Engineering Mathematics: Quadratic Equation, Root Finding Techniques, Differential Equations

Engineering Mathematics 1
Engineering Mathematics
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Engineering Mathematics 2
Task 10
From the graph, the roots of the quadratic are -2 and 4 because the graph is cutting the x-axis
through these points.
So f(x) = (x + 2) (x – 4)
This can be expanded as:
f(x) = x2 – 2x – 8
To determine whether this is the correct equation of the parabolic graph, we substitute x = 0 to
see if the value of y = -8
f(0) = 0 – 0 – 8 = -8
Therefore the correct equation of the graph is
f(x) = x2 – 2x – 8
Alternatively, the equation of this graph can be determined using all the three points on the
general quadratic equation as follows:
y = ax² + bx + c
Substituting (-2, 0)
0 = a(-2)2 + b(-2) + c = 4a – 2b + c
Substituting (4, 0)
0 = a(4)2 + b(4) + c = 16a + 4b + c
Substituting (0, -8)

Engineering Mathematics 3
-8 = a(0)2 + b(0) + c = c
Therefore c = -8
Substituting c is the equation 4a – 2b + c = 0 gives 4a – 2b – 8 = 0
Substituting c in the equation 16a + 4b + c = 0 gives 16a + 4b – 8 = 0
These two can be solved simultaneously as follows:
(4a – 2b = 8) x 2
16a + 4b = 8
8a – 4b = 16
16a + 4b = 8
Adding the above equations gives:
24a = 24; a = 1
Substituting a = 1 into one of the equations to find b gives
16(1) + 4b = 8; 4b = -8; b = -2
So the quadratic equation of the graph becomes f(x) = x2 – 2x – 8
Domain are the x values while range are y values of the graph.
The domain of the graph are all real numbers −∞< x <∞
The range of the graph are the y values: range is y ≥−9
Task 11