# How do you find the vertex and intercepts for #y = (–¼)x^2#?

Explained below

Vertex of this parabola is (0,0). There are no other x or y intercepts. It is a vertical parabola opening downwards.

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To find the vertex and intercepts for the equation y = (-1/4)x^2:

Vertex:

- The vertex of a parabola in the form y = ax^2 is given by the point (0, c), where c is the y-coordinate of the vertex.
- In this case, since there is no linear term, the x-coordinate of the vertex is always 0.
- Substitute x = 0 into the equation y = (-1/4)x^2 to find the y-coordinate of the vertex.
- Therefore, the vertex is (0, 0).

x-intercepts:

- To find the x-intercepts, set y = 0 in the equation y = (-1/4)x^2 and solve for x.
- 0 = (-1/4)x^2
- Since x^2 can't be negative, the only solution is x = 0.
- Therefore, the x-intercept is (0, 0).

y-intercept:

- To find the y-intercept, set x = 0 in the equation y = (-1/4)x^2 and solve for y.
- y = (-1/4)(0)^2
- y = 0
- Therefore, the y-intercept is (0, 0).

In summary:

- Vertex: (0, 0)
- x-intercept: (0, 0)
- y-intercept: (0, 0)

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