PARABOLA

In this presentation, I will construct graphs for the parabola

for different values of a, b, and c. (a, b, c can be any rational numbers).

Before plotting the graphs, I'd like to follow a different approach: COMPLETING TO SQUARE. In this way we can see how far a parabola is being translated from the origin. We can even rotate a parabola! But for the moment, let's only translate it:

y = a x ^ 2 + b x + c = a [ x ^ 2 + ( b / a ) x + (c / a) ] = a ( x - r ) ^ 2 + k , where , r = - b / 2a , and , k = ( 4 a c - b ^ 2 ) / 4 a

With this notation, r = - b / 2a is the symmetry axis of the parabola , and , k represents its extremum value. Now the parabola has the form y = a ( x - r ) ^ 2 + k which can also be written as ( y - k ) = a ( x - r ) ^ 2 . Translation procedure is complete! We translated the parabola to a new coordinate system where the origin is ( r , k ). This new origin is nothing but the extremum point (or vertex) of the parabola. The following plots summarize this paragraph. When a > 0 , the parabola looks up , and when a < 0 , it looks down.

CASE ANALYSIS

(1) r = 0 => - b / 2a = 0 => b = 0

In this case, the extremum point (vertex) of the parabola is on the y - axis , AND , the curve is symmetric with respect to the y - axis:

(2) k = 0 => ( 4 a c - b ^ 2 ) / 4 a = 0 = > 4 a c = b ^ 2

In this case, the discriminant of the 2nd degree equation is zero. Equivalently, y = a ( x - r ) ^ 2 . Therefore, y = 0 has only one real solution, and that solution is x = r . From this, we can deduce that the point ( r , 0 ) is the extremum point (vertex) of the parabola AND it is on the x - axis : THE PARABOLA IS TANGENT TO THE X - AXIS.

(3) r = 0 AND k = 0 => - b / 2a = 0 AND b^2 - 4ac = 0 => b = 0 AND b^2 = 4ac

In this case, the conditions in case 1 and 2 are both satisfied. This results in y = a x^2. Therefore, the extremum point (vertex) of the parabola is at the origin ( 0 , 0 ). If a > 0, then the parabola looks up. If a < 0 , it looks down.

(4) If | a | increases, then the arms of the parabola narrow. They get close to each other.

What are the possibilities when the parabola has the form ? Then we have slightly different cases now:

* The point ( 0 , c ) is the y - intercept of the parabola
* If b = 0 => y - axis is the symmetry axis of the parabola
* If c = 0 => One of the arms of the parabola passes through the origin
* If b = 0 AND c = 0 => The extremum point is at the origin

These slightly different cases can be seen in the figure below:

In this way, I covered all the possible cases starting from a different approach. The following little tables summarize the cases I analyzed:

(i) When the parabola has the form y = a ( x - r ) ^ 2 + k

(ii) When the parabola has the form

APPLICATION

Can we determine parabola's equation from its graph?

1st Solution:

y = a ( x + 1 ) ( x - 4 ) and we can also see from the figure that c = - 4 . Therefore the point ( 0 , - 4 ) satisfies y = a ( x + 1 ) ( x - 4 ). After substitution, we get - 4 = a ( 1 ) ( - 4 ) => a = 1 => y = ( x + 1 ) ( x - 4 ) => y = x^2 - 3x - 4.

2nd solution

y = a x^2 + b x + c AND c = - 4 => y = a x^2 + b x - 4. From the graph, (-1,0) and (4,0) satisfy the equation. Substitution gives:
x = -1 => a - b - 4 = 0 => a - b = 4
x = 4 => 16 a + 4 b - 4 = 0 => 4 a + b = 1
Solving these two equations simultaneously, we get a = 1 and b = -3 and hence y = x^2 - 3x - 4.

3rd solution

c = - 4
Product of the roots = c / a => (-1) (4) = c / a => a = 1
Sum of the roots = - b / a => -1 + 4= - b / 1 => b = -3
Hence y = x^2 - 3x - 4.