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Graphing a Horizontal Parabola Using Properties

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Graphing $x = -4(y+1)^2 + 4$

To graph the horizontal parabola $x = -4(y+1)^2 + 4$ using its properties, follow these steps: First, identify the constants $a$ , $h$ , and $k$ . Here, $a = -4$ , $h = 4$ , and $k = -1$ . Since $a = -4$ , the parabola opens to the left. The axis of symmetry is $y = k$ , which is $y = -1$ . The vertex is $(h, k)$ , which is $(4, -1)$ . Find the $x$ -intercept by substituting $y = 0$ into the equation: $x = -4(0+1)^2 + 4 = 0$ , so the $x$ -intercept is $(0, 0)$ . The point symmetric to $(0, 0)$ across the axis of symmetry is $(0, -2)$ . Find the $y$ -intercepts by substituting $x = 0$ into the equation: $0 = -4(y+1)^2 + 4$ , which simplifies to $-4 = -4(y+1)^2$ , and then $1 = (y+1)^2$ . Solving $y+1 = \pm 1$ gives $y = 0$ and $y = -2$ . The $y$ -intercepts are $(0, 0)$ and $(0, -2)$ . Finally, graph the parabola using the vertex, axis of symmetry, and intercept points.

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Updated 2026-05-26

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OpenStax Intermediate Algebra

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