Example

Graphing y=x28x+12y = x^2 - 8x + 12

To graph the parabola y=x28x+12y = x^2 - 8x + 12, follow the standard graphing steps:

The parabola opens upward because a=1a = 1 is positive. The axis of symmetry is x=b2a=82(1)=4x = -\frac{b}{2a} = -\frac{-8}{2(1)} = 4. The vertex is found by substituting x=4x = 4: y=(4)28(4)+12=4y = (4)^2 - 8(4) + 12 = -4, making the vertex (4,4)(4, -4). The yy-intercept is found by setting x=0x = 0, yielding (0,12)(0, 12), and its symmetric point across the axis of symmetry x=4x = 4 is (8,12)(8, 12). To find the xx-intercepts, set y=0y = 0: 0=x28x+120 = x^2 - 8x + 12. Factoring yields 0=(x2)(x6)0 = (x - 2)(x - 6), giving xx-intercepts of (2,0)(2, 0) and (6,0)(6, 0). Plotting the vertex, the intercepts, and the symmetric point, and connecting them produces the graph of the parabola.

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Updated 2026-04-21

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