What is the vertex of 3x^2+6x+7?

Question

anonymous · Answer

Do you know the standard form of a parabola?

anonymous · Answer

a^2+b^2=c^2?

anonymous · Answer

No, that is a circle. For a parabola, it's y = a(x - h)^2 + k

anonymous · Answer

Where (h,k) is the vertex

anonymous · Answer

ahh okay thanks!

anonymous · Answer

Do you know how to put the equation into standard form?

anonymous · Answer

no sir

anonymous · Answer

Okay, do you know how to complete the square?

anonymous · Answer

so far all I have written down is a=3 b=6 and c=7 sorry, i'm really bad at math..

anonymous · Answer

I think you should brush up on your quadratics, but to complete the square, you have to first remove the coefficient in front of the leading term which is x^2. I would factor it out:

$$3x^2+6x+7 = 3(x^2 + 2x + \frac{7}{3})$$ complete the square: $$3[(x^2 + 2x +1) + \frac{7}{3} - 1] = 3[(x + 1)^2 + \frac{4}{3}]$$expand with the distributive property to get into the standard form: $$3[(x + 1)^2 + \frac{4}{3}] = 3(x+1)^2 + 4$$ and now you can identify your vertex (h,k) quite easily: (-1,4)