Assignment 3:

Varying
Coefficients in a Quadratic

by

Jenny Johnson

We are going to explore the graph of a quadratic
equation y = ax²+ bx + c when a = 1 and c = 1. Thus, we have the equation y = x²+
bx + 1. On an xy plane, the graph
of this equation is a parabola with a y-intercept of 1. A movie of the parabola with b values
ranging from -5 to 5 is shown below.

Depending on the value of
b, the parabola sometimes has two real roots, sometimes it has one real root,
and sometimes it has no real roots.
A root of the equation is defined as the value of x when y = 0, thus a
real root of the equation is the x-value when the graph intersects the x-axis.

**What
can we learn when the equation is graphed on the xb plane?**

** **Let
us now consider the graph of y = x^{2 }+ bx + 1 when it is graphed on
the xb plane instead of the xy plane.
This graph is shown below.

Assume
now that b = 4, which is a horizontal line through 4 on the b-axis. This line intersects the graph in two
places, as shown below.

On
the xy-plane, the graph of the parabola for y = x^{2 }+ 4x + 1 (when b
= 4) crosses the x-axis in two places.
We can find those two values, or roots, by using the quadratic formula.

The decimal approximations of these two roots are x
= -.268 and x = -3.732. If we
trace the graph on the xb-plane, we discover that the intersections of y = x^{2
}+ bx + 1 and b = 4 have the same x-values of the real roots of the equation in the
xy-plane. See the graphs below.

We can then make the conjecture that the x-values of
the intersection(s) of the line b = n with the equation for y = x^{2
}+ bx + 1 on the xb-plane will yield the same x-values as the real roots
of the equation. We can see from
the picture that when b > 2, there will be two negative real roots to the quadratic. Below is a picture of several lines
when b > 2. Each of them
intersects the curve in two places with negative corresponding x-values.

Let us examine what is happening in the quadratic
equation when b > 2. Since a =
1 and c = 1, then the value inside the square root (which we call the
discriminant) will always be: b^{2} – 4. Thus, if b > 2, then this value will
always be positive and the square root will always be a real number. This then yields two roots because
–b ± a real number will give us two real numbers (and
then we would divide them by 2).
Similarly, if b < -2, then the discriminant will always be positive
and the equation will have two positive real roots. Below is a movie showing b varying from -5 to 5.

We also can observe from the graph that if b = 2,
there will be one negative real root.
In the case that b = 2, the discriminant will be 2^{2} –
4, which is always 0. Since the
square root of 0 is 0, we will get one root and it will always be
–b/2. Similarly, if b = -2, there
will be one positive root, x = b/2.

If -2 < b < 2, then the graphs of b = n do not
cross the graph of y = x^{2} + bx + 1 and thus there will be no real
roots. This makes sense with the
quadratic equation since the discriminant is negative if -2 < b < 2 because
b^{2} – 4 < 0.
Since the square root of a negative number is imaginary, there will be
no real roots.

**What can we observe when the
value of c is changed?**

** **Let us change the value of c
from 1 to -1 graph the resulting equation on the xb-plane.

We can see that no matter what horizontal line we
draw (b = n), there will always be two real roots for the equation. This makes sense with the quadratic
formula since a value of c = -1 yields a discriminant of b^{2 }- (- 4)
= b^{2} + 4, which will always be positive no matter the value of
b. Thus, there will always be two
roots to the equation. For what
other values of c will the equation always yield two real roots? Since we understand the discriminant,
there will always be two roots when c < 0. Let us draw several graphs on the xb-plane when c < 0.

The
gray line is the line formed when c = 0.
We can see that for any c ² 0, the parabola will
always have two real roots no matter the value of b.

If
the value of c is greater than 0, the number of roots will depend on b. Several graphs with c > 0 are shown
below on the xb-plane.

Notice
that for each of these graphs there are certain values of b that will yield no
real roots.

**What is the significance of
the line 2x + b = 0?**

** **Let us graph 2x + b = 0 on
the same xb-plane as the quadratic y = x^{2} + bx + 1 and b = 4.

It appears that the intersection of this line 2x + b
= 0 and the line b = 4 is equidistant from the two intersections of b = 4 and y
= x^{2} + bx + 1. In other
words, the x-value of the intersection of the line and the horizontal line b =
4 is equidistant from the roots of the equation when b = 4. It appears that this is true for any
value of b. So, the x-value of the
intersection of 2x + b = 0 and b = n is equidistant from the two roots of the
parabola when b = n. Let us
observe this line as we vary c in the equation.

Well, the intersection of 2x + b = 0 and b = n will
always occur when 2x + n = 0 or when x = -n/2, or in other words when x =
-b/2. This makes sense when we
consider the parabola of the equation y = x^{2} + bx + c. The x-value of the vertex will always
be –b/2 and it is completely independent of the value of c. Also, the line x = -b/2 is the line of
symmetry for the parabola which means it is equidistant horizontally from the
roots of the parabola. Thus, the
line 2x + b = 0 is just the line connecting all the vertices of the parabolas
of the form y = x^{2} + bx + c.

We have already discussed that when the discriminant
is 0, then the parabola has exactly one root. The quadratic formula for all quadratics of the form y = x^{2}
+ bx + c will give us x = -b/2 when the discriminant is 0. Thus, the line formed by x = -b/2 goes
through the points where the parabola has exactly one root, which is the
vertex. Thus, the line x = -b/2
(or 2x – b = 0) is the line that intersects all of the vertices of the
parabolas of the form y = x^{2} + bx + c.