Math CentralQuandaries & Queries


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
y = -6 is a tangent to the curve. Find the values of p and q?

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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