What is the vertex of  y=-3x^2-4x+2 ?

Dec 4, 2015

$\left(- \frac{2}{3} , \frac{10}{3}\right)$

Explanation:

The vertex of a quadratic equation can be found through the vertex formula:

$\left(- \frac{b}{2 a} , f \left(- \frac{b}{2 a}\right)\right)$

The letters represent the coefficients in the standard form of a quadratic equation $a {x}^{2} + b x + c$.

Here:
$a = - 3$
$b = - 4$

Find the $x$-coordinate of the vertex.

$- \frac{b}{2 a} = - \frac{- 4}{2 \left(- 3\right)} = - \frac{2}{3}$

The $y$-coordinate is found by plugging $- \frac{2}{3}$ into the original equation.

$- 3 {\left(- \frac{2}{3}\right)}^{2} - 4 \left(- \frac{2}{3}\right) + 2 = - 3 \left(\frac{4}{9}\right) + \frac{8}{3} + 2$
$= - \frac{4}{3} + \frac{8}{3} + \frac{6}{3} = \frac{10}{3}$

Thus, the vertex is located at the point $\left(- \frac{2}{3} , \frac{10}{3}\right)$.

This can also be found through putting the quadratic into vertex form $y = a {\left(x - h\right)}^{2} + k$ by completing the square.

y=-3(x^2+4/3x+?)+2

$y = - 3 \left({x}^{2} + \frac{4}{3} x + \textcolor{b l u e}{\frac{4}{9}}\right) + 2 + \textcolor{b l u e}{\frac{4}{3}}$

$y = - 3 {\left(x + \frac{2}{3}\right)}^{2} + \frac{10}{3}$

Again, the vertex is located at the point $\left(- \frac{2}{3} , \frac{10}{3}\right)$.