2004froggy
Answered

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What is the maximum vertical distance between the parabola [tex]y = 3 - x^2[/tex] and the line [tex]y = x + 1[/tex] for -2 ≤ x ≤ 1?

Sagot :

LammettHash

The (unsigned) vertical distance between the two curves is

|(3 - x²) - (x + 1)| = |x² + x - 2|

Since

x² + x - 2 = (x + 2) (x - 1)

we know the distances between the curves at the endpoints of the interval [-2, 1] are both zero.

For all other x in (-2, 1), the quadratic x² + x - 2 is negative. (Consider x = 0, for instance, which gives a value of -2 < 0.) So by definition of absolute value, we have

|x² + x - 2| = - (x² + x - 2) = 2 - x - x²

Complete the square:

2 - x - x² = 2 + 1/4 - (1/4 + x + x²)

2 - x - x² = 9/4 - (1/2 + x)²

Then the maximum distance between the curves is 9/4, which occurs at x = -1/2.

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