1. The problem asks to identify which function corresponds to the given graph. 2. The graph shows symmetry about the y-axis, two peaks at $x=\pm 1$ with value 1, and a minimum at $x=0$ with value $-2$. 3. Let's analyze each function: - $f(x) = -x^2 - 2$: This is a downward parabola shifted down by 2. Its vertex is at $(0,-2)$, but it has only one peak (the vertex), not two peaks. - $f(x) = |-x^2 - 2|$: This is the absolute value of $-x^2 - 2$. Since $-x^2 - 2$ is always negative or zero, its absolute value will be positive or zero, creating a "W" shape with minimum at zero. The graph does not match this because the graph dips below zero. - $f(x) = -|x^2 - 2|$: This is the negative of the absolute value of $x^2 - 2$. The expression $x^2 - 2$ is zero at $x=\pm \sqrt{2} \approx \pm 1.414$. The absolute value creates a "V" shape with minimum zero at these points, and the negative sign flips it downward, creating two peaks at $x=\pm 1.414$ and a minimum at $x=0$. - $f(x) = -|x^2| - 2$: Since $|x^2| = x^2$, this is $-x^2 - 2$, same as the first function. 4. The graph shows two peaks at $x=\pm 1$ with value 1, and a minimum at $x=0$ with value $-2$. The function $f(x) = -|x^2 - 2|$ matches this behavior. 5. To verify, calculate $f(0) = -|0 - 2| = -| -2| = -2$, matching the minimum. 6. Calculate $f(1) = -|1 - 2| = -| -1| = -1$, but the graph shows peak at 1, so let's check carefully. 7. The graph peaks at 1 at $x=\pm 1$, so let's check $f(x) = -|x^2 - 2|$ at $x=\pm 1$: $$f(\pm 1) = -|1 - 2| = -| -1| = -1$$ But the graph shows peaks at 1, so this is inconsistent. 8. Check $f(x) = |-x^2 - 2|$ at $x=\pm 1$: $$f(\pm 1) = |-1 - 2| = |-3| = 3$$ No match. 9. Check $f(x) = -|x^2| - 2 = -x^2 - 2$ at $x=\pm 1$: $$f(\pm 1) = -1 - 2 = -3$$ No match. 10. Check $f(x) = -x^2 - 2$ at $x=\pm 1$: $$f(\pm 1) = -1 - 2 = -3$$ No match. 11. The graph shows peaks at 1 at $x=\pm 1$, so the function must be positive 1 at these points. 12. Try $f(x) = |-x^2 - 2|$ again: At $x=0$, $f(0) = |-0 - 2| = 2$, but the graph shows minimum at $-2$. 13. Try $f(x) = -|x^2 - 2|$: At $x=0$, $f(0) = -|0 - 2| = -2$, matches minimum. At $x=\pm 1$, $f(\pm 1) = -|1 - 2| = -1$, graph shows 1, so no. 14. Try $f(x) = -|x^2| - 2 = -x^2 - 2$: At $x=0$, $f(0) = -0 - 2 = -2$, matches minimum. At $x=\pm 1$, $f(\pm 1) = -1 - 2 = -3$, no match. 15. The graph shows two peaks at 1 and a minimum at -2, so the function must be $f(x) = -|x^2 - 2|$ but with peaks at $-1$ not $1$. 16. The graph's peaks at $x=\pm 1$ with value 1 suggest the function is $f(x) = -|x^2 - 2|$ but shifted or scaled. 17. Since the graph matches $f(x) = -|x^2 - 2|$ best, the answer is: **$f(x) = -|x^2 - 2|$**