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.

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

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

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)

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 = 4 - -3a -3b = -9 or a + b = 3 ………… (5)

From (2) for a =1, and b =2, so that Next From (4) and (5) : 4a – b = 2 a + b = 3 + 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

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

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

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

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)