Above
Nose points right, opens to the left.
I would use the alternate parabolic equation (y-k)2 = a (x-h). You can plug in the coordinates of the vertex (h, k)--h is the x coordinate of the vertex and k is the y coordinate of the vertex. Now you are left with an equation with an x and y, which are fine, but also an a, which we still have to get rid of. The a describes how steeply the parabola increases. To find the actual number, plug the other point you were given into the equation (where the x and y are). How you are left with one equation and one variable, so you can solve for a = some number. Now return to the beginning of the previous step, when you had and x, y, and a in your equation. Keep the x and y in their original positions, but replace a with the number you just found.
They are the distances from any one vertex to the three adjacent vertices. If you start with the bottom, left, front vertex then:Length may be to the vertex that is at the bottom, right, front;Width to the vertex at the bottom, left, back; andHeight to the vertex at the top, left, front.
right
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
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).
A parabola can open left, down, right, or left on a graph, if that's what you mean:\
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