In this write-up, I investigate the quadratic equation y =
ax^{2} + bx + c and analyze what happens to the graph
of the equation as two values are fixed and one value varies.
I note that the characteristics of a parabola are its x-intercepts,
y-intercept, axis of symmetry, vertex, and whether it opens up
or down. Therefore, first I fix the values of "b" and
"c" at 1 and examine what happens to the graph as I
**vary the values of "a"**.
Then, I fix the values of "a" and "c" at 1
and examine what happens to the graph as I **vary
the values of "b"**. Finally, I fix the values
of "a" and "b" at 1 and examine what happens
to the graph as I **vary the values of
"c"**.

In this section, I examine the effects on varying the values
of "a" using the equation y = ax^{2}+ x + 1
(notice the values of "b" and "c" are fixed
at one).

1) This graph was created using "a" values of -4
(**red**), -2 (**green**),
0 (**turquoise**), 2 (**blue**), and 4 (**purple**).

2) This graph was created using "a" values of -9
(**red**), -7(**green**),
-5 (**turquoise**), -3 (**blue**), and -1 (**purple**).

3) This graph was created using "a" values of 1 (**red**), 3 (**green**),
5 (**turquoise**), 7 (**blue**), and 9 (**purple**).

To see a QuickTime movie of y = ax^{2} + x + 1 as the
values of "a" vary from -10 to 10 and back, click **here**.

Based on the example graphs, it seems obvious that the value
of "a" has some effect on the x-intercepts of the graph.
I can check this by completing the square on the equation ax^{2}
+ x + 1 = 0. First, I divide by "a," so x^{2}
+ x/a + 1/a = 0. Therefore, x^{2} + x/a + 1/4a^{2}
= -1/a + 1/4a^{2}. Factoring on the left gives (x + 1/2a)^{2}
= (-4a + 1)/4a^{2}. Taking the square root of each side
leaves x + 1/2a = +/- sqrt(1 - 4a)/2a. Finally, x = (-1 +/- sqrt(1
- 4a))/2a. Since the variable "a" is left in the equation,
I know that its value affects the location of the x-intercepts.

It is also easy to notice that the y-intercept of each graph
remains constant. Therefore, I can assume that the value of "a"
does not affect the y-intercept of the graph. This should make
sense, because to find the y-intercept of a graph, we let x =
0. If x = 0 in the above situation, then the equation y = ax^{2}
+ x + 1 reduces to y = 1. Therefore, the x^{2} term has
disappeared, and so the value of "a" has no bearing
on the y-intercept.

I conjecture that the value of "a" does have effects on the axis of symmetry. The equation of the axis of symmetry is found by averaging the x-coordinates of the x-intercepts. Therefore, [((-1 + sqrt(1 - 4a))/2a) + ((-1 - sqrt(1 - 4a))/2a)]/2 = (-2/2a)/2 = (-1/a)/2 = -1/2a. So the equation of the axis of symmetry is x = -1/2a. Thus, the value of "a" affects the location of the axis of symmetry. Furthermore, if "a" is positive, the equation of the axis of symmetry is negative, and so the axis of symmetry (and vertex) will be to the left of the y-axis. If "a" is negative, the equation of the axis of symmetry is positive, and so the axis of symmetry (and vertex) will be to the right of the y-axis. Note that when a = 0, the axis of symmetry is undefined (the equation actually becomes linear, so the graph is a line in this case).

Since the vertex of the parabola is on the axis of symmetry,
it follows that the vertex must be affected by the value of "a."
The x-coordinate of the axis of symmetry is -1/2a, and the y-coordinate
can be found by substituting this value into the equation y =
ax^{2} + x + 1. By doing so, y = a(-1/2a)^{2}
+ (-1/2a) + 1 = a(1/4a^{2}) - 1/2a + 1 = 1/4a - 2/4a +
4a/4a = (4a - 1)/4a.

Finally, when "a" is positive, the parabola opens up and has a minimum. When "a" is negative, the parabola opens down and has a maximum.

In this section, I examine the effects on varying the values
of "b" using the equation y = x^{2}+ bx + 1
(notice the values of "a" and "c" are fixed
at one).

1) This graph was created using "b" values of -4
(**red**), -2 (**green**),
0 (**turquoise**), 2 (**blue**), and 4 (**purple**).

2) This graph was created using "b" values of -5
(**red**), -4(**green**),
-3 (**turquoise**), -2 (**blue**), and -1 (**purple**).

3) This graph was created using "b" values of 1 (**red**), 2 (**green**),
3 (**turquoise**), 4 (**blue**), and 5 (**purple**).

To see a QuickTime movie of y = x^{2} + bx + 1 as the
values of "b" vary from -10 to 10 and back, click **here**.

