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

Nov 22, 2017

y-int $\left(0 , - 1\right)$
x-int $\left(- 1 \pm \frac{2}{3} \sqrt{3} , 0\right)$
vertex $\left(- 1 , - 4\right)$

#### Explanation:

to find the intercepts

let x = 0, in the equation for y-intercept
$3 \cdot {0}^{2} + 6 \cdot 0 - 1$
$= {3}^{2} - 1$
$= 0 - 1$
$= - 1$
$y -$int =>$\left(0 , - 1\right)$

let y = 0, in the equation for x-intercepts
since this equation is not factorable, use uadratic fomula to solve for $x$
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$3 {x}^{2} + 6 x - 1 = 0$
a=3
b=6
c=-1
$x = - 1 \pm \frac{- 2}{3} \sqrt{3}$

for parabola $a {x}^{2} + b x + c$ the vertex's x equals $- \frac{b}{2 a}$
$x = - \frac{6}{2 \cdot 3}$
$x = - 1$

plug in -1 to find y value

$y = 3 {\left(- 1\right)}^{2} + 6 \left(- 1\right) - 1$
$y = - 4$

vertex is $\left(- 1 , - 4\right)$