Example

Graphing y=3x2+12x12y = -3x^2 + 12x - 12

To graph the parabola y=3x2+12x12y = -3x^2 + 12x - 12, apply the standard graphing procedure:

Step 1: The coefficient a=3a = -3 is negative, so the parabola opens downward.

Step 2: Find the axis of symmetry using x=b2ax = -\frac{b}{2a}. Since b=12b = 12 and a=3a = -3, x=122(3)=126=2x = -\frac{12}{2(-3)} = -\frac{12}{-6} = 2. The axis of symmetry is the line x=2x = 2.

Step 3: Find the vertex by substituting x=2x = 2 into the equation: y=3(2)2+12(2)12=12+2412=0y = -3(2)^2 + 12(2) - 12 = -12 + 24 - 12 = 0. The vertex is the point (2,0)(2, 0).

Step 4: Find the yy-intercept by setting x=0x = 0: y=3(0)2+12(0)12=12y = -3(0)^2 + 12(0) - 12 = -12. The yy-intercept is (0,12)(0, -12). The point symmetric to the yy-intercept across the axis of symmetry x=2x = 2 is (4,12)(4, -12).

Step 5: Find the xx-intercepts by setting y=0y = 0: 0=3x2+12x120 = -3x^2 + 12x - 12. Factoring out the greatest common factor gives 0=3(x24x+4)0 = -3(x^2 - 4x + 4), which is a perfect square trinomial: 0=3(x2)20 = -3(x - 2)^2. Solving yields x=2x = 2. The only xx-intercept is (2,0)(2, 0), which is the same as the vertex.

Step 6: Graph the parabola by plotting the vertex, the yy-intercept, and the symmetric point, then connect them with a smooth, downward-opening curve.

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Updated 2026-04-21

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