# How do you find the vertex and the intercepts for  -3x^2 - 3x + 1?

Apr 17, 2016

Vertex $\left(\frac{- 1}{3} , \frac{5}{3}\right)$

Y-intercept

$\left(1 , 0\right)$

X-Intercepts

$\left(\frac{3 + \sqrt{21}}{- 6} , 0\right)$
$\left(\frac{3 - \sqrt{21}}{- 6} , 0\right)$

#### Explanation:

Given -

$y = - 3 {x}^{2} - 3 x + 1$

Find the vertex.

x- co-ordinate of the vetex

$x = \frac{- \left(- b\right)}{2 a} = \frac{- \left(- 3\right)}{2 \times \left(- 3\right)} = \frac{3}{- 6} = \frac{- 1}{3}$

Y-co-ordinate

$y = - 3 \left(\frac{- 1}{3}\right) - 3 \left(\frac{- 1}{3}\right) + 1$
$y = - 3 \left(\frac{1}{9}\right) + \frac{3}{3} + 1$
$y = - \frac{1}{3} + 1 + 1$
$y = \frac{5}{3}$

Vertex $\left(\frac{- 1}{3} , \frac{5}{3}\right)$

Find Intercepts

Y- intercept at $x = 0$

$y = - 3 \left(0\right) - 3 \left(0\right) + 1$

$\left(1 , 0\right)$

X Intercept, at $\left(y = 0\right)$

$- 3 {x}^{2} - 3 x + 1 = 0$

$x = \frac{\left(- b\right) \pm \sqrt{{\left(- b\right)}^{2} - \left(4 a c\right)}}{2 a}$
$x = \frac{\left(- \left(- 3\right)\right) \pm \sqrt{{\left(- 3\right)}^{2} - \left(4 \times \left(- 3\right) \times 1\right)}}{2 \times \left(- 3\right)}$
$x = \frac{3 \pm \sqrt{9 - \left(- 12\right)}}{- 6}$
$x = \frac{3 \pm \sqrt{21}}{- 6}$
$\left(\frac{3 + \sqrt{21}}{- 6} , 0\right)$
$\left(\frac{3 - \sqrt{21}}{- 6} , 0\right)$