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

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)