# How do you graph the parabola y = (x + 2) (3x + 2) using vertex, intercepts and additional points?

Dec 6, 2017

$x$-intercepts: $\left(- 2 , 0\right)$ and $\left(- \setminus \frac{2}{3} , 0\right)$

$y$-intercept: $\left(0 , 4\right)$

Vertex: $\left(- \setminus \frac{4}{3} , - \setminus \frac{4}{3}\right)$

#### Explanation:

Convert it to standard form by FOILing the expression:

$\left(x + 2\right) \left(3 x + 2\right) \textcolor{red}{\setminus \implies} 3 {x}^{2} + 2 x + 6 x + 4 \textcolor{red}{\setminus \implies} 3 {x}^{2} + 8 x + 4$

Now that’s it in the form $a {x}^{2} + b x + c$, find the $x$-coordinate of the vertex using $- \setminus \frac{b}{2 a}$:

$- \setminus \frac{b}{2 a} \textcolor{red}{\setminus \implies} - \setminus \frac{8}{2 \setminus \cdot 3} \textcolor{red}{\setminus \implies} - \setminus \frac{4}{3}$

Plug that into the standard form equation to find the $y$-coordinate:

$3 {x}^{2} + 8 x + 4 \textcolor{red}{\setminus \implies} 3 {\left(- \setminus \frac{4}{3}\right)}^{2} + 8 \left(- \setminus \frac{4}{3}\right) + 4 \textcolor{red}{\setminus \implies} \setminus \frac{48}{9} - \setminus \frac{32}{3} + 4 \textcolor{red}{\setminus \implies} \setminus \frac{48}{9} - \setminus \frac{96}{9} + \setminus \frac{36}{9} \textcolor{red}{\setminus \implies} - \setminus \frac{4}{3}$

$\textcolor{red}{\setminus \therefore}$ the vertex is $\left(- \setminus \frac{4}{3} , - \setminus \frac{4}{3}\right)$

To find the $x$-intercepts, set each parenthetical expression of the original form of the equation (in the question) to $0$, and solve for $x$:

$x + 2 = 0 \setminus q \quad , \setminus q \quad 3 x + 2 = 0$

$\textcolor{red}{\setminus \implies} x = - 2 \setminus q \quad , \setminus q \quad x = - \setminus \frac{2}{3}$

$\textcolor{red}{\setminus \therefore}$ the $x$-intercepts are $\left(- 2 , 0\right)$ and $\left(- \setminus \frac{2}{3} , 0\right)$.

Finally, the $y$-intercept is $\left(0 , c\right) \textcolor{red}{\setminus \implies} \left(0 , 4\right)$

Here’s the graph: