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

Feb 27, 2017

$y$-intercept = 24
no $x$-intercepts
vertex: $\left(- 3 , 6\right)$, the minimum value for y

#### Explanation:

Sorry if the explanation is very long (it's actually very simple), I had to make sure people would be able to fully understand what was going on. If it was shorter, there could be lots of confusion.

The $x$-intercept(s) can be found by making $y = 0$ then solving for $x$;

$2 {\left(x + 3\right)}^{2} + 6 = 0$
$2 {\left(x + 3\right)}^{2} = - 6$
${\left(x + 3\right)}^{2} = - 3$

At this stage you'll notice that negative numbers do not have square roots, and so there is it is impossible to find a value for $x$ (when $y = 0$), meaning there are no $x$-intercepts, and the graph will not cross the $x$-axis at any point.

To make sure, you could check by finding the discriminant ($\Delta$) of the line;
$\Delta = {b}^{2} - 4 a c$
Where $a$ is the coefficient of ${x}^{2}$, $b$ is the coefficient of ${x}^{1}$, and $c$ is the constant (coefficient of ${x}^{0}$). Make sure you simplify the equation fully before taking the coefficients.

If $\Delta > 0$, there are two $x$-intercepts, and the graph is said to have distinct roots.
If $\Delta = 0$, there is one $x$-intercept and the graph is said to have repeating roots.
If $\Delta < 0$, there are no $x$-intercepts, and the graph is said to have imaginary roots.

In this case;
$\Delta = {\left(12\right)}^{2} - 4 \left(2\right) \left(24\right)$
$\Delta = - 48$
$\Delta < 0$ and so $x$ has imaginary roots. Confirming our judgment.

The $y$-intercept(s) can be found by making $x = 0$ then solving for $y$;

$y = 2 {\left(0 + 3\right)}^{2} + 6$
$y = 18 + 6$
$y = 24$

The vertex can be found by carrying out either of two methods;

1. Memorizing and using this expression:

Vertex is $\left(- b / 2 a , 4 a c - {b}^{2} / 4 a\right)$

Where as stated above, $a$, $b$ and $c$ are the coefficients of indices of $x$ in the simplified equation.
(I'd rather not use this method though.)
2. Completing the square, then taking the coordinate values from it.

In the equation $y = a {\left(x + p\right)}^{2} + q$, the vertex is $\left(- p , q\right)$

So in our case, $p$ is $3$ and $q$ is $6$, which means the vertex is $\left(- 3 , 6\right)$.

$a$ will tell you if the quadratic curve is positive or negative, and so if the vertex is a maximum or minimum value for $y$ in the curve. This is referred to as the "nature of the vertex" in exam-type questions.
If $a$ is positive (like it is in out case), then the graph is positive and the vertex is at the minimum value for $y$. The graph doesn't have a maximum value for $y$. The reason for this is obvious if you observe the shape of a positive quadratic graph.

The equation you submitted is already in completed square format and so this is obviously the easier method of the two.