1. **Problem statement:** Given the function $$h(x) = \frac{a}{x+p} + 4$$ with vertical asymptote at $$x=8$$ and horizontal asymptote at $$y=5$$, and a point on the curve at $$(8,4)$$, find the values of $$a$$ and $$p$$, domain, line of symmetry, new asymptote after transformations, and values of $$m$$ and $$n$$ for a shifted function. 2. **Find $$a$$ and $$p$$:** - Vertical asymptote at $$x=8$$ means denominator zero there: $$x+p=0 \Rightarrow p=-8$$. - Horizontal asymptote at $$y=5$$ means $$\lim_{x\to\infty} h(x) = 5$$. - Since $$h(x) = \frac{a}{x-8} + 4$$, horizontal asymptote is $$y=4$$, but given is $$y=5$$, so the constant term must be 5 instead of 4. - The problem states $$h(x) = \frac{a}{x+p} + 4$$ but horizontal asymptote is $$y=5$$, so the constant term must be 5, not 4. - Adjust function to $$h(x) = \frac{a}{x+p} + 5$$. - Given point $$(8,4)$$ lies on the curve: $$4 = \frac{a}{8+p} + 5 \Rightarrow \frac{a}{8+p} = -1$$. - Since vertical asymptote at $$x=8$$, $$8+p=0 \Rightarrow p=-8$$. - Substitute $$p=-8$$: $$\frac{a}{8-8} = \frac{a}{0}$$ undefined, but point is at $$x=8$$, so point cannot be on the curve at vertical asymptote. - The point is at $$(8,4)$$ on the curve, but vertical asymptote at $$x=8$$ means function undefined there, so point must be near asymptote, not on it. - Re-examine problem: horizontal asymptote is $$y=5$$, vertical asymptote $$x=8$$, point on curve at $$(8,4)$$. - Since function undefined at $$x=8$$, point cannot be on curve at $$x=8$$. - Possibly point is at $$(7,4)$$ or $$(9,4)$$, or problem states point at $$(8,4)$$ on curve, meaning function value at $$x=8$$ is 4. - To resolve, assume function is $$h(x) = \frac{a}{x+p} + 4$$ with horizontal asymptote $$y=5$$, so constant term is 4, but asymptote is 5, so constant term must be 5. - Given asymptote $$y=5$$, so function is $$h(x) = \frac{a}{x+p} + 5$$. - Vertical asymptote at $$x=8$$ means $$p=-8$$. - Use point $$(8,4)$$: $$4 = \frac{a}{8-8} + 5$$ undefined. - Contradiction. - Possibly point is at $$(7,4)$$ or $$(9,4)$$. - Assume point is $$(7,4)$$: $$4 = \frac{a}{7-8} + 5 = \frac{a}{-1} + 5 \Rightarrow \frac{a}{-1} = -1 \Rightarrow a=1$$. - So $$a=1$$, $$p=-8$$. 3. **Domain of $$h$$:** - Domain excludes vertical asymptote $$x=-p=8$$. - So domain is $$\{x \in \mathbb{R} : x \neq 8\}$$. 4. **Line of symmetry with negative gradient:** - For hyperbola $$h(x) = \frac{a}{x+p} + 5$$, line of symmetry is vertical line through asymptote $$x=8$$ or horizontal line through horizontal asymptote $$y=5$$. - Negative gradient line of symmetry is the line through the center of hyperbola. - Center is at $$( -p, 5 ) = (8,5)$$. - Line with negative gradient passing through $$(8,5)$$ is $$y = -x + b$$. - Find $$b$$: $$5 = -8 + b \Rightarrow b=13$$. - So line of symmetry is $$y = -x + 13$$. 5. **New horizontal asymptote after reflection and shift:** - Reflect $$h$$ in y-axis: replace $$x$$ by $$-x$$, so asymptote $$x=8$$ becomes $$x=-8$$. - Shift 4 units down: horizontal asymptote $$y=5$$ becomes $$y=5-4=1$$. - New horizontal asymptote is $$y=1$$. 6. **Find $$m$$ and $$n$$ for $$k(x) = h(x+m) + n$$ with lines of symmetry intersecting at $$(5,-2)$$:** - Lines of symmetry intersect at center of hyperbola. - Original center at $$(8,5)$$. - After transformation $$x \to x+m$$ and vertical shift $$+n$$, center moves to $$(8 - m, 5 + n)$$. - Given center is $$(5,-2)$$, so: $$8 - m = 5 \Rightarrow m = 3$$ $$5 + n = -2 \Rightarrow n = -7$$ **Final answers:** - $$a=1$$, $$p=-8$$ - Domain: $$x \neq 8$$ - Line of symmetry: $$y = -x + 13$$ - New horizontal asymptote after reflection and shift: $$y=1$$ - Values: $$m=3$$, $$n=-7$$