# How do you identify the vertex for the graph of y = −3x^2 + 6x − 4?

May 4, 2017

$\left(1 , - 1\right)$

#### Explanation:

$\text{for the standard quadratic function } y = a {x}^{2} + b x + c$

"the x-coordinate of the vertex " x_(color(red)"vertex")=-b/(2a)

$\text{for } y = - 3 {x}^{2} + 6 x - 4$

$a = - 3 , b = 6 \text{ and } c = - 4$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{6}{- 6} = 1$

$\text{substitute this value into function and evaluate for y}$

${y}_{\textcolor{red}{\text{vertex}}} = - 3 {\left(1\right)}^{2} + 6 \left(1\right) - 4 = - 1$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(1 , - 1\right)$
graph{-3x^2+6x-4 [-10, 10, -5, 5]}