# How do you find the vertex and intercepts for f(x) = 3x^2 + 6x + 1?

Oct 13, 2017

Vertex (-1,-2)
y-intercept (0,1)
x-intercept 1.8165, 0.1835

#### Explanation:

$y = a \cdot {\left(x - h\right)}^{2} + k$ is the equation of parabola, with vertex (h,k).
$y = 3 {x}^{2} + 6 x + 1 = 3 \cdot \left({x}^{2} + 2 x + 1\right) - 2$
$\therefore y = 3 \cdot {\left(x + 1\right)}^{2} - 2$
Vertex (h,k) = $\left(- 1 , - 2\right)$

To find y-intercept, let $x = 0$
$y = 3 - 2 = 1$;

To find x-intercept, let $y = 0$
$0 = 3 {\left(x + 1\right)}^{2} - 2$
${\left(x + 1\right)}^{2} = \frac{2}{3}$
$x = 1 - \left(\pm \sqrt{\frac{2}{3}}\right)$
$x = 1 + \sqrt{\frac{2}{3}} , 1 - \sqrt{\frac{2}{3}} = 1.8165 , 0.1835$