SOLUTION: I am graphing a parabola and I have all of my points done but I'm having a problem getting the x & y vertex's....the equation is 3x²-x-1=0 ...I have 1.5 for x but I'm told this is

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Question 111223: I am graphing a parabola and I have all of my points done but I'm having a problem getting the x & y vertex's....the equation is 3x²-x-1=0 ...I have 1.5 for x but I'm told this is wrong and I have no idea how to solve it for the y vertex. Please help!
Thanks so much!
B. Cook
Found 2 solutions by solver91311, checkley71:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
In general, for ax%5E2%2Bbx%2Bc, vertex at (h, k), h=-b%2F%282a%29, and k=ah%5E2%2Bbh%2Bc.
:
So, for your particular problem, b = -1 and a = 3. That makes h=%28-%28-1%29%29%2F%282%2A3%29=1%2F6, and
:
k=3%281%2F6%29%5E2-%281%2F6%29-1
k=3%2F36-1%2F6-1
k=3%2F36-6%2F36-36%2F36
k=-39%2F36
:
Putting the vertex squarely on the point (1/6,-39/36)
:
Send me a note if there is any part of that you don't understand.
:
John

Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
Y=3X^2-X-1
+graph%28+300%2C+200%2C+-6%2C+5%2C+-10%2C+10%2C+y+=+3x%5E2+-x+-1%29+ (graph 300x200 pixels, x from -6 to 5, y from -10 to 10, y = 3x^2 -x -1).
x=-b/2a
x=-(-1/2*3)
x=-(-1/6)
x=1/6 answer for the x coordinate of the vertex.
y=3(1/6)^2-1/6-1
y=3(1/36)-1/6-1
y=3/36-1/6-1
y=1/12-1/6-1
y=(1-2-12)/12
y=-13/12 the y coordinate of the vertex.
vertex=(1/6,-13/12)