How do you find the quadratic function with vertex (6,6) and point (61/10, 3/2)?

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Answer 1 · 2017-07-26T12:28:30

The quadratic equation is # y= -450x^2 + 5400x -16194 #

The quadratic equation in vertex form is #y=a(x-h)^2+k ; (h,k)#

being vertex . So the equation with vertex as #6,6# is

#y=a(x-6)^2+6# . The point #61/10=6.1 , 3/2=1.5 or (6.1,1.5)#

lies on the equation. So the point will satisfy the equation.

#:. 1.5 = a(6.1-6)^2+6 or 1.5 = 0.01*a +6 or 0.01*a = 1.5-6 # or

#0.01*a= -4.5 or a = -4.5/0.01 = -450 #

The quadratic equation is #y=-450(x-6)^2+6 = -450(x^2-12x+36)+6#

or # y= -450x^2 + 5400x -16200 +6# or

# y= -450x^2 + 5400x -16194 # [Ans]