1Cademy - Graphing $$y = -2x^2

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Graphing a Quadratic Equation in Two Variables

Example

Graphing $y = -2x^2 - 1$

To graph the parabola $y = -2x^2 - 1$ , apply the systematic graphing procedure:

Step 1: The coefficient $a = -2$ is negative, so the parabola opens downward.

Step 2: Find the axis of symmetry using $x = -\frac{b}{2a}$ . Since $b = 0$ and $a = -2$ , $x = -\frac{0}{2(-2)} = 0$ . The axis of symmetry is the $y$ -axis, $x = 0$ .

Step 3: Find the vertex by substituting $x = 0$ into the equation: $y = -2(0)^2 - 1 = -1$ . The vertex is the point $(0, -1)$ .

Step 4: Find the $y$ -intercept by setting $x = 0$ , which also yields $(0, -1)$ . Since the $y$ -intercept is the vertex itself, there is no distinct symmetric point across the axis of symmetry.

Step 5: Find the $x$ -intercepts by setting $y = 0$ : $0 = -2x^2 - 1$ . This simplifies to $2x^2 = -1$ , or $x^2 = -\frac{1}{2}$ . Since the square of a real number cannot be negative, there are no real solutions, meaning there are no $x$ -intercepts.

Step 6: To complete the graph, find additional points. For example, when $x = 1$ , $y = -2(1)^2 - 1 = -3$ , giving the point $(1, -3)$ . By symmetry, when $x = -1$ , $y = -3$ , giving $(-1, -3)$ . Plot the vertex and these additional points, and connect them with a smooth, downward-opening curve.

Updated 2026-04-21

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OpenStax Elementary Algebra

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