# Finding the vertex of a parabola from an equation

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.

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 thats 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