What is the vertex of $3y= 2(x-7)^2 -5$ ?

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Answer 1 · 2018-05-31T06:48:04

Transform the function into vertex form, and match the values.

The vertex form is: $y = a(x-h)^2 + k$ , where $(h,k)$ is the location of the vertex.

To convert the original equation into this form, we can divide both sides of the equation by 3: $y = (2/3)(x-7)^2 - 5/3$

Reading from this equation we can see that $h = 7$ and $k = -5/3$ , and therefore the vertex is located at $(7,-5/3)$ .

What is the vertex of 3y= 2(x-7)^2 -53y= 2(x-7)^2 -5?