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)(x9), but you have to find the coefficient, using the vertex.
Y= a(x+5)(x9)
plugging in the vertex for x and y,
2= a(2+5)(29)
2= a(7)(7)
2= 49a
a= (2/49)
So y = (2/49)(x+5)(x9)
In standard form, y= (2/49)(x^24x45)
Y = (2/49)x^2(8/49)x90/49
Hoping this helps!
Y= a(x+5)(x9)
plugging in the vertex for x and y,
2= a(2+5)(29)
2= a(7)(7)
2= 49a
a= (2/49)
So y = (2/49)(x+5)(x9)
In standard form, y= (2/49)(x^24x45)
Y = (2/49)x^2(8/49)x90/49
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*(x2)^2  2
The 2 in the parentheses is due to the vertex being moved to +2 on the xaxis
The 2 outside the parentheses is due to the vertex being moved down to 2 with respect to the yaxis.
Now plug in either of the roots to solve for a. I'll use (9,0)
0 = a*(92)^2  2
0 = 49a  2
49a = 2
a = 2/49
So the final equation is:
y = (2/49)*(x2)^2  2
y = a*(x2)^2  2
The 2 in the parentheses is due to the vertex being moved to +2 on the xaxis
The 2 outside the parentheses is due to the vertex being moved down to 2 with respect to the yaxis.
Now plug in either of the roots to solve for a. I'll use (9,0)
0 = a*(92)^2  2
0 = 49a  2
49a = 2
a = 2/49
So the final equation is:
y = (2/49)*(x2)^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