1Cademy - Graphing $$x = -2(y+3)^2 + 2$$

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Graphing a Horizontal Parabola Using Properties

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Graphing $x = -2(y+3)^2 + 2$

To graph the horizontal parabola $x = -2(y+3)^2 + 2$ using its properties, identify the constants $a$ , $h$ , and $k$ . Here, $a = -2$ , $h = 2$ , and $k = -3$ . Since $a = -2$ is negative, the parabola opens leftward. The axis of symmetry is $y = -3$ , and the vertex is $(2, -3)$ . To find the $x$ -intercept, set $y = 0$ : $x = -2(0+3)^2 + 2 = -2(9) + 2 = -16$ . The $x$ -intercept is $(-16, 0)$ . The symmetric point to this across the axis of symmetry $y = -3$ is $(-16, -6)$ . Find the $y$ -intercepts by setting $x = 0$ : $0 = -2(y+3)^2 + 2$ , which simplifies to $1 = (y+3)^2$ . Solving this yields $y+3 = \pm 1$ , giving $y = -2$ and $y = -4$ . The $y$ -intercepts are $(0, -2)$ and $(0, -4)$ . These key features are then plotted to form the graph.

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Updated 2026-06-17

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OpenStax Intermediate Algebra

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