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

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- Thread starter DLxX
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- #1

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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:

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 ?

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- #3

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Plug in the point values to solve for k.

What kind of parabola is this? Sideways or vertical? (hint)

What kind of parabola is this? Sideways or vertical? (hint)

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would be a sideways since the y is squared, but how would i find the h in (h,k)?Jameson said:Plug in the point values to solve for k.

What kind of parabola is this? Sideways or vertical? (hint)

- #5

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Show me the equation in the form of x =. Start from there. Did you solve for k?

- #6

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x = y^2 + 4y + k. So to solve for k do I then plug in 12 and 1 for x and y?Jameson said:Show me the equation in the form of x =. Start from there. Did you solve for k?

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Alright so that would give me the k value, but what about the h value?DeathKnight said:Yup thats what you have to do.

- #8

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plug in values , then just complete the square to put it in the formDLxX said:

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

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

and b,-a is the vertex

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