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.
(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
What are the possibilities when the parabola
has the form ? Then we have slightly different cases
* 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
|r = 0||The extremum point of the parabola is on the y - axis AND the parabola is symmetric with respect to y - axis|
|k = 0||The parabola is tangent to the x - axis|
|r = 0 AND k = 0||The extremum point of the parabola is at the origin|
|| a | increasing||The arms of the parabola narrow.|
(ii) When the parabola has the form
|What is c ?||The point ( 0 , c ) is the y - intercept of the parabola|
|b = 0||y - axis is the symmetry axis of the parabola|
|c = 0||One of the arms of the parabola passes through the origin|
|b = 0 AND c = 0||The extremum point is at the origin|
Can we determine parabola's equation from its graph?
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.
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.
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.