Detailed Solutions for Algebraic Equations and Graphing Tasks

This document provides detailed solutions to a series of algebra problems involving linear and quadratic equations, as well as polynomial curves. The solutions include analyzing the order of equations, graphing functions, and exploring how changes in coefficients and constants affect the shape and position of the graphs. Tasks cover topics such as linear equations with varying slopes and intercepts, quadratic equations with different coefficients, and factorization of polynomial expressions to find x-intercepts. The analysis includes observations on how increasing or decreasing coefficients impacts the curves and their positions on the coordinate plane, with specific examples provided for each task. The final task involves analyzing a degree 5 polynomial curve and identifying its key features, such as intercepts and inflection points. Desklib offers this assignment solution to aid students in understanding and mastering algebraic concepts.

Q1)
x -6 -4 -2 0 2 4 6
f: Y = -4x-1 23 15 7 -1 -9 -17 -25
g: Y = -2x-1 11 7 3 -1 -5 -9 -13
h: Y = 2x-1 -13 -9 -5 -1 3 7 11
i: Y = 4x-1 -25 -17 -9 -1 7 15 23
a) Order is zero
When the equations are differentiated the highest power of the value of x will be 0, hence
having an order of 0.
b) i. When the value of m is negative the line curve is inversely proportional
ii. When the value of m is positive the line curve is directly proportional
Task 2

x -6 -4 -2 0 2 4 6
f: Y = 3x -2 -20 -14 -8 -2 4 10 16
g: Y = 3x + 2 -16 -10 -4 2 8 14 20
h: Y = 3x + 4 -14 -8 -2 4 10 16 22
a) An increase in the value of c increase the values of y
b) i. When the value of m = 0 and c = 3, the resultant value is a point of value 3
ii. When the value of m = 5 and c = 0, the value of y increases and form a directly
proportional line graph.
Task 3

x -6 -4 -2 0 2 4 6
c: Y = 2x2 72 32 8 0 8 32 72
d: Y = 5x2 180 80 20 0 20 80 180
a) The graphs have order of 1

When the equations are differentiated the highest power of the value of x will be 1, hence
having an order of 1.
b) Increase in the coefficient from 2 to 5 increased the value of y, and the curve will broaden
too.
Task 4
x -6 -4 -2 0 2 4 6
c: Y = -2x2 -72 -32 -8 0 -8 -32 -72
d: Y = -5x2 -180 -80 -20 0 -20 -80 -180

a) The graphs have an order of 1
When the equations are differentiated the highest power of the value of x will be 1, hence
having an order of 1.
b) Decreasing in the coefficient from -2 to -5 decreases the value of y, and the curve
broaden to this decrease.

Task 5
x -6 -4 -2 0 2 4 6
c: Y = 3x2 +
1
109 49 13 1 13 49 109
d: Y = 3x2 +
6
114 54 18 6 18 54 114
e: Y = 3x2 -7 101 41 5 -7 5 41 101

Increasing the value c from -7, to 1 to 6 increases the value of y respectively, while the curve
becomes smaller, as it move to the positive side of the y axis.
Task 6
x -6 -4 -2 0 2 4 6
c: Y =
x2 + 8x
+ 15
3 -1 3 15 35 63 99

x -6 -4 -2 0 2 4 6
c: Y = x2 - 8x + 15 99 63 35 15 3 -1 3

a) The curves are equal but when the value is 8x the curve is position on the negative sides
of the x- axis, while when the value is -8x the curve is position on the positive sides of x
– axis.
b)
Factorize the equation Y = x2 + 8x + 15
8x = 3x + 5x.
The equation becomes: (x2+3x)+(5x +15).
Factorizing each of the expressions in the parentheses: x(x + 3) +5(x+3)

= (x +5)(x+3)
0 = (x +5)(x+3)
X = 5 and x = 3
Factorize the equation Y = x2 -8x + 15
8x2 = -3x -5x.
The equation becomes: (x2-3x)-(5x -15).
Factorizing each of the expressions in the parentheses: x(x - 3) -5(x-3)
= (x -5)(x-3)
0 = (x -5)(x-3)
X = 5 0r x = 3
Comment
The curve intersect the x – axis at -5 and -3, as shown too on the graph
The curve intersect the x – axis at 5 and 3, as shown too on the graph
Task 7
x -8 -6 -4 -2 0 2 4 6 8
c: Y = x3 + 4x2 -
12x -160 0 48 32 0 0 80 288 672

Factorize the equation Y = x3 + 4x2 -12x
4x2 = 2x2 + 2x2.
The equation becomes: (x3+2x2)+(2x2 – 12x).
Factorizing each of the expressions in the parentheses: x2(x - 2) +2x (x+6)
= (x2 +2x)(x+2)(x-6)
0= (x2 +2x)(x-2)(x+6)
X =0 and x = -6, and x = 2

Comment
The curve intersect the x – axis at 0, -6 and 2, as shown too on the graph
Task 8
x -6 -4 -2 0 2 4 6
Y = 2x5 – 5x4 +10x3
-
2419
2
-
3968
-
224 0 64 1408 11232

The graph shows a degree of 5 polynomials curve, it intersects x and y axis at origin point (0, 0)
and form an inflection point at x= 0 and y = 0