How do you find the vertex of a parabola $y= 1/3[x-1]^2+2$ ?

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Answer 1 · 2015-07-18T22:46:37

The vertex is at $color(red)((1,2)$ .

$y = -1/3(x-1)^2+2$

The vertex form for the equation of a parabola is

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

By comparing the two equations, we see that $h = 1$ and $k = 2$ .

The vertex is at ( $1,2$ ).

graph{-1/3(x-1)^2+2 [-5, 6, -5, 5]}

How do you find the vertex of a parabola y= 1/3[x-1]^2+2 y= 1/3[x-1]^2+2 ?