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Find the vertex of a parabola with an equation of 3x^2+9x-y-3=0.

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Answer:

Explanation:

We will have to convert this to vertex form.

First we add y to both sides.

y = 3x^2 + 9x - 3

The vertex form is y = a(x – h)^2 + k

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

We need to complete the square now, and half if 3 is 1.5, and 1.5^2 is 2.25, so we add 2.25 inside the parentheses but we have to be careful, because if we subtract 2.25 from outside the parentheses it wouldn't work because the 2.25 inside the parentheses is being multiplied by 3, so we have to subtract 2.25*3=6.75 from the outside.

y=3(x^2 + 3x + 2.25) - 9.75

we can simplify this now to

y=3(x + 1.5)^2 - 9.75

and now this is vertex form. From here it is incredibly easy to get the vertex. Just take the x value required to make the total in the parentheses zero, so in this problem it would be -1.5 (Because (-1.5+1.5)^2 is zero) and then you have your x coordinate. From there we just take k, which in this problem is -9.75, and it is the y value of our vertex.

In conclusion, the vertex is (-1.5. -9.75)

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