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The graph of a quadratic function is y = ax + bx + c or y = a(x - h) + k, where (h, k) is the vertex.

Since x2 - 12x + 40 cannot be factored, then I will complete the square which will give me the vertex. So I will draw the parabola (which opens upward) by using the vertex and the y-intercept, (0, 40).

y = x2 - 12x + 40 add and subtract 62

y = x2 - 12x + 62 - 36 + 40

y = (x - 6)2 + 4 so the vertex is (6, 4) which shows that the parabola does not intersect the x-axis.

Since the axis of symmetry is x = 6, and the y-intercept point is (0, 40), I have another point (12, 40). I can also find other points by letting x to be 2, or 3, or 4 and finding the corresponding y-values. For example, if x = 3, then y = 13. So the point (3, 13) will give me its symmetric point (9, 13), and so on. Thus, I can use all these points to draw the parabola.

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12y ago
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Q: How do you graph x2-12x 40?
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