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

May 21, 2018

$v e r t e x \left(\frac{1}{3} , - \frac{11}{3}\right)$

#### Explanation:

${a}^{2} + b x + c$

y= -3x^2 + 2x − 4

a = -3
b=2
c=-4

vertex is $\left(h , k\right)$

$h = \frac{- b}{2 a}$

$k = f \left(h\right)$ i.e. put what you found for h back into your function as x and solve for y.

$h = \frac{- 2}{2 \cdot - 3} = \frac{1}{3}$

k = (-3*1/3)^2 + 2*1/3 − 4 = -11/3

$v e r t e x \left(\frac{1}{3} , \frac{1}{3}\right)$

May 21, 2018

$\text{vertex } = \left(\frac{1}{3} , - \frac{11}{3}\right)$

#### Explanation:

"given the equation in standard form ";y=ax^2+bx+c

$\text{then the x-coordinate of the vertex is}$

•color(white)(x)x_(color(red)"vertex")=-b/(2a)

$y = - 3 {x}^{2} + 2 x - 4 \text{ is in standard form}$

$\text{with "a=-3,b=2" and } c = - 4$

$\Rightarrow {x}_{\text{vertex}} = - \frac{2}{- 6} = \frac{1}{3}$

$\text{substitute this value into the equation for y}$

$y = - 3 {\left(\frac{1}{3}\right)}^{2} + \frac{2}{3} - 4$

$\textcolor{w h i t e}{y} = - \frac{1}{3} + \frac{2}{3} - \frac{12}{3} = - \frac{11}{3}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{1}{3} , - \frac{11}{3}\right)$