1. COMPOSING QUADRATIC FUNCTION
TO FIND THAT FUNTION, DEPEND ON THE CONDITION OF THE GRAPH.
THE FORMULA WHICH USED TO FIND IT :
Y = a (x–x1)(x-x2), When known x intercepts one point. X1 and x2 are x – intercepts
Y = a (x-p)2 + q, When known Vertex and one point. And the vertex is V (p,q)
Y = ax2 + bx +c generally, When known 3 point.

2. Example
1. Find The Quadratic of the following figure,
at the x- intercepts (-1/2, 0) and (-3,0), and
passes through ( -1, 2)
Solution :
Let Y = a (x –x1)(x-x2)
By x- intercepts, so Y = a(x+1/2)(x+3)
The curve passes trough (-1,2)
So Y = a (x+1/2 )(x+3)
2 = a (-1+ ½ )(-1+3)
2 = a ( -1/2 )(2)
a = -2
So QE is Y = -2(x + ½ )(x +3) → Y = -2x2 -7x -3

3. Next Example
2. A Parabola has its vertex at (-2, -10). And
passes trough at the point (2,6). Write its
Function.
Solution : → V (-2,-10) → p = -2, q=-10
Let QE Y = a (x-p)2 + q
Y = a (x+2)2 -10
Curve passes trough (3,2) → x=2, Y=6
Y = a (x+2)2 -10 → 6 = a (2+2)2 -10, a =1
a =1, → the QE is → y = x2 +4x-6

4. Next Example 3. A parabola passes trough at the point
(-3,1), (-1,-5) and (2,4). Write its function!
Solution :
Let the Function is Y = ax2 + bx + c
At the point (-3,-1) → -1 = 9a -3b +c ….. (1)
(-1,-5) → -5 = a –b +c ….….(2)
(2,4) → 4 = 4a + 2b + c ….(3)

5. From (1) and (2): 9a -3b + c = -1 a -b +c = -5 -
Next
From (1) and (2):
9a -3b + c = -1
a -b +c = -5 -
8a – 2b = 4 → or 4a –b = 2 …….. (4)
From (2) and (3)
a – b + c = -5
4a + 2b + c =
-3a -3b = -9 or a + b = 3 ………… (5)

6. From (2) for a =1, and b =2, so that
Next
From (4) and (5) :
4a – b = 2
a + b =
5a = 5 or a = 1
From ( 5), for a = 1
a + b = 3 ↔ 1 + b = 3 ↔ b = 2
From (2) for a =1, and b =2, so that
a – b + c = -5 ↔ 1 – 2 + c = -5 , → c = -4
So The QE is Y = ax2 + bx + c
Y = x2 + 2x -4

7. Next Example 4. A Function parabola has maximum value 4 at x
=1. The value of function is 0 when x = 3. Write its
function!
Solution :
Vertex V ( 1, 4 ) → so p =1 and q = 4
Passes trough point (3,0) → x = 3 , Y = 0
Let QF : Y = a (x-p)2 + q → Y = a (x-1)2 + 4
by point (3,0) → 0 = a (3-1)2+4, 4a = -4
a= -1, than QF is Y = -1(x-1)2 + 4
Y = -1(x2-3x +1)+4
Y = -x2 + 3x -1 +4
Y = -x2 + 3x + 3

8. Next Example
5. Write the Equation Quadratic Function Y
●(3,1) x
(2, -5)

9. Solution
Vertex V ( 2, -5), and Passes Trough point at (3,1) Here, p =2, q= -5, and than x= 3, y=1
Let Y = a (x-p)2 + q
↔ Y = a (x-2)2 -5
↔ 1 = a (3-2)2 -5
↔ 1 = a (1) - 5 ↔ a = 6
So its Function : Y = a (x-2)2 -5
Y = a (x2-4x +4) -5
Y = 6 (x2-4x +4) -5
Y = 6x2 -24x +24-5
Y = 6x2-24x + 19

10. Exercise
1. a parabola has the vertex (-3, 1) and
passes through point (-5, 2). Write its
Function
2. Curve of Quadratic function by x
intercepts at the point (4,0) and ( 6,0). If
the curve passes through (3 -9). Write its
Function!
3. Write the quadratic function , if are
known, Passes through (0,2), (2,0) and
(4,0)