The discussion revolves around finding the vertex of the parabola defined by the equation y² - x + 4y + k = 0, given that it passes through the point (12,1). Participants explore various methods to approach the problem, including differentiation, completing the square, and identifying the type of parabola.
Participants generally agree on the need to find k and the vertex, but there are multiple approaches suggested, and no consensus on a single method or solution has been reached.
There are unresolved steps regarding the calculation of k and the transformation of the equation, as well as the implications of the parabola's orientation.
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.
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)
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?
Alright so that would give me the k value, but what about the h value?DeathKnight said:Yup that's what you have to do.
plug in values , then just complete the square to put it in the formDLxX 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.