What is the vertex form of $y = 3 x^{2} - 11 x + 6$ $y= 3x^2-11x+6$ ?

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Answer 1 · 2017-08-11T07:08:36

$(\frac{11}{6}, - \frac{49}{12})$ $(11/6, -49/12)$

The $x$ $x$ value of the axis of symmetry is the same as the $x$ $x$ value of the vertex.

Use the axis of symmetry formula $x = - \frac{b}{2 a}$ $x=-b/(2a)$ to find the $x$ $x$ value of the vertex.

$x = \frac{- (- 11)}{2 (3)}$ $x = (-(-11))/(2(3))$

$x = \frac{11}{6}$ $x = 11/6$

Substitute $x = \frac{11}{6}$ $x = 11/6$ into the original equation for the $y$ $y$ value of the vertex.

$y = 3 {(\frac{11}{6})}^{2} - 11 (\frac{11}{6}) + 6$ $y = 3(11/6)^2 - 11(11/6) + 6$

$y = - \frac{49}{12}$ $y = -49/12$

Therefore, the vertex is at $(\frac{11}{6}, - \frac{49}{12})$ $(11/6, -49/12)$ .

What is the vertex form of y=3x2−11x+6y= 3x^2-11x+6 ?