1. **State the problem:** We need to find which values of $x$ from the given list make the function $f(x) = x^2 - 5x - 36$ equal to zero. 2. **Formula and rule:** A zero of the function is a value of $x$ such that $f(x) = 0$. We substitute each $x$ into the function and check if the result is zero. 3. **Calculate $f(x)$ for each $x$:** - For $x=0$: $$f(0) = 0^2 - 5\times0 - 36 = -36 \neq 0$$ - For $x=2$: $$f(2) = 2^2 - 5\times2 - 36 = 4 - 10 - 36 = -42 \neq 0$$ - For $x=-2$: $$f(-2) = (-2)^2 - 5\times(-2) - 36 = 4 + 10 - 36 = -22 \neq 0$$ - For $x=6$: $$f(6) = 6^2 - 5\times6 - 36 = 36 - 30 - 36 = -30 \neq 0$$ - For $x=-6$: $$f(-6) = (-6)^2 - 5\times(-6) - 36 = 36 + 30 - 36 = 30 \neq 0$$ - For $x=-4$: $$f(-4) = (-4)^2 - 5\times(-4) - 36 = 16 + 20 - 36 = 0$$ - For $x=4$: $$f(4) = 4^2 - 5\times4 - 36 = 16 - 20 - 36 = -40 \neq 0$$ - For $x=9$: $$f(9) = 9^2 - 5\times9 - 36 = 81 - 45 - 36 = 0$$ - For $x=-9$: $$f(-9) = (-9)^2 - 5\times(-9) - 36 = 81 + 45 - 36 = 90 \neq 0$$ 4. **Conclusion:** The zeros of the function are $x = -4$ and $x = 9$ because these values make $f(x) = 0$. 5. **Explanation:** By substituting each $x$ value into the function and simplifying, we identified which values yield zero. This method directly tests the definition of zeros of a function. Final answer: The zeros are $-4$ and $9$.