What is the vertex of # y= (x+2)^2-3x^2-4x-4#?

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Answer 1 · 2016-08-21T17:06:21

Vertex is at the origin #(0,0)#

This is a somewhat unusual format for a parabola! Simplify first to see what we are working with..

#y = x^2 +4x +4 -3x^2 -4x -4 = -2x^2 #

What does an equation tell us about the parabola?

The standard form is #y = color(red)(a)x^2 + color(blue)(b)x + color(magenta)(c)#

#color(red)(a)# changes the shape of the parabola - whether it is narrow or wide, or open upwards or downwards.

#color(blue)(b)x# moves the parabola to the left or right

#color(magenta)(c) # gives the y-intercept. It moves the parabola up or down.

In #y = -2x^2# there is no x-term, and #c = 0#

This means that the parabola has not moved to the left or right, nor has it moved up or down, although it is 'upside-down' with a maximum TP.

Its vertex is at the origin #(0,0)#

Changing it to vertex form will give #y =-2(x+0)^2 + 0#
graph{-2x^2 [-4.92, 5.08, -3.86, 1.14]}