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What is the vertex form of #y=6x^2+17x+12 #?

Answer 1

Isaiah Anthony

#6(x+17/32)^2 + 5277/512# This is the required vertex form. Vertex is #(-17/32, 5277/512)#

It is #y= 6(x^2 +(17x)/6) +12#

= #6(x^2 + (17x)/16 + 289/1024 -289/1024)+12#

= #6(x + 17/32)^2 + 12 -6(289/1024)#

=#6(x+17/32)^2 + 5277/512# This is the required vertex form. Vertex is #(-17/32, 5277/512)#

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Answer 2

Eli Assouline

The vertex form of ( y = 6x^2 + 17x + 12 ) is ( y = 6(x + \frac{17}{12})^2 - \frac{17^2}{12} + 12 ).

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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.