Vertex of Parabola: Find y^2-12=12x Solution

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Homework Statement


I'm trying to find the vertex of this parabola: y^2-12=12x

The Attempt at a Solution


Initially I tried to take the derivative and find the vertex by finding where it the function is zero, but apparently that does not work in this situation. Is it because it is not in standard form? I've been staring at this problem for hours. I need a hint.

Thanks.
 
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Sarah Kenney said:

Homework Statement


I'm trying to find the vertex of this parabola: y^2-12=12x

The Attempt at a Solution


Initially I tried to take the derivative and find the vertex by finding where it the function is zero, but apparently that does not work in this situation. Is it because it is not in standard form? I've been staring at this problem for hours. I need a hint.

Thanks.
You could have made a sketch of this equation in about 5 minutes and figured out where the vertex is located.

Remember, since the parabola is in y2, setting dy/dx = 0 isn't going to work for finding the vertex.
 
Ok, so I figured it out.
It's:
y^2=12x+12
y^2=12(x+1)
x=-1
So y= 0
I don't know why that took me so long. Math brain.
Thanks.
 
Sarah Kenney said:
Ok, so I figured it out.
It's:
y^2=12x+12
y^2=12(x+1)
x=-1
So y= 0
I don't know why that took me so long. Math brain.
Thanks.
In future posts, please show what you have tried, rather than describe it.
 
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