# How do you find the vertex and intercepts for y = -4(x + 3)^2 - 6?

Feb 23, 2016

x_("intercepts") -> "none"
${y}_{\text{intercepts}} \to - 42$

$\text{vertex } \to \left(x , y\right) \to \left(- 3 , - 6\right)$

#### Explanation: Given: $\text{ } y = - 4 {\left(x + 3\right)}^{2} - 6$

This equation is quadratic in Vertex Form.

Notice that if you expand the brackets you would have $- 4 {x}^{2}$
As the coefficient of ${x}^{2}$ is negative it means that you have a graph of the form $\cap$

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$\textcolor{b l u e}{\text{To find the vertex}}$

Consider the content of the brackets. You have +3.

Multiply this by negative 1 and you have the vertex x-value

$\textcolor{b l u e}{{x}_{\text{vertex}} = \left(- 1\right) \times 3 = - 3}$
Substitute -3 for $x$ in the equation and you have

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

color(blue)(y_("vertex")= 0-6" "=" "-6)

$\textcolor{b r o w n}{\text{vertex " ->" "(x,y)" "->" "(-3,-6)" }}$(below the x-axis)

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$\textcolor{b l u e}{\text{To find the x-intercepts}}$

And is of shape $\cap$ then the curve does not cross the x=axis
so there is no intercepts on the x-axis

$\textcolor{b l u e}{\text{None}}$
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$\textcolor{b l u e}{\text{To find the y-intercept}}$

The y-axis intercept is when $x = 0$

${y}_{\text{intercept}} = - 4 {\left(0 + 3\right)}^{2} - 6$

$\textcolor{b l u e}{{y}_{\text{intercept}} = - 42}$