The discussion focuses on finding the vertex of the parabola defined by the equation y² - x + 4y + k = 0, which passes through the point (12, 1). Participants confirm that the parabola is sideways since the y variable is squared. To find the vertex, users suggest solving for k by substituting the point values into the equation and then completing the square to express the equation in the form x = (y + a)² + b, where the vertex can be derived as (-b, -a).
PREREQUISITESStudents, educators, and anyone interested in mastering the concepts of quadratic equations and parabolas, particularly in determining their vertices and understanding their orientations.
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.