1. The problem asks: How many gallons of oil are in the gun when it is full? 2. The equation given is $$y = -\frac{1}{32}x + 6$$ where $y$ is the gallons of oil remaining and $x$ is the number of times oil is dispensed. 3. When the gun is full, no oil has been dispensed yet, so $x=0$. 4. Substitute $x=0$ into the equation: $$y = -\frac{1}{32} \times 0 + 6 = 0 + 6 = 6$$ 5. Therefore, the gun holds 6 gallons of oil when full. 6. The correct answer is C. 6.