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Graphing a Horizontal Parabola Using Properties

Example

Graphing $x = 2y^2 + 12y + 17$

To graph the horizontal parabola $x = 2y^2 + 12y + 17$ , first rewrite the equation in standard form, $x = a(y - k)^2 + h$ , by completing the square. Factor $2$ from the $y$ terms: $x = 2(y^2 + 6y) + 17$ . Complete the square inside the parentheses by adding $9$ , and subtract $18$ (since $2 \cdot 9 = 18$ ) on the outside to keep the equation balanced: $x = 2(y^2 + 6y + 9) + 17 - 18$ . This simplifies to standard form: $x = 2(y + 3)^2 - 1$ .

Now, identify the properties. The constants are $a = 2$ , $h = -1$ , and $k = -3$ . Because $a = 2$ , which is positive, the parabola opens to the right. The axis of symmetry is $y = k$ , so $y = -3$ . The vertex is $(h, k)$ , which is $(-1, -3)$ . Find the $x$ -intercept by setting $y = 0$ : $x = 2(0 + 3)^2 - 1 = 2(9) - 1 = 17$ , giving the point $(17, 0)$ . Find the symmetric point across the axis of symmetry, which is $(17, -6)$ . Find the $y$ -intercepts by setting $x = 0$ : $0 = 2(y + 3)^2 - 1$ , so $\frac{1}{2} = (y + 3)^2$ . Solving gives $y + 3 = \pm \frac{\sqrt{2}}{2}$ , or $y = -3 \pm \frac{\sqrt{2}}{2}$ . This yields approximate $y$ -intercepts at $y \approx -2.3$ and $y \approx -3.7$ . Finally, plot these points and draw a smooth curve to graph the parabola.

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Updated 2026-05-25

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OpenStax Intermediate Algebra

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