In which of the following intervals, the function \(f(x) = -x^2 - 2x + 15\) is decreasing?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \((-1,\infty)\)
A function is decreasing where its derivative is negative.
\(f(x) = -x^2 - 2x + 15\)
\(f'(x) = -2x - 2 = -2(x + 1)\)
Since \(-2 \lt 0\), \(f'(x) \lt 0\) exactly when \(x + 1 \gt 0\):
\(x \gt -1\)
So f is decreasing on \((-1, \infty)\).
Shortcut: The graph of f is a downward parabola (coefficient of \(x^2\) is negative) with vertex at \(x = -\frac{b}{2a} = -\frac{-2}{2(-1)} = -1\). A downward parabola rises up to its vertex and falls after it, so f decreases on \((-1, \infty)\).
Why the other options are incorrect
Option (2) \((-\infty,-1)\) — Incorrect. Here \(x + 1 \lt 0\), so \(f'(x) \gt 0\): f is increasing on this interval.
Option (3) \((-\infty,2)\) — Incorrect. It contains points to the left of \(-1\), where f is increasing; for example, \(f(-3) = 12\) but \(f(-1) = 16\).
Option (4) \((0, -1)\) — Incorrect. As printed, this is not an interval: its left end 0 is greater than its right end \(-1\).