It is (y - b)^2 = ax + c
The equation of a parabola that opens left or right with its vertex at the point ((h, v)) is given by ((y - v)^2 = 4p(x - h)), where (p) is the distance from the vertex to the focus. If (p > 0), the parabola opens to the right, and if (p < 0), it opens to the left.
An equation for a sideways parabola can be expressed in the form ( y^2 = 4px ) for a parabola that opens to the right, or ( y^2 = -4px ) for one that opens to the left. Here, ( p ) represents the distance from the vertex to the focus. The vertex of the parabola is at the origin (0,0), and the axis of symmetry is horizontal. If the vertex is not at the origin, the equation can be adjusted to ( (y-k)^2 = 4p(x-h) ), where ((h, k)) is the vertex.
A parabola opens upward when its leading coefficient (the coefficient of the (x^2) term in the quadratic equation (y = ax^2 + bx + c)) is positive. This means that as you move away from the vertex of the parabola in both the left and right directions, the values of (y) increase. Consequently, the vertex serves as the minimum point of the parabola.
Above
The value of ( b ) in a quadratic equation of the form ( y = ax^2 + bx + c ) affects the position and shape of the parabola. Specifically, it influences the location of the vertex along the x-axis and the direction in which the parabola opens. A larger absolute value of ( b ) can make the parabola wider or narrower depending on the value of ( a ), while the sign of ( b ) can shift the vertex left or right. Overall, these changes alter how the parabola intersects with the x-axis and its symmetry.
Above
when you have y=+/-x2 +whatever, the parabola opens up y=-(x2 +whatever), the parabola opens down x=+/-y2 +whatever, the parabola opens right x=-(y2 +whatever), the parabola opens left so, your answer is up
x= ay² + by + c Apex :3
Solution 1Start by putting the parabola's equation into the form y = ax2 + bx + c if it opens up or down,or x = ay2 + by + c if it is opens to the left or right,where a, b, and c are constants.The x-value for the vertex is -(b/2a). You can use this x-value to solve for the y-value by substituting the x value in the original quadratic equation.Solution 2Put the parabola's equation into this form: y - k = 4p(x - h)2or x - h = 4p(y - k)2You just need to simplify the equation until it looks like this. The vertex is located at the coordinates (h,k). (p is for the focus, but that isn't important as long as you know h and k.)
Nose points right, opens to the left.
The extreme point is called a vertex.
It is a quadratic equation that normally has two solutions
A parabola can open left, down, right, or left on a graph, if that's what you mean:\
The derivative if a function is basically it's slope, or its rate of change. An example is the function y = 4x - 6. This is a line with a slope of 4. The derivative is y' = 4. Another example is the function y = 3x2. This is a parabola with a vertex at (0,0). Its derivative is y' = 6x. At x = 0, the slope of the parabola is 6*0, which is 0, since this is the vertex of the parabola. To the left, at x is -4 for example, the derivative (and therefore slope) is negative. To the right, at x = 5 for example, the derivative is positive. The farther away from the vertex, the greater the value of the derivative so the the slope of the function increases as you move away from the vertex (it gets steeper).
left
Let's say you want the standard form of the equation x2 + 10x + y + 20 = 0. x2 + 10x + y + 20 = 0 (add 5 and subtract y to both sides) x2 + 10x + 25 = -y + 5 (form the square to the left, and factor out -1 to the right) (x + 5)2 = -(y - 5) which is in the standard form (x - h)2 = 4p(y - k), where (h, k) = (-5, 5) is the vertex, and 4p = -1 yields p = -1/4, so the parabola opens downward.
your answer would actually be x=a(y-v)2+h