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

Answer 1

The vertex is #(4/3,-47/3)#

#y=-x^2-4x-3-2(x-3)^2#

Since this is not yet in vertex form, we must finish the square, expand and arrange the quadratic, and then find the vertex.

Expand: #y=-x^2-4x-3-2(x^2-6x+9)#

#y=-x^2-4x-3-2x^2+12x-18#

Organize: #y=-3x^2+8x-21#

Complete the square: #y=-3[x^2-(8x)/3+7]#

#y=-3[(x-4/3)^2-16/9+7]#

#y=-3[(x-4/3)^2+47/9]#

#y=-3(x-4/3)^2-3(47/9)#

#y=-3(x-4/3)^2-47/3#

Determine vertex: Vertex form is #y=a(x-color(red)(h))^2+color(blue)(k)# where #(color(red)(h),color(blue)(k))# is the vertex of the parabola.

The vertex is therefore at #(color(red)(4/3),color(blue)(-47/3))#.

Use the graph: graph{y=-x^2-4x-3-2(x-3)^2 [-30, 30, -30, 5]} to double check.

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Answer 2

Nathan Axel

The vertex of the function y = -x^2 - 4x - 3 - 2(x - 3)^2 is (-3, -9).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.