Respuesta :

1)Rewrite in standard form and use that to find the vertex, which would be (-2,15)   so second option
2) second option



(both of the attachments are for the second problem, I was too lazy to write it out :P)

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irspow
vertex of -3x^2-12x+3

dy/dx=-6x-12, dy/dx=0 when:

-6x-12=0

-6x=12

x=-2

y(2)=-3x^2-12x+3=-12+24+3

y(2)=15

So the vertex is the point (-2, 15)

...

y=3x^2+12x-6

dy/dx=6x+12, dy/dx=0 when

6x+12=0

6x=-12

x=-2

y(-2)=3x^2+12x-6=12-24-6=-18

So it is the second one which has its vertex and absolute minimum at the point (-2, -18)

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