SOLUTION: I have sent in my graph and solutions to my school and they sent my work back saying my vertex was wrong. The equation is -3x^2-6x-5. I had (0,-5) as my vertex and I can't figure

Question 836304: I have sent in my graph and solutions to my school and they sent my work back saying my vertex was wrong. The equation is -3x^2-6x-5. I had (0,-5) as my vertex and I can't figure out why its wrong or what else it could. Please help.

Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
In general, if you have y = ax^2 + bx + c, the axis of symmetry is x = -b/(2a)

In this case, a = -3, b = -6

x = -b/(2a)

x = -(-6)/(2(-3))

x = 6/(-6)

x = -1

The axis of symmetry, and the x coordinate of the vertex, is x = -1

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Plug this into the given equation to find the y coordinate of the vertex

y = -3x^2-6x-5

y = -3(-1)^2-6(-1)-5

y = -3(1)-6(-1)-5

y = -3 + 6 - 5

y = -2

So the vertex is (-1,-2)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I have sent in my graph and solutions to my school and they sent my work back saying my vertex was wrong. The equation is -3x^2-6x-5. I had (0,-5) as my vertex and I can't figure out why its wrong or what else it could.
Note: (0,-5) is the y-intercept.
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Your Problem:
Vertex occurs where x = -b/(2a) = 6/(-6) = -1
f(-1) = -3(-1)^2-6(-1)-5 = -3+6-5 = -2
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Vertex is (-1,-2)
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Cheers,
Stan H.
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