1. **State the problem:** We are given the function $$y = -2^{-\frac{1}{2}x+1} + 5$$ and need to find its domain, range, x-intercept, y-intercept, and asymptotes. 2. **Rewrite the function for clarity:** The function is $$y = -2^{\left(-\frac{1}{2}x + 1\right)} + 5$$. 3. **Domain:** The base of the exponential function is 2, which is positive and defined for all real numbers. Therefore, the domain is all real numbers: $$\text{Domain} = (-\infty, \infty)$$. 4. **Range:** Since $$2^{\left(-\frac{1}{2}x + 1\right)} > 0$$ for all real $$x$$, then $$-2^{\left(-\frac{1}{2}x + 1\right)} < 0$$. Adding 5 shifts the graph up by 5 units. So the range is: $$y < 5$$ More precisely, $$\text{Range} = (-\infty, 5)$$. 5. **Y-intercept:** Set $$x=0$$: $$y = -2^{\left(-\frac{1}{2}(0) + 1\right)} + 5 = -2^{1} + 5 = -2 + 5 = 3$$. So the y-intercept is at $$(0, 3)$$. 6. **X-intercept:** Set $$y=0$$ and solve for $$x$$: $$0 = -2^{\left(-\frac{1}{2}x + 1\right)} + 5$$ $$2^{\left(-\frac{1}{2}x + 1\right)} = 5$$ Take logarithm base 2: $$-\frac{1}{2}x + 1 = \log_2 5$$ $$-\frac{1}{2}x = \log_2 5 - 1$$ $$x = -2(\log_2 5 - 1) = -2\log_2 5 + 2$$ 7. **Asymptotes:** The horizontal asymptote is the value that $$y$$ approaches as $$x \to \pm \infty$$. - As $$x \to \infty$$, $$-\frac{1}{2}x + 1 \to -\infty$$, so $$2^{\text{large negative}} \to 0$$, thus $$y \to 0 + 5 = 5$$. - As $$x \to -\infty$$, $$-\frac{1}{2}x + 1 \to \infty$$, so $$2^{\text{large positive}} \to \infty$$, thus $$y \to -\infty + 5 = -\infty$$. So the horizontal asymptote is: $$y = 5$$. **Final answers:** - Domain: $$(-\infty, \infty)$$ - Range: $$(-\infty, 5)$$ - Y-intercept: $$(0, 3)$$ - X-intercept: $$\left(-2\log_2 5 + 2, 0\right)$$ - Horizontal asymptote: $$y = 5$$