Example

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

To graph the horizontal parabola x=4(y+2)2+4x = -4(y+2)^2 + 4 using its properties, first identify the constants aa, hh, and kk. Here, a=4a = -4, h=4h = 4, and k=2k = -2. Since aa is negative, the parabola opens to the left. The axis of symmetry is the horizontal line y=2y = -2. The vertex is (4,2)(4, -2). To find the xx-intercept, substitute y=0y = 0: x=4(0+2)2+4=4(4)+4=12x = -4(0+2)^2 + 4 = -4(4) + 4 = -12. The xx-intercept is (12,0)(-12, 0). The point symmetric to (12,0)(-12, 0) across the axis of symmetry is (12,4)(-12, -4). To find the yy-intercepts, set x=0x = 0: 0=4(y+2)2+40 = -4(y+2)^2 + 4, which gives 1=(y+2)21 = (y+2)^2. Solving yields y+2=±1y+2 = \pm 1, so y=1y = -1 and y=3y = -3. The yy-intercepts are (0,1)(0, -1) and (0,3)(0, -3). Plot these points to complete the graph.

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Updated 2026-05-25

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