Graph the function. y=2x2

1 answer

The function you're dealing with is \( y = 2x^2 \), which is a quadratic function. Here's a step-by-step guide to understanding and graphing this function:

  1. Identify the Basic Shape:

    • Since the coefficient of \( x^2 \) is positive (2), the parabola opens upwards.
  2. Vertex:

    • The vertex of the parabola \( y = ax^2 + bx + c \) (in standard form) is at the point (0, 0) since there are no linear or constant terms.
  3. Intercepts:

    • Y-intercept: Set \( x = 0 \) to find the y-intercept:
      \( y = 2(0)^2 = 0 \).
    • X-intercept: Set \( y = 0 \) to find the x-intercepts:
      \( 0 = 2x^2 \) results in \( x = 0 \). Therefore, the only intercept is at the origin (0, 0).
  4. Other Points:

    • Choose some values for \( x \) to determine \( y \):
      • For \( x = 1 \): \( y = 2(1)^2 = 2 \) → point (1, 2)
      • For \( x = -1 \): \( y = 2(-1)^2 = 2 \) → point (-1, 2)
      • For \( x = 2 \): \( y = 2(2)^2 = 8 \) → point (2, 8)
      • For \( x = -2 \): \( y = 2(-2)^2 = 8 \) → point (-2, 8)
  5. Draw the Graph:

    • Plot the points (0, 0), (1, 2), (-1, 2), (2, 8), and (-2, 8) on the graph.
    • Sketch a smooth curve through these points to form the parabola.

The graph of the function will look like this:

      |
      |          *
      |       *     *
 2    |    *         *
      | *               *
      |*                 *
      |____________________
     -3 -2 -1  0  1  2  3

The parabola is symmetric about the y-axis, indicating that its left and right sides are mirror images. The vertex at (0,0) is the lowest point of the curve, representing the minimum value of the function.