a dot
the y-axis is vertical, and the x- axis is horizontal
If a figure has line symmetry about the y-axis, then for every point (x, y) on the figure, there is a corresponding point (-x, y). Given that one vertex is at (-1, -3), its symmetric counterpart across the y-axis would be at (1, -3). Thus, the coordinates of another vertex of the figure are (1, -3).
It will look like a straight vertical line that cuts the x-axis at the point (100, 0).
it will be pointing to the right side of the plane
An x2 parabola will always have one vertex, but depending on the discriminant of the function (b2-4ac) the parabola will either have 2 roots (it crosses the x-axis twice), 1 repeating root (the parabola meets the x-axis at a single point), or no real roots (the parabola doesn't meet the x-axis at all)
the x-axis... obviously! the x-axis... obviously!
the y-axis is vertical, and the x- axis is horizontal
axis of symmetry is x=0 Vertex is (0,0) So the answer is : YES
y2 = 32x y = ±√32x the vertex is (0, 0) and the axis of symmetry is x-axis or y = 0
The axis of symmetry is x = -2.
It is (-1, 3).
If a figure has line symmetry about the y-axis, then for every point (x, y) on the figure, there is a corresponding point (-x, y). Given that one vertex is at (-1, -3), its symmetric counterpart across the y-axis would be at (1, -3). Thus, the coordinates of another vertex of the figure are (1, -3).
2
It would look like a straight vertical line, i.e. parallel to the y-axis, passing through the point on the x-axis where x=3.
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
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)
Vertex = (0,0) Line of symmetry = y axis You should of known that as this function is only X^2