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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