1. We are asked to find the quadratic function $f(x) = ax^2 + bx + c$ that passes through the points $(3,0)$, $(-2,0)$, and $(1,7)$. 2. Since the points $(3,0)$ and $(-2,0)$ are zeros of the function, the quadratic can be expressed as: $$f(x) = a(x - 3)(x + 2)$$ 3. Expand the factors: $$f(x) = a(x^2 - x - 6) = a x^2 - a x - 6a$$ 4. Use the point $(1,7)$ to find $a$: $$f(1) = a(1)^2 - a(1) - 6a = a - a - 6a = -6a$$ Set equal to 7: $$-6a = 7 \implies a = -\frac{7}{6}$$ 5. Substitute $a$ back into the function: $$f(x) = -\frac{7}{6} x^2 + \frac{7}{6} x + 7$$ 6. This matches option C: $$f(x) = -\frac{7}{6} x^2 - \frac{7}{6} x + 7$$ Note: The sign of the $x$ term in option C is negative, but from our expansion it is positive. Re-examining step 3, the expansion is $a(x^2 - x - 6)$, so the $x$ term is $-a x$. Since $a = -\frac{7}{6}$, $-a x = -(-\frac{7}{6}) x = +\frac{7}{6} x$. So the $x$ term is positive. Therefore, the correct function is: $$f(x) = -\frac{7}{6} x^2 + \frac{7}{6} x + 7$$ This corresponds to option B. Final answer: Option B.