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Graphing a Horizontal Parabola Using Properties

Example

Graphing $x = 3y^2 + 6y + 7$

To graph the horizontal parabola $x = 3y^2 + 6y + 7$ , first rewrite the equation in standard form by completing the square. Factor out the $3$ from the $y$ terms: $x = 3(y^2 + 2y) + 7$ . Complete the square inside the parentheses by adding $1$ , and subtract $3$ outside to balance the equation (since $3 \cdot 1 = 3$ ): $x = 3(y^2 + 2y + 1) + 7 - 3$ . This simplifies to the standard form: $x = 3(y + 1)^2 + 4$ .

From this standard form, $x = a(y - k)^2 + h$ , identify the constants: $a = 3$ , $h = 4$ , and $k = -1$ . Because $a = 3$ , the parabola opens to the right. The axis of symmetry is the horizontal line $y = -1$ , and the vertex is $(4, -1)$ . Next, find the intercepts. Substitute $y = 0$ to find the $x$ -intercept: $x = 3(0 + 1)^2 + 4 = 7$ , giving the point $(7, 0)$ . Its symmetric point across the axis of symmetry is $(7, -2)$ . To find any $y$ -intercepts, set $x = 0$ : $0 = 3(y + 1)^2 + 4$ , which leads to $(y + 1)^2 = -\frac{4}{3}$ . Since the square of a real number cannot be negative, there are no real solutions, meaning the graph has no $y$ -intercepts. Finally, plot the vertex, the $x$ -intercept, and its symmetric point to draw the parabola.

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Updated 2026-05-25

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OpenStax Intermediate Algebra

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