X-Ploration of the quadratic equation (parabola's)

by Frater Elijah

This exploration will examine the effects the quadratic equation:

y = ax2 + bx + c for various a, b, c real

We will explore the effects of each parameter by fixing the other two and noticing the effects.

For the first case we will see what the parameter a does while fixing b = c = 0.

If a = 1 then y = x2. This corresponds to the below graph:

 This is a parabola.

The next graph corresponds to the overlapping of a = 2 & a = 1while b = c = 0.

We can see that an increase in a leads to an increase in the "steepness" or slope of the parabola.

The next graph corresponds to a = 1, 2, 3, 5:

As a is getting larger, the parabola's are getting narrower. What do you think will happen if 0 < a < 1?

Let try a = 1/2 and compare it to a =1:  y = 1/2x2 y = x2

Interesting, it appears that the parabola has gotten wider.

Letís try smaller values and see what happens. Say a = 1, .5, .3, .1, .01 (from inside to outside respectively)

It is as we thought. The parabola is getting wider for a smaller and more narrow for a larger.

The following graph shows a = 100, 4, 1, .4, .04

 Thinking


What about a negative?

Usually negative signs cause some form of reflection in Euclidean space.

Letís think about this, if y = -(some #) x2, then x2 is always positive, and then we are taking the negation of this.

So for each x value both positive and negative, are y value will be a negative number (except of course for x = 0).

So the parabola will be flipped over the x-axis.


The following corresponds to y = -x2:

Wow, our thinking about this was right. Now what do you think would happen if a were a large negative number as opposed to if -1 < a < 0?

Well the only difference that the negative sign made on y = x2 was to flip it over the x-axis.

It may be safe to predict that a large negative number for a will increase the steepness of the slope, and a fractional value for a, will widen the parabola or make the slope smaller.

Lets check,

y = -x2 and y = -5x2

It seems that this is our day, as all of our predictions about a have been verified.

The following graph corresponds to a = 5, .01, 1, -.01, -1,-5


This was for b = c = 0 lets. The reason we chose b = c = 0 is because x2 is the highest degree term in our equation

and will dominate all the other terms for large x, that is, this term will "govern" what the function looks like as x

goes to infinity & a is itís coefficient.


Let us explore some graphs with other possible variances of fixing b and c while changing a.

y = ax2 + x + 1 (a = -2, -1, 0, 1, 2 b = 1, c = 1)

We can easily notice the straight line on this graph. This line is when the value of a = 0. This is a linear equation in

the form y = mx + b (in this case y = x + 1).

Before we investigate what b and c do let us look at one more overlapping graph:

y = ax2 + (sin x) x + 1 in this case ( a = -2, -1, 0, 1, 2 c = 1 but b = sin x is a function).

This investigation will only cover cases where a, b, and c are constants, but exploring some cases where the coefficients are functions can be of great importance! Notice that the picture still has the general resemblance of a parabola. This is because of the leading degree (power) as discussed before. Let us now look at what the value of b does to our graphs.

Let us first examine the case of a = c = 0 and b = 1 or y = x.

Is this a good graph to investigate?

Not really, because when a = c = 0 we just have a straight line which does not really tell us how the overall function will behave.

Let us start with a something more interesting like a = c = 1 and vary b.

Say b = 1 and 2 separately and then overlapped.




Overlapped: It appears that ourparabola has moved!


For now let us keep the parameter c = 0 and look at different cases of y = x2 +bx (for a =1 b = 0, 1, 2, 3, 4, 5 c = 0).

This is because I want to see how just the addition (or subtraction) of bx onto x2 affects x2.

Wow! We see the y = x2 equation, and then for each successive b value increased the parabola has descended below the x-axis so that itís ëleft-mostí x-intercept corresponds to the negative b parameter, where as the right most intercept has remained fixed at the origin. B seems to be moving the vertex of our parabola.

Let me make a table of the location of the vertex and the values of b & c.


Letís think about this:

    1) We are looking at the graphs of quadratic equations.
 
    2) When I think of quadratic equations I think of the quadratic formula:

This will give us the roots for x, that is, where the parabola will intercept the x-axis.


Now this will happen either once, twice or not at all. The discriminant will tell us this.

The discriminant is d = b2 ó 4ac:

    If d is negative the parabolaís vertex lies above the x-axis and there are two distinct conjugate complex roots (but

    no real roots).

    If d is positive the parabolaís vertex lies below the x-axis and there are two real roots (call these zero's x1, x2 ).

    If d is zero then the parabolas vertex lies on the x-axis and there is one real zero (a double root).

    Now back to finding our vertex. It is easy to see that when d=0 our x roots correspond to -b/2a.

    Which gives us a double root for x.


    Now to find the y value of the vertex let us plug this root back into our equation.

    If f(x) = ax2 + bx + c and (b2 - 4ac = 0) then the vertex lies at the point (-b /2a, 0).

    If f(x) = ax2 + bx + c and (b2 ó 4ac > 0) then the vertex lies below the x axis and has x intercepts of x1, x2

    which correspond to real roots roots of the quadratic equation and our vertex lies at [(x1 +x2)/2, f((x1 +x2)/2)].

    If f(x) = ax2 + bx + c and (b2 ó 4ac < 0) then the vertex lies above the x-axis and has no x-intercepts!

    This is interesting how will we find the vertex?
    Let us try a similar method that we used for finding the vertex beneath the x-axis.
    We will find where our parabola intersects the y-axis (which it must because it is not a hyperbola, it does this at
    x = 0) and call this point y.

    At y define the horizontal line y = x1. Now follow y to itís other intersection of the parabola (we can do this by

    plugging in y into our equation and solving for x) call this other intersection x2.

    Now we have two distinct points on our parabola and the vertex can be found as V = [(x1 +x2)/2, f((x1 +x2)/2)].

    X-ploration 3 covers more on this.



The following graph corresponds to y = x2 -2x +3.

Notice the intersections with the vertical lines. Can you think of any other ways to do this?

So now we know what the a parameter does (it modifies the slope) and what the b parameter does (it moves the vertex). We also have another term called the discriminant which we have used to help us. Now we need to find out what c does. Letís look at the following graph:

This graph corresponds to y = x2 + c (a=1, b=0 and c=2, 1, 0, -1, -2)

C is the easiest of them all it changes the vertical distance of the parabola.


This makes sense because if all the "x's" were to be zero, then our equation would just be y = some number be it positive or negative or zero. We are just adding on a scalar quantity to y. Notice again how all of our parabolas have a similar shape.

Definition: Scalar ó Usually, a real number, particularly in operations with vectors.

Definition: Parabola ó A conic with eccentricity 1; it is defined as the locus of points P whose distance from a fixed point (the vertex) is equal to itís distance from a fixed line (the directrix). The parabola is symmetrical about it ís axis, which is the line through the vertex and perpendicular to the directrix; the point of intersection of the axis with the curve is the vertex V.


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