How do you find the vertex for $y=4/3x^(2)-8x+6$ ?

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Answer 1 · 2015-09-22T13:02:16

It is point $(3,-6)$

This is a parabola . The vertex of a parabola is the point where the parabola crosses its axis of symmetry.

The vertex formula for a parabola is

$y=a*(x-h)^2+k$ where $(h,k)$ is the vertex. Hence rewriting

the given parabola as

$y=4/3*(x-3)^2+(-6)$

we see that the vertex is $(3,-6)$

graph{4/3*(x^2)-8x+6 [-20, 20, -10, 10]}

How do you find the vertex for y=4/3x^(2)-8x+6y=4/3x^(2)-8x+6?