Example

Graphing y=x2+2x8y = x^2 + 2x - 8

To graph the parabola y=x2+2x8y = 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+2x80 = x^2 + 2x - 8, which factors to 0=(x+4)(x2)0 = (x + 4)(x - 2), giving x=4x = -4 and x=2x = 2. The xx-intercepts are (4,0)(-4, 0) and (2,0)(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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