What is the vertex of $y = (\frac{1}{2}) {(3 x - 1)}^{2} + 3$ $y = (1/2)(3x - 1)^2 + 3$ ?

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Answer 1 · 2015-11-09T16:45:10

vertex $= (\frac{1}{3}, 3)$ $= (1/3, 3)$

If there is a coefficient in front of the x variable, always factor it out first. In this problem, factor out the 3:

$y = (\frac{1}{2}) (3^{2}) {(x - \frac{1}{3})}^{2} + 3$ $y=(1/2)(3^2)(x-1/3)^2+3$

Now, this is in vertex form:

vertex $= (\frac{1}{3}, 3)$ $= (1/3, 3)$

hope that helped

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