Mathematics Investigation - QuadraticsUse of the program Efofex. Proofing and justifying. Inclusion of inequalities where necessary.

Q.no.1
a) Graph is shown below
b) Algebraically we substitute y value and find x and hence the point.
When y =3, we have
So points are (0,3) and (4,3)
When y =-1, we get
So point is (2,-1)
When y =-10, we have
Here since discriminant is negative =-36, there is no real solution or there is no point of intersection.
c) Positive value is y =2 and negative value is y =-2
When y =2,
So intersecting points are (3.732, 2) and (0.268,0)
When y =-2, we have
In this discriminant=
No real roots or there is no point of intersection
d) The points of intersection are shown in the graph below:
e) From the parts above we find that
y=m
i) Has two points of intersection if y >2
ii) One point of intersection if...