iv. sketch the graph of each of the following functions. a f(x)=(1)/(x-4) b g(x)=(x+2)/(x^2)+5x+6 C h(x)=(x^2+1)/(x^2)-4x+4 d, k(x)=(x+6)/(x^2)+4 Let The ratio of boys to girls in a certain high school

To sketch the graphs of the given functions, we analyze their properties such as domain, asymptotes, intercepts, and behavior. Here's a breakdown for each function:---### **a) \( f(x) = \frac{1}{x-4} \)**- **Domain**: (denominator cannot be zero).- **Vertical Asymptote**: (where the denominator becomes zero).- **Horizontal Asymptote**: (as , \( f(x) \to 0 \)).- **Behavior**: The graph is a hyperbola with branches in quadrants I and III, centered around the vertical asymptote .---### **b) \( g(x) = \frac{x+2}{x^2 + 5x + 6} \)**- Factorize the denominator: \( x^2 + 5x + 6 = (x+2)(x+3) \).- **Domain**: (denominator cannot be zero).- **Vertical Asymptotes**: and .- **Simplification**: The numerator cancels with one factor in the denominator, leaving \( g(x) = \frac{1}{x+3} \).- **Horizontal Asymptote**: (as , \( g(x) \to 0 \)).- **Behavior**: The graph resembles a shifted hyperbola with a vertical asymptote at .---### **c) \( h(x) = \frac{x^2 + 1}{x^2 - 4x + 4} \)**- Factorize the denominator: \( x^2 - 4x + 4 = (x-2)^2 \).- **Domain**: (denominator cannot be zero).- **Vertical Asymptote**: .- **Horizontal Asymptote**: Since the degrees of the numerator and denominator are equal, divide the leading coefficients: .- **Behavior**: The graph approaches as , with a vertical asymptote at .---### **d) \( k(x) = \frac{x+6}{x^2 + 4} \)**- **Domain**: All real numbers ( for all ).- **Vertical Asymptote**: None (denominator never equals zero).- **Horizontal Asymptote**: Since the degree of the numerator is less than the degree of the denominator, .- **Behavior**: The graph has no vertical asymptotes and approaches as .---For the second part of your question regarding the ratio of boys to girls in a high school, please provide more details or clarify the context so I can assist further!