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| Math Central | Quandaries & Queries |
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Question from Chin, a parent: How do you solve this problem? The quadratic equation y = (x-p)^2 + q cuts the y axis at y = 10. The straight line |
Hello Chin.
The quadratic in this case is a parabola opening upwards. The line you have provided is horizontal. The only place a horizontal line can be tangent to an upwards-opening parabola is at the vertex of the parabola.
Since the standard form for a parabola is y = A(x - h)2 + k where (h, k) is the vertex, your question is actually asking for the vertex of the parabola.
You know the y value of the vertex, because the vertex lies on the horizontal line. That's q. You also know that when x = 0 (that's the y axis), the intercept is 10, so y = 10. That means (0, 10) solves
the equation:
10 = (0-p)2 + (-6)
Can you solve for p now? Hint: there are two solutions!
Stephen La Rocque.
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