Saturday, November 17, 2012

Suppose we wish to graph the equation y = -3(x - 3)^2 + 5.   We recognize this is a parabola with the vertex at (3, 5) because we can rewrite this as y - 5 = -3(x - 3)^2, which is in the form y - k = a(x - h)^2 (vertex at (h, k)).

The parabola opens up or down, and since a is negative, the parabola opens down. 

Interesting, we can use calculus to determine whether (3, 5) is a maximum or a minimum. If it's a maximum, the parabola opens down, if it's a minimum, the parabola opens up.

y' = -6(x - 3) = -6x + 18

We now find the derivative for a value of x < 3 and one for a value for x > 3

for x = 2, y' = -6(2-3) = -6(-1) = 6

for x = 4, y' = -6(4-3) = -6(1) = -6

Since the slope of the curve at x = 2 is positive and the slope of the curve at x = 4 is negative we have a maximum at x = 3.  Therefore (3, 5) is a maximum and the parabola opens downward.

Just another way to determine how a parabola opens.

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