Answer: the maximum distance is \(\frac{289}{4}\) and can be found at x = \(\frac{1}{2}\)
Let's call:
f(x) = x + 72
g(x) = x²
A point belonging to the line will be L(x, x+72) and a point on the parabola will be P(x, x²). It can be easily seen that in the interval -8 ≤ x ≤ 9 the line is above the parabola (it's enough to graph them or plug in some numbers), therefore their distance at any point will be:
d(x) = f(x) - g(x) = - x² + x + 72
The function d(x) is a parabola that opens downward, therefore the maximum will be the vertex; given a parabola
y(x) = ax² + bx + c
the coordinates of the vertex will be
\(V(\frac{-b}{2a}, y(\frac{-b}{2a}))\)
Therefore:
\(V(\frac{1}{2}, \frac{289}{4})\)
Hence, the maximum distance is \(\frac{289}{4}\) = 72.25 and can be found at x = \(\frac{1}{2}\)
Another way to find the maximum is to use calculus to find the first derivative of the distance:
d'(x) = -2x + 1
and set it equal to zero:
-2x + 1 = 0
\(x = \frac{1}{2}\)
Since the second derivative:
d"(x) = -2
is negative, the point is a maximum.
Then, substitute this value in the equation for the distance:
\(d(\frac{1}{2}) = -(\frac{1}{2})^{2} + \frac{1}{2} + 72\)
\(d(\frac{1}{2}) = \frac{289}{4} = 72.25\)
Hence, the maximum distance is \(\frac{289}{4}\) = 72.25 and can be found at x = \(\frac{1}{2}\)
