Take the derivative: y' = 2x + 4. Set equal to zero: 2x + 4 = 0 --> x = -2. Substitute into original and y = -3.
Axis of symmetry is x = -2. Vertex is (-2,-3)
It is a turning point. It lies on the axis of symmetry.
y2 = 32x y = ±√32x the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
In the form y = ax² + bx + c the axis of symmetry is given by the line x = -b/2a The axis of symmetry runs through the vertex, and the vertex is given by (-b/2a, -b²/4a + c). For y = 2x² + 4x - 10: → axis of symmetry is x = -4/(2×2) = -4/4 = -1 → vertex = (-1, -4²/(4×2) - 10) = (-1, -16/8 - 10) = (-1, -12)
It is (-1, 3).
2
Vertex = (0,0) Line of symmetry = y axis You should of known that as this function is only X^2
axis of symmetry is x=0 Vertex is (0,0) So the answer is : YES
It is a turning point. It lies on the axis of symmetry.
y2 = 32x y = ±√32x the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
In the form y = ax² + bx + c the axis of symmetry is given by the line x = -b/2a The axis of symmetry runs through the vertex, and the vertex is given by (-b/2a, -b²/4a + c). For y = 2x² + 4x - 10: → axis of symmetry is x = -4/(2×2) = -4/4 = -1 → vertex = (-1, -4²/(4×2) - 10) = (-1, -16/8 - 10) = (-1, -12)
The axis of symmetry is x = -2.
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
How about y = (x - 2)2 = x2 - 4x + 4 ? That is the equation of a parabola whose axis of symmetry is the vertical line, x = 2. Its vertex is located at the point (2, 0).
There is no equation (nor inequality) in the question so there can be no graph - with or without an axis of symmetry.
K
y=x2+2x+1 -b -2 2a= 2= -1 = axis of symmetry is negative one.
It is (-1, 3).