zeros are at (-5,0) and (9,0), the vertex is (2,-2)
what is the factored form of this parabola? and standard form?
what is the factored form of this parabola? and standard form?
-
The zeros tell you that it factors as (x+5)(x-9), but you have to find the coefficient, using the vertex.
Y= a(x+5)(x-9)
plugging in the vertex for x and y,
-2= a(2+5)(2-9)
-2= a(7)(-7)
-2= -49a
a= (2/49)
So y = (2/49)(x+5)(x-9)
In standard form, y= (2/49)(x^2-4x-45)
Y = (2/49)x^2-(8/49)x-90/49
Hoping this helps!
Y= a(x+5)(x-9)
plugging in the vertex for x and y,
-2= a(2+5)(2-9)
-2= a(7)(-7)
-2= -49a
a= (2/49)
So y = (2/49)(x+5)(x-9)
In standard form, y= (2/49)(x^2-4x-45)
Y = (2/49)x^2-(8/49)x-90/49
Hoping this helps!
-
The fact that the vertex is at (2, -2) means that the equation of the parabola is of some from:
y = a*(x-2)^2 - 2
The -2 in the parentheses is due to the vertex being moved to +2 on the x-axis
The -2 outside the parentheses is due to the vertex being moved down to -2 with respect to the y-axis.
Now plug in either of the roots to solve for a. I'll use (9,0)
0 = a*(9-2)^2 - 2
0 = 49a - 2
49a = 2
a = 2/49
So the final equation is:
y = (2/49)*(x-2)^2 - 2
y = a*(x-2)^2 - 2
The -2 in the parentheses is due to the vertex being moved to +2 on the x-axis
The -2 outside the parentheses is due to the vertex being moved down to -2 with respect to the y-axis.
Now plug in either of the roots to solve for a. I'll use (9,0)
0 = a*(9-2)^2 - 2
0 = 49a - 2
49a = 2
a = 2/49
So the final equation is:
y = (2/49)*(x-2)^2 - 2
-
factored form is:
(x + 5) (x - 9) = 0
x^2 - 4x - 45 = 0
y = x^2 - 4x - 45 is the equation
(x + 5) (x - 9) = 0
x^2 - 4x - 45 = 0
y = x^2 - 4x - 45 is the equation