I am used to going the other way. We will try this. (3,1) = vertex.
(X - 3)^2 + 1 = 0
X^2 - 6X + 9 + 1 = 0
X^2 - 6X + 10 = 0
b = -6
c = 10
The vertex has a minimum value of (-4, -11)
(-3, -5)
The vertex of the positive parabola turns at point (-2, -11)
It is a parabola with its vertex at the origin and the arms going upwards.
20 and the vertex of the parabola is at (3, 20)
The vertex has a minimum value of (-4, -11)
(-3, -5)
The vertex of the positive parabola turns at point (-2, -11)
The minimum value of the parabola is at the point (-1/3, -4/3)
It is a parabola with its vertex at the origin and the arms going upwards.
20 and the vertex of the parabola is at (3, 20)
The points at which the parabola intersects the x axis are 3-sqrt(10)/2 and 3+sqrt(10)/2. The X position of the vertex is in the middle, at 3. The y position, from there, is simply found by substituting 2 for x in the equation. As a result, the vertex is at (3, 5).
The vertex of a parabola is the minimum or maximum value of the parabola. To find the maximum/minimum of a parabola complete the square: x² + 4x + 5 = x² + 4x + 4 - 4 + 5 = (x² + 4x + 4) + (-4 + 5) = (x + 2)² + 1 As (x + 2)² is greater than or equal to 0, the minimum value (vertex) occurs when this is zero, ie (x + 2)² = 0 → x + 2 = 0 → x = -2 As (x + 2)² = 0, the minimum value is 0 + 1 = 1. Thus the vertex of the parabola is at (-2, 1).
Interpreting that function as y=x2+2x+1, the graph of this function would be a parabola that opens upward. It would be equivalent to y=(x+1)2. Its vertex would be at (-1,0) and this vertex would be the parabola's only zero.
The given equation is not that of a parabola.
By completing the square y = (x+3)2+1 Axis of symmetry and vertex: x = -3 and (-3, 1) Note that the parabola has no x intercepts because the discriminant is less than zero
-2-5