Find the Quadratic Equation of the Graph of Quadratic Functions


Previously, it was discussed how to draw a parabolic graph of quadratic functions based on equations with intersection values, extreme points and discrimination. But the times that will be discussed are the opposite of the previous process, which is Determining the Quadratic Equation Based on the known Quadratic Function Graph. Where the graph of quadratic functions has or certain properties that you must understand before determining which formula is right to use



1.  The graph of the quadratic function intersects the X axis at (x1, 0) and (x2, 0) and through any point (x3, y3) on the graph, then the equation of the quadratic function can be determined by the formula,

y = f(x) = a(x- x1)(x – x2)


The value of a is determined by entering the arbitrary point to x and y.

Where does this formula come from,
Y = f(x) = ax2 + bx + c


Because

then, 


2.  The graph of the quadratic function alludes to the x axis at (x1, 0) and through one arbitrary point, then the equation of the quadratic function can be determined by the formula,
y = f(x) = a(x- x1)2


The formula y = f(x) = a(x- x1)2 is formed from the previous formula namely y = f(x) = a(x- x1)(x – x2) but because the values ​​of x1 and x2 are the same then the formula becomes,
y = f(x) = a(x- x1)(x – x1) = a(x- x1)2
The value of a is determined by entering the arbitrary point to x and y.

3.  The graph of the quadratic function through the cusp or turning point (Xp, Yp) and through a certain point, then the equation of the quadratic function can be determined by the formula,
y = f(x) = a(x- xp)2 + yp


how this formula is formed is


Note the factoring results below,

4.   Graph of quadratic functions through three arbitrary points, namely (x1, y1), (x2, y2), and (x3, y3), then the equation of the quadratic function can be determined by the formula.
y = f(x)  = ax2 + bx + c
with values ​​a, b and c determined after

Example 1.
Find the quadratic function equation from the parabolic chart sketch in figure-3 below,


If we look at Figure-3, the known value is,
The intersection of the X axis: (-2, 0) and through a certain point, namely
Extreme points: (1, -4) = (xp, yp) = (peak point, turning point / extreme)
So to find the values ​​a and b, use the formula
Y = ax2 + bx + c is to use a formula
y = f(x) = a(x- xp)2 + yp
Substitute the value of x, xp and yp to the formula equation of the quadratic function,
0 = a ((- 2) - 1) 2 + (-4)
0 = a (-3) 2-4
0 = 9a - 4
a = 4/9
the parabolic equation sought is


Example 2.
Find the parabolic equation through points (1, 3), (2, -1), and (4, 3). Because the line through three points is arbitrary, the formula used is
y = f(x)  = ax2 + bx + c
Through points (1, 3) the equation obtained is
3= a(1)2 + b(1) + c
a + b + c = 3 ………………….(1)
Through point (2, -1) the equation obtained is
-1 = a(2)2 + b(2) + c
4a + 2b + c = -1………………(2)
Through points (4, 3) the equation obtained is
3 = a(4)2 + b(4) + c
16a + 4b + c = 3 …………….(3)    
Do the solution to equations (1), (2), and (3) using the elimination method so the results obtained are
a = 2 b = -10 c = 11
The parabolic equation you are looking for is
y = f(x)  = ax2 + bx + c
y = f(x)  = 2x2 -10x + 11


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