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Graphing a Horizontal Parabola Using Properties

Example

Graphing $x = -4y^2 - 16y - 12$

To graph the horizontal parabola $x = -4y^2 - 16y - 12$ , begin by writing the equation in standard form via completing the square. Factor $-4$ from the $y$ terms: $x = -4(y^2 + 4y) - 12$ . Complete the square inside the parentheses by adding $4$ . To balance the equation, subtract the equivalent value on the outside; since $-4 \cdot 4 = -16$ , you add $16$ outside: $x = -4(y^2 + 4y + 4) - 12 + 16$ . This simplifies to standard form: $x = -4(y + 2)^2 + 4$ .

Identify the properties from the standard form $x = a(y - k)^2 + h$ : $a = -4$ , $h = 4$ , and $k = -2$ . Since $a = -4$ , the parabola opens to the left. The axis of symmetry is $y = -2$ , and the vertex is $(4, -2)$ . Find the $x$ -intercept by setting $y = 0$ : $x = -4(0 + 2)^2 + 4 = -4(4) + 4 = -12$ , which gives the point $(-12, 0)$ . Its symmetric point is $(-12, -4)$ . Find the $y$ -intercepts by setting $x = 0$ : $0 = -4(y + 2)^2 + 4$ , which gives $(y + 2)^2 = 1$ . Solving this yields $y + 2 = \pm 1$ , so $y = -1$ and $y = -3$ . The $y$ -intercepts are $(0, -1)$ and $(0, -3)$ . Plot these points and draw a smooth curve to complete the graph.

Updated 2026-06-03

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OpenStax Intermediate Algebra

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