Based on the example graphs, it seems obvious that the value
of "b" has some effect on the x-intercepts of the graph.
I can check this by completing the square on the equation x^{2}
+ bx + 1 = 0. First, x^{2} + bx + b^{2}/4 = -1
+ b^{2}/4. Factoring on the left gives (x + b/2)^{2}
= (-4 + b^{2})/4. Taking the square root of each side
leaves x + b/2 = +/- sqrt(b^{2} - 4)/2. Finally, x = (-b
+/- sqrt(b^{2} - 4))/2. Since the variable "b"
is left in the equation, I know that its value affects the location
of the x-intercepts.

Again, it is easy to notice that the y-intercepts of each graph
remain constant. Therefore, the value of "b" does not
affect the y-intercept of the graph. This should make sense, because
to find the y-intercept of a graph, we let x = 0. If x = 0 in
the above situation, then the equation y = x^{2} + bx
+ 1 reduces to y = 1. Therefore, the x term has disappeared, and
so the value of "b" has no bearing on the y-intercept.

I conjecture that the value of "b" does have effects
on the axis of symmetry. The equation of the axis of symmetry
is found by averaging the x-coordinates of the x-intercepts. Therefore,
[((-b + sqrt(b^{2} - 4))/2) + ((-b - sqrt(b^{2}
- 4))/2)]/2 = (-2b/2)/2 = -b/2. So the equation of the axis of
symmetry is x = -b/2. Thus, the value of "b" affects
the location of the axis of symmetry. Furthermore, if "b"
is positive, the equation of the axis of symmetry is negative,
and so the axis of symmetry (and vertex) will be to the left of
the y-axis. If "b" is negative, the equation of the
axis of symmetry is positive, and so the axis of symmetry (and
vertex) will be to the right of the y-axis.

Since the vertex of the parabola is on the axis of symmetry,
it follows that the vertex must be affected by the value of "a."
The x-coordinate of the axis of symmetry is -b/2, and the y-coordinate
can be found by substituting this value into the equation y =
x^{2} + bx + 1. By doing so, y = (-b/2)^{2} +
b(-b/2) + 1 = b^{2}/4 - b^{2}/2 + 1 = b^{2}/4
- 2b^{2}/4 + 4/4 = (-b^{2} + 4)/4.

Finally, the value of "b" seems to have no bearing on whether the parabola opens up or down.

In this section, I examine the effects on varying the values
of "c" using the equation y = x^{2}+ x + c (notice
the values of "a" and "b" are fixed at one).

1) This graph was created using "c" values of -2
(**red**), -1 (**green**),
0 (**turquoise**), 1 (**blue**), and 2 (**purple**).

2) This graph was created using "c" values of -5
(**red**), -4(**green**),
-3 (**turquoise**), -2 (**blue**), and -1 (**purple**).

3) This graph was created using "c" values of 1 (**red**), 2 (**green**),
3 (**turquoise**), 4 (**blue**), and 5 (**purple**).

To see a QuickTime movie of y = x^{2} + x + c as the
values of "c" vary from -10 to 10 and back, click **here**.

Based on the example graphs, it seems obvious that the value
of "c" has some effect on the x-intercepts of the graph.
I can check this by completing the square on the equation x^{2}
+ x + c = 0. First, x^{2} + x + 1/4 = -c + 1/4. Factoring
on the left gives (x + 1/2)^{2} = (-4c + 1)/4. Taking
the square root of each side leaves x + 1/2 = +/- sqrt(1 - 4c)/2.
Finally, x = (-1 +/- sqrt(1 - 4c))/2. Since the variable "c"
is left in the equation, I know that its value affects the location
of the x-intercepts.

Obviously, the value of "c" affects the y-intercept
of the graph. This should make sense, because to find the y-intercept
of a graph, we let x = 0. If x = 0 in the above situation, then
the equation y = x^{2} + x + c reduces to y = c. Therefore,
the value of "c" __is__ the y-intercept of the graph.

I conjecture that the value of "c" does __not__
have effects on the axis of symmetry. The equation of the axis
of symmetry is found by averaging the x-coordinates of the x-intercepts.
Therefore, [((-1 + sqrt(1 - 4c))/2) + ((-1 - sqrt(1 - 4c))/2)]/2
= (-2/2)/2 = -1/2. So the equation of the axis of symmetry is
x = -1/2. Thus, the value of "c" has no effect on the
location of the axis of symmetry.

But I do submit that the vertex must is affected by the value
of "c." The x-coordinate of the axis of symmetry is
-1/2, and the y-coordinate can be found by substituting this value
into the equation y = x^{2} + x + c. By doing so, y =
(-1/2)^{2} + (-1/2) + c = 1/4 - 1/2 + c = 1/4 - 2/4 +
4c/4 = (4c - 1)/4.

Finally, the value of "c" seems to have no bearing on whether the parabola opens up or down.