The equation is linear and so has no vertex.
The vertex is at the point (0, 4).
y = 3x2+2x-1 Line of symmetry: x = -1/3 Vertex coordinate: (-1/3, -4/3)
We need to complete the square to find the vertex of this parabola. X^2 + 2X + 5 = 0 X^2 + 2X = -5 halve the coefficient of the linear term ( 2X ) and square it perfectly and add to other side (X + 1)^2 = -5 + 1 (X + 1)^2 = -4 (X + 1)^2 + 4 = 0 Vertex is.... X = -1 Y = 4
y = x2 + 14x + 21 a = 1, b = 14 x = -b/2a = -14/2*1 = -7
The equation is linear and so has no vertex.
The vertex is at the point (0, 4).
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
(1/2, 71 and 3/4)or(0.5, 71.75)
y = 3x2+2x-1 Line of symmetry: x = -1/3 Vertex coordinate: (-1/3, -4/3)
We need to complete the square to find the vertex of this parabola. X^2 + 2X + 5 = 0 X^2 + 2X = -5 halve the coefficient of the linear term ( 2X ) and square it perfectly and add to other side (X + 1)^2 = -5 + 1 (X + 1)^2 = -4 (X + 1)^2 + 4 = 0 Vertex is.... X = -1 Y = 4
y = x2 + 14x + 21 a = 1, b = 14 x = -b/2a = -14/2*1 = -7
x = -3y = -14
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 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).
It is the average of -2 and +6, that is, x = 2
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.