1. The problem asks for the range of the function defined by the equation $$y = x^2 - 3x + 5$$. 2. This is a quadratic function in the form $$y = ax^2 + bx + c$$ where $$a=1$$, $$b=-3$$, and $$c=5$$. 3. The range of a quadratic function depends on whether the parabola opens upwards or downwards. Since $$a=1 > 0$$, the parabola opens upwards, so the function has a minimum value but no maximum. 4. To find the minimum value, we use the vertex formula for the x-coordinate: $$x = -\frac{b}{2a} = -\frac{-3}{2 \times 1} = \frac{3}{2}$$. 5. Substitute $$x=\frac{3}{2}$$ back into the function to find the minimum y-value: $$y = \left(\frac{3}{2}\right)^2 - 3 \times \frac{3}{2} + 5 = \frac{9}{4} - \frac{9}{2} + 5 = \frac{9}{4} - \frac{18}{4} + \frac{20}{4} = \frac{11}{4} = 2.75$$. 6. Since the parabola opens upwards and the minimum value is $$2.75$$, the range is all real numbers $$y$$ such that $$y \geq 2.75$$. 7. Therefore, the range is not all real numbers, but all real numbers greater than or equal to $$2.75$$. 8. None of the provided options exactly match this range, but the closest understanding is that the range excludes values less than the minimum. Final answer: The range of $$y = x^2 - 3x + 5$$ is $$[2.75, \infty)$$.