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What is the vertex of # y= 3x^2+x+6+3(x-4)^2#?

Answer 1

#(23/12, 767/24)#

Our best option to solve this problem is to expand everything and write the equation in standard form: hmm, this parabola isn't in standard form or vertex form.

#f(x) = ax^2+bx+c#

where #a,b,# and #c# are constants and #((-b)/(2a), f((-b)/(2a)))# is the vertex.

#y = 3x^2+x+6+3(x^2-8x+16)#

#y = 3x^2+x+6+3x^2-24x+48#

#y = 6x^2-23x+54#

Now we have the parabola in standard form, where #a=6# and #b=-23#, so the #x# coordinate of the vertex is:

#(-b)/(2a) = 23/12#

Finally, we need to plug this #x# value back into the equation to find the #y# value of the vertex.

#y = 6(23/12)^2-23(23/12)+54#

#y = 529/24 - 529/12 + 54#

#y = -529/24 + (54*24)/24#

#y = (1296-529)/24 = 767/24#

So the vertex is #(23/12, 767/24)#

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Answer 2

Eva Adamson

The vertex of the quadratic function ( y = 3x^2 + x + 6 + 3(x-4)^2 ) is (4, 6).

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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.