Finding the vertex of a parabola from an equation

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To find the vertex of the parabola given by the equation y^2 - x + 4y + k = 0, it is determined that the parabola is sideways since the y variable is squared. To solve for the constant k, the point (12,1) is substituted into the equation. After finding k, the equation can be rearranged into the form x = (y + a)^2 + b. Completing the square will then allow for the identification of the vertex, which is located at the coordinates (-b, -a). The discussion emphasizes the importance of plugging in values and completing the square to derive the vertex accurately.
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I need help with the following question.

The parabola y^2 - x + 4y + k = 0 passes through the point (12,1). Find the vertex of the parabola.
 
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DLxX said:
I need help with the following question.

The parabola y^2 - x + 4y + k = 0 passes through the point (12,1). Find the vertex of the parabola.


by differentiating ? or the line of symmetry between the roots of the equation ?



or completin the square ?
 
Last edited:
Plug in the point values to solve for k.

What kind of parabola is this? Sideways or vertical? (hint)
 
Jameson said:
Plug in the point values to solve for k.

What kind of parabola is this? Sideways or vertical? (hint)
would be a sideways since the y is squared, but how would i find the h in (h,k)?
 
Show me the equation in the form of x =. Start from there. Did you solve for k?
 
Jameson said:
Show me the equation in the form of x =. Start from there. Did you solve for k?
x = y^2 + 4y + k. So to solve for k do I then plug in 12 and 1 for x and y?
 
DeathKnight said:
Yup that's what you have to do.
Alright so that would give me the k value, but what about the h value?
 
DLxX said:
I need help with the following question.

The parabola y^2 - x + 4y + k = 0 passes through the point (12,1). Find the vertex of the parabola.
plug in values , then just complete the square to put it in the form

f(y) = x = (y+a)^2+b

and b,-a is the vertex
 

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