Quadratic Functions
I will examine the standard
equation, , and
the effect of different values of a, b, and c on its roots.
Let y = ax^2 + x+ 1. The
following shows the graph of this function when b and c remain constant while a
varies.
a = 3, 2, 1, 0, 1, 2, 3
In all cases, the graph
intersects the y-axis at y=1. When a>0, this function has no real roots
because the function opens up. However, when a<0, the function has two real
roots, because the function opens down and when a=0, the function has one real
root. Also, if we imagine a starting from 0, then as the value of a increases, the parabola gets more narrow and as a decreases the function gets wider. Of course, the
number of roots will change depending on if the vertex is above or below the x
axis.
Now, let y=x^2 –x + c
and letŐs see what happens when a and b remain constant and c is varied. c=3,
2, 1, 0, -1, -2, -3
Notice that when c=0 or
c>0, this function has two real roots. If c=0, one of its roots is zero and
the other is -1. If c >0, one of its roots is negative and the other is
positive.
Finally, let y=x^2 +bx+1. The
following shows the graphs of functions when a and c are constant, but b
varies. c= 3, 2, 1, 0, -1, -2, -3
It looks like, when |b| =2,
that the function will have only one root. When |b|>2, the function will
have two real roots and when |b|<2 , the function will have no real roots.
LetŐs consider the locu of
the vertices of the set of parabolas graphs from y=x^2 +bx +1. Notice that the
graph of this set of vertices is the parabola y=-x^2+1.
What about if a is negative,
for the same values of b?
Notice this time that the
graph of this set of vertices is the parabola y=x^2+1.
Now, I can generalize that
the set of vertices of all parabola of the form y=x^2+bx+c, is the parabola,
y=-x^2+c.