Question

consider the graph shown. which of the following is the function represented by the graph? $y = \frac{1}{(x - 3)} - 5$ $y = \frac{1}{(x + 3)} - 5$ $y = \frac{1}{(x - 5)} + 3$ $y = \frac{1}{(x + 5)} - 3$

Explanation:

Step1: Identify Vertical Asymptote

The vertical asymptote of a rational function \( y = \frac{1}{(x - h)} + k \) is at \( x = h \). From the graph, the vertical asymptote is at \( x = 3 \)? Wait, no, looking at the graph, the vertical asymptote is at \( x = 3 \)? Wait, the graph has a vertical asymptote where the curve approaches a vertical line. Looking at the graph, the vertical asymptote is at \( x = 3 \)? Wait, no, the left curve is near \( x = 3 \)? Wait, no, let's check the options. Wait, the standard form is \( y=\frac{1}{x - h}+k \), vertical asymptote at \( x = h \), horizontal asymptote at \( y = k \).

From the graph, the horizontal asymptote is \( y = -5 \)? Wait, no, the horizontal line the graph approaches is \( y = -5 \)? Wait, the graph has a horizontal asymptote. Let's check the horizontal asymptote: the horizontal asymptote of \( y=\frac{1}{x - h}+k \) is \( y = k \). From the graph, the horizontal asymptote is \( y = -5 \)? Wait, no, the graph's horizontal asymptote: looking at the graph, the left and right curves approach \( y = -5 \)? Wait, no, the graph shows a horizontal asymptote. Wait, the options have \( k = -5 \) or \( +3 \) or \( -3 \).

Now vertical asymptote: the vertical asymptote is at \( x = 3 \) (from the graph, the vertical line the curve approaches is \( x = 3 \)). So in the function \( y=\frac{1}{x - h}+k \), \( h = 3 \), so \( x - 3 \) in the denominator. And horizontal asymptote \( y = -5 \), so \( k = -5 \). So the function is \( y=\frac{1}{(x - 3)} - 5 \). Wait, but let's check the options. The first option is \( y=\frac{1}{(x - 3)} - 5 \), second is \( y=\frac{1}{(x + 3)} - 5 \) (vertical asymptote \( x = -3 \)), third is \( y=\frac{1}{(x - 5)} + 3 \) (vertical \( x = 5 \), horizontal \( y = 3 \)), fourth is \( y=\frac{1}{(x + 5)} - 3 \) (vertical \( x = -5 \), horizontal \( y = -3 \)).

Wait, maybe I made a mistake. Let's re-examine the graph. The vertical asymptote: the graph has a vertical asymptote at \( x = 3 \)? Wait, the graph's vertical asymptote: the right curve is near \( x = 3 \), so vertical asymptote at \( x = 3 \), so \( h = 3 \), so denominator \( x - 3 \). Horizontal asymptote: the graph approaches \( y = -5 \), so \( k = -5 \). So the function is \( y=\frac{1}{(x - 3)} - 5 \), which is the first option.

Step2: Verify Horizontal Asymptote

The horizontal asymptote of \( y=\frac{1}{(x - 3)} - 5 \) is \( y = -5 \), which matches the graph's horizontal asymptote (the graph approaches \( y = -5 \)).

Step3: Verify Vertical Asymptote

The vertical asymptote of \( y=\frac{1}{(x - 3)} - 5 \) is \( x = 3 \), which matches the graph's vertical asymptote (the graph has a vertical asymptote at \( x = 3 \)).

Answer:

\( y = \frac{1}{(x - 3)} - 5 \) (the first option, assuming the first option is labeled as, e.g., A. \( y = \frac{1}{(x - 3)} - 5 \))