What is the vertex of #y=9(5x-2)^2+3#?

1 Answer
Konstantinos Michailidis
Feb 4, 2016

The vertex form is written as #y = a(x - h)^2 + k#, where the a value determines the direction of opening (if a > 0, then it opens upwards and if a < 0, then it opens downward); h represents the x-coordinate of the vertex; k represents the y-coordinate of the vertex; x and y represent the x- and y-coordinates of a point on the parabola, other than the vertex.

Hence in our case we write this as

#y=9*(5(x-2/5))^2+3=>y=9*5^2*(x-2/5)^2+3#

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