# How do you find the vertex and the intercepts for y = 3x^2 + 12x?

$\textcolor{red}{\text{Vertex at } \left(- 2 , - 12\right)}$
color(red)("The intercepts are " (0, 0) and (-4, 0)

#### Explanation:

From the given equation
$y = 3 {x}^{2} + 12 x$

We transform the equation by Completing the square method

$y = 3 {x}^{2} + 12 x$
factor out 3

$y = 3 \left({x}^{2} + 4 x\right)$

The coefficient 4 will be divided by 2 and the result will be squared giving 4. This is the number to be added and subtracted inside the grouping symbol.

$y = 3 \left({x}^{2} + 4 x\right)$

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

Now notice the ${x}^{2} + 4 x + 4$ is a perfect square trinomial reducible to ${\left(x + 2\right)}^{2}$, so that

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

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

Distribute the 3 back

$y = 3 {\left(x + 2\right)}^{2} - 12$

transpose the -12 to the left of the equation

$y + 12 = 3 {\left(x + 2\right)}^{2}$

divide both sides by 3

$\frac{1}{3} \left(y + 12\right) = {\left(x + 2\right)}^{2}$

We now have the Vertex Form

${\left(x - - 2\right)}^{2} = \frac{1}{3} \left(y - - 12\right)$

with $\textcolor{red}{\text{Vertex at } \left(- 2 , - 12\right)}$

To solve for the intercepts, we will use the given general form
$y = 3 {x}^{2} + 12 x$

Intercepts are points of intersection of the curve with the axes.

To solve for the x-intercept, we set $y = 0$ then find $x$ values
$y = 3 {x}^{2} + 12 x$
$0 = 3 {x}^{2} + 12 x$
We can solve by factoring
$0 = 3 x \left(x + 4\right)$
We have two factors 3x and x+4 to be equated to 0.

$3 x = 0$

$x = \frac{0}{3}$
$x = 0$

the other one

$x + 4 = 0$
$x = - 4$

The x-intercepts are $\left(0 , 0\right)$ and $\left(- 4 , 0\right)$

Let us now solve the y-intercept by setting $x = 0$ this time, and solving for $y$ values

$y = 3 {x}^{2} + 12 x$
$y = 3 {\left(0\right)}^{2} + 12 \left(0\right)$
$y = 0 + 0$
$y = 0$

Therefore, the y-intercept is the same point $\left(0 , 0\right)$.

color(red)("The intercepts are " (0, 0) and (-4, 0)
$\textcolor{red}{\text{Vertex at } \left(- 2 , - 12\right)}$

Kindly inspect the graph of the equation and check the points that we solved

graph{(x--2)^2=1/3(y--12)[-30,30,-15,15]}

God bless....I hope the explanation is useful.