2
It is (-1, 3).
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)
We will be able to identify the answer if we have the equation. We can only check on the coordinates from the given vertex.
Because the base of a cone is round, it is the only three dimensional figure with only one vertex.
I'm assuming that you meant y = 2(x^2) +4. If it were only y = 2x +4, then this would be a linear equation and not a parabola. Anyways, use the equation x = -b/2a to find the x-value of your vertex AND your axis of symmetry. (Given the standard equation y = a(x^2) + bx + c) So, x = -0/2(2) - x = 0 (Axis of Symmetry) Now plug 0 back into your equation to find your y-value of your vertex. y = 2(0^2)+4 y=0 + 4 y = 4 Therefore Vertex = (0,4)
It is (-1, 3).
true
Because they represent a pair of coordinates
i think that the range and the domain of a parabola is the coordinates of the vertex
Only equilateral triangles and some irregular polygons of (3n) sides have 3 lines of symmetry. A regular polygon with n sides (or vertices) has n lines of symmetry. If n is even, there are n/2 lines of symmetry from vertex to opposite vertex and another n/2 from the middle of a side to the middle of the opposite side. If n is odd, there are n lines of symmetry from vertex to the midpoint of the opposite side.
False
y = 2x2 + 3x + 6 Since a > 0 (a = 2, b = 3, c = 6) the graph opens upward. The coordinates of the vertex are (-b/2a, f(-b/2a)) = (- 0.75, 4.875). The equation of the axis of symmetry is x = -0.75.
True. A triangle is said to be inscribed in another figure if each vertex of the triangle lies on the boundary of that figure. For example, a triangle inscribed in a circle has all its vertices touching the circumference of the circle.
To determine the coordinates of the preimage of vertex M, I would need additional information about the transformation that was applied to vertex M, such as the type of transformation (e.g., translation, rotation, reflection, scaling) and the coordinates of M itself. If you provide the coordinates of M and the details of the transformation, I can help you find the preimage coordinates.
It depends on what the coordinates of the first three vertices are!
To rotate a point or figure 90 degrees clockwise about the origin, you can use the transformation formula: for a point (x, y), the new coordinates after rotation will be (y, -x). Apply this transformation to each vertex of the figure. After calculating the new coordinates for all points, plot them to visualize the rotated figure.
The vertex is at the origin of coordinates ... the point (0, 0).