step to provide a clear understanding of the concepts and problem-solving techniques involved.

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Question 1
Slope is given by y2-y1/x2-x1
=given the points (9,-2) and (7,2)
The slope = 2—2/7-2 = 4/-2 = -2
Question 2
Slope intercept equation
The line equation is given by y =mx + c
The slope , m = ¼
Pass point is given as (6, -3)
Therefore;
-3 = ¼ *6 + c
-3 = 3/2 + c
-3-3/2 =c
C = -9/4
Y= 1/4x -9/4
Question 3
The midpoint of segment is given by
M = ¿)

Given the two end points as (-5,6) and (-6,4)
= ¿ )
The midpoint = (-11/2, 5)
Question 4
Graph y = |x| + 3
The graph equation can be written as f(x) = -x+ 3
The function of f(-x) = x+ 3 = -(x-3) for –f(x)
Therefore f(x) ≠ f(-x)
This is an odd function meaning that the graph is symmetrical with the origin
Question 5
Domain of a function f(x) = 𝑓(𝑥) = (3𝑥−3)/ (𝑥 2−𝑥−2)
= (3𝑥−3)/ (𝑥 2−𝑥−2)= 3(x-1) (x – 2)(x + 1)
Question 6
The domain of the piecewise is -2 to 2 while the range is -1 to 3
Question 7
5x2- 2x =2 = 0 therefore a = 5, b=-2 and c =2
Using quadratic equation;
x=−b ± √ b2−4 ac
2 a

x=−−2 ± √22−4 x 5 x 2
2 x 5
x= 2± √4−40
10
Question 8
F(x) = 3x+8 and g(x) = 2x2 + 4
Therefore f(g(x)) = (2x2 +4)(3x+8) = 6x3 + 16x2 + 12x + 32
Question 9
F(x) = x2 +4x+5
A=1, b=4 and c=5
X = -b/2a = -4/2*1 = -2
F(x) = -2*-2 + 4*-2 + 5
= 4-8+5
= 1
Vertex = (-2,1)
Question 10
|9-x|≥ 5

9-x ≥ 5
9-5 ≥ x
4≥ x
Question 11
√5−4 x =x
Squaring the both sides;
5 – 4x = x2
X2 + 4x-5 = 0
(x-1)(x+5) = 0
X = 1 or -5
Question 12
Vertical asymptotes
f ( X )= 4 x +3
x2−1
Setting the denominator to zero
= x2 -1 = 0
= x2 = 1
= x = ±1 which represent the vertical asymptotes

Horizontal asymptotes turn out to be zero since the numerator and denominator do not have a
common degree.
Question 13
Zeros of polynomial 3x3 + x2 -3x -1
The factors of constant are ±1
Factors of leading coefficient are ±1 and ±3
Possible rational zeros include ± 1
1 ∧± 1
3
Question 14
5/9 = 2/(x+2)
5(x+2) = 2*9
5x + 10 = 18
5x = 18-10
5x = 8
X = 5/8
Question 15
x+ 1
x+6 ≤ 0
X + 1 ≤ 0(x+6)

X+1≤ 0
x≤ -1
Question 16
Inverse of function f(x) = 5x -4
Replace f(x) with y.
y=5x−4
Interchange the variables.
x=5y−4
Solve for y
y=4/5+x/5
Solve for yy and replace with f−1(x)
f−1(x)=4/5+x/5
Question 17
The leading coefficient for the graph function f(x) = 5x2 –x4 + x+ 7 is 5 which is positive
The leading degree of the function is 4 which is even
The behavior for the combination of positive coefficient and even degree is a rises in left and
right side.
There C is the correct symbol representing the function behavior.
Question 18
The interval is always increasing at levels where the f(x) ≥ 0
These include (-5 to -2) and (2 to 5)

Question 19
A