How do you graph #y= -1/4x^2# by plotting points?

Question

Eva Abramson · Accepted Answer

See explanation below.

#y# is a parabola with a critical point at #(0,0)# Since the coefficient of #x^2# is negative #y# has a single maximum value. Hence, #(0,0)# is an absolute maximum. and, #y# has no other intercepts on the #x# or #y# axes. To assist plotting #y# by points it will be necessary to construct a table of points, bearing in mind that #y# is symetric about the #y# axis. E.g. #(-4,-4), (-2,-1), (0,0), (2,-1), (-4,-4)# This can be seen from the graph of #y# below: graph{-1/4x^2 [-11.01, 11.49, -8.865, 2.385]}

Kennedy Adams · Answer

undefined To graph the equation y = -1/4x^2 by plotting points, you can choose various x-values, calculate the corresponding y-values using the equation, and then plot those points on a coordinate plane. Here are some steps to follow:

1. Choose a set of x-values. For example, you could choose x-values like -4, -3, -2, -1, 0, 1, 2, 3, and 4.
2. Substitute each x-value into the equation y = -1/4x^2 to find the corresponding y-values.
3. Plot the points (x, y) on the coordinate plane.
4. Connect the points to create a smooth curve.

Here's a table showing some x-values and their corresponding y-values:

| x    | y = -1/4x^2 |
|------|-------------|
| -4   | -4          |
| -3   | -2.25       |
| -2   | -1          |
| -1   | -0.25       |
| 0    | 0           |
| 1    | -0.25       |
| 2    | -1          |
| 3    | -2.25       |
| 4    | -4          |

Plotting these points and connecting them will result in a downward-opening parabola.