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)#