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

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Graphing a Horizontal Parabola Using Properties

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Graphing $x = 2y^2$

To graph the horizontal parabola $x = 2y^2$ using its properties, follow these steps: Since the coefficient $a = 2$ is positive, the parabola opens to the right. To find the axis of symmetry, use $y = -\frac{b}{2a}$ . Since there is no $y$ term, $b = 0$ , so the axis of symmetry is the horizontal line $y = 0$ (the $x$ -axis). The vertex lies on the axis of symmetry. Substituting $y = 0$ into the equation gives $x = 2(0)^2 = 0$ , so the vertex is $(0, 0)$ . Because the vertex is at the origin, both the $x$ - and $y$ -intercepts are at the point $(0, 0)$ . To draw the parabola, select additional values for $y$ to find more points. For instance, when $y = 1$ , $x = 2(1)^2 = 2$ , giving the point $(2, 1)$ . When $y = 2$ , $x = 2(2)^2 = 8$ , giving the point $(8, 2)$ . Next, plot the points symmetric to these across the axis of symmetry ( $y = 0$ ). These symmetric points are $(2, -1)$ and $(8, -2)$ . Finally, plot the vertex and these points, and connect them with a smooth curve opening to the right.

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Updated 2026-05-26

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OpenStax Intermediate Algebra

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