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Algebra 2 Parabola Project
71
You already have your x- intercepts, because your equation is in intercept form. So the roots of this quadratic equation are (3,0) and (-6,0).
Algebra 2 :
Parabola Project
Y= -(x- 3)( x+6)
- This parabola opens downward because a<0 .
Axis of Symmetry:
The equation to find the axis of symmetry in intercept form : X = (p+q)/2 (-3/2) = (3+ (-6))/2
The Graph
By: Erin Walker
Y= -(x- 3)( x+6)
Vertex
Now that you know the x value you can plug it back into the equation to get your Y value.
y = -(-3/2 - 3) (-3/2 + 6)
y = (3/2 + 3) (-3/2 + 6)
y= (9/2) ( 9/2)
y= (81/4)
The vertex is located at (-3/2, 81/4) or (-1.5, 20.25)
Y= -(x- 3)( x+6)
Maximum/ Minimum
Knowing that the parabola opens down we also know that the vertex will be highest point or the max. So the max. of the Quadratic Function will be y = (81/4).
Y-Intercept:
y= -(0-3)(0+6)
y= (3)(6)
y=18
(0,18)
Y- intercept
To find the y- in. one must substitute the "X" for zero, distribute the negative to the 3 and multiply.
Extra Points
y= -( x-3) ( x+6)
(1, 14) - ( 1-3)(1+6)
(-1+3) *( 1+6)
(2)(7) =14
( -2, 20) -( -2-3) (-2+6)
(2+3)(-2+6)
(5)( 4)= 20
(2, 8) -(2-3)(2+6)
(-2+3)(2+6)
(1)(8)= 8
The axis symmetry cuts the parabola into two equal halves so the point below I will show you will be equidistant from the Y- intercept.
(-3,18) -(-3-3) ( -3+6)
(3+3) ( -3+6)
(6)( 3) = 18
Y= -(x- 3)( x+6)
X- Intercepts
Graph
(-3,18)
0= x-3 0= x+6
+3 +3 -6 -6
3=x x=-6
(3,0) (-6,0)