# How do you find the axis of symmetry and vertex point of the function: Y=3x^2 + 8x + 4?

Oct 23, 2015

The axis of symmetry is $- \frac{4}{3}$.
The vertex is $\left(- \frac{4}{3} , - \frac{4}{3}\right)$.

#### Explanation:

$y = 3 {x}^{2} + 8 x + 4$ is a quadratic equation in the form $a {x}^{2} + b x + c$, where $a = 3 , b = 8 , c = 4$.

Axis of Symmetry
An imaginary vertical line that divides the parabola into to equal halves.

Equation: $x = \frac{- b}{2 a}$

$x = \frac{- 8}{2 \cdot 3} = - \frac{8}{6} = - \frac{4}{3}$

Axis of symmetry is $x = - \frac{4}{3}$.

Vertex
The maximum or minimum point of the parabola. Since $a$ is a positive number, the parabola opens upward, so the vertex is the minimum point.

The $x$ value of the point is the axis of symmetry, $x = - \frac{4}{3}$. To find the $y$ value, substitute $- \frac{4}{3}$ for $x$ in the equation and solve for $y$.

$y = 3 {x}^{2} + 8 x + 4 =$

$y = 3 {\left(- \frac{4}{3}\right)}^{2} + 8 \left(- \frac{4}{3}\right) + 4 =$

$y = 3 \left(\frac{16}{9}\right) - \frac{32}{3} + 4 =$

$y = \frac{48}{9} - \frac{32}{3} + 4 =$

The LCD is $9$. Make each term have a denominator of $9$.

$y = \frac{48}{9} - \frac{32}{3} \cdot \frac{3}{3} + 4 \cdot \frac{9}{9} =$

$y = \frac{48}{9} - \frac{96}{9} + \frac{36}{9}$

$y = \frac{48 - 96 + 36}{9} =$

$y = - \frac{12}{9}$

Simplify.

$y = - \frac{4}{3}$

The vertex is $\left(- \frac{4}{3} , - \frac{4}{3}\right)$

graph{y=3x^2+8x+4 [-10, 10, -5, 5]}