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What is the vertex of #y=-2(x+3)^2+12x+4 #?

Answer 1

Abigail Allen

vertex #(0, -14)#

Given -

#y=-2(x+3)^2+12x+4#

#y=-2(x^2+6x+9)+12x+4#

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

#y=-2x^2-14#

#x# term is missing in the expression #-2x^2-14# Let us supply it.

#y=-2x^2+0x-14#

#x=(-b)/(2xxa)=0/(2xx (-2))=0#

At #x=0#

#y=-2(0)^2-14=-14#

vertex #(0, -14)#

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Answer 2

Nathan Axel

To find the vertex of the quadratic function (y = -2(x + 3)^2 + 12x + 4), we'll first rewrite it in vertex form: (y = a(x - h)^2 + k). Then, we'll identify the values of (h) and (k), which represent the coordinates of the vertex.

Given function: (y = -2(x + 3)^2 + 12x + 4)

Vertex form: (y = -2(x - (-3))^2 + 4)

Comparing with vertex form: (h = -3) and (k = 4)

Thus, the vertex of the given function is ((-3, 4)).

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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.