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

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Answer 1 · 2016-02-23T11:19:15

$x_("intercepts") -> "none"$
$y_("intercepts") -> -42$

$"vertex "->(x,y)->(-3,-6)$

Tony B

Given: $" "y=-4(x+3)^2-6$

This equation is quadratic in Vertex Form.

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

'~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("To find the vertex")$

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

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

$color(blue)(x_("vertex")=(-1)xx3=-3)$
Substitute -3 for $x$ in the equation and you have

$y_("vertex")=-4(-3+3)^2-6$

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

$color(brown)( "vertex " ->" "(x,y)" "->" "(-3,-6)" " )$ (below the x-axis)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$color(blue)("To find the x-intercepts")$

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

$color(blue)("None")$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("To find the y-intercept")$

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

$y_("intercept") =-4(0+3)^2-6$

$color(blue)(y_("intercept") =-42)$

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