Mathematics Problems with Solutions | Desklib

This page contains solutions to various mathematics problems including slope, line equations, midpoint, graphing, domain, piecewise functions, quadratic equations, function composition, vertex, inequalities, asymptotes, polynomial zeros, and more. It also provides access to Desklib's online library for solved assignments, essays, and dissertations related to mathematics. The content is suitable for students studying mathematics at various levels in different colleges and universities.

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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