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.
The vertex of a parabola is the point where the parabola reaches its maximum or minimum value. It is also known as the "turning point" of the parabola.
To find the vertex of a parabola from its equation, you can use the formula x = -b/2a, where a and b are the coefficients of the quadratic equation in the form of ax^2 + bx + c. Substituting the value of x in the equation will give you the y-coordinate of the vertex.
Yes, you can still find the vertex of a parabola if the equation is not in standard form. You can use the process of completing the square to convert the equation into standard form and then use the formula x = -b/2a to find the vertex.
The a coefficient in a quadratic equation represents the coefficient of the x^2 term. It determines the shape of the parabola and whether it opens upwards or downwards.
Yes, the vertex is always the highest or lowest point on a parabola. If the parabola opens upwards, the vertex will be the minimum point, and if it opens downwards, the vertex will be the maximum point.