Example

Graphing y=x2+2x−8y = x^2 + 2x - 8

To graph the parabola y=x2+2x−8y = x^2 + 2x - 8, apply the standard graphing procedure:

First, determine the direction of the opening. Since a=1a = 1 is positive, the parabola opens upward. Next, find the axis of symmetry: x=−b2a=−22(1)=−1x = -\frac{b}{2a} = -\frac{2}{2(1)} = -1. Then, find the vertex by substituting x=−1x = -1: y=(−1)2+2(−1)−8=−9y = (-1)^2 + 2(-1) - 8 = -9. The vertex is (−1,−9)(-1, -9). Find the yy-intercept by setting x=0x = 0, yielding y=−8y = -8, so the yy-intercept is (0,−8)(0, -8). A point symmetric to it across the axis of symmetry x=−1x = -1 is (−2,−8)(-2, -8). Find the xx-intercepts by setting y=0y = 0: 0=x2+2x−80 = x^2 + 2x - 8, which factors to 0=(x+4)(x−2)0 = (x + 4)(x - 2), giving x=−4x = -4 and x=2x = 2. The xx-intercepts are (−4,0)(-4, 0) and (2, 0). Finally, plot the vertex, intercepts, and the symmetric point, and connect them with a smooth curve to sketch the parabola.

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Updated 2026-04-21

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