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Thank you for helping me :)

 May 11, 2021
 #1
avatar+32833 
+2

Vertex is   4, 0

vertex form   =   y = a (x-4)^2   + 0      sub in the given point  (6,2)  to calculate 'a'

                          2 = a ( 6-4)^2

                              a = 1/2

 

your equation is then     y = 1/2 (x-4)^2 + 0

 May 11, 2021
 #2
avatar+498 
+1

General Equation of Parabola: $y=a(x-h)^2 + k$ where $(h,k)$ is the vertex of the parabola.

We have $y=a(x-4)^2 + 0$

We can’t the vertex $(4,0)$ again to plug in for $(x,y)$ because it will yield: $0=0+0.$

Let’s use the other point given(namely $(6,2)$): $2=a(6-4)^2 + 0$

We have $a=\frac{2}{4}=\frac{1}{2}.$

Thus the equation is $y=\frac{1}{2}(x-4)^2$

 May 11, 2021

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