Question

match the function with its graph.
$y = \frac{1}{x - 4} + 1$
graphs are shown, but ocr text only includes the function and the instruction to match with its graph

Explanation:

Step1: Identify Vertical Asymptote

The function is \( y = \frac{1}{x - 4} + 1 \). The vertical asymptote occurs where the denominator is zero, so \( x - 4 = 0 \) gives \( x = 4 \). So we look for graphs with a vertical asymptote at \( x = 4 \).

Step2: Identify Horizontal Asymptote

For rational functions, as \( |x| \to \infty \), \( \frac{1}{x - 4} \to 0 \), so \( y \to 0 + 1 = 1 \). Thus, the horizontal asymptote is \( y = 1 \).

Step3: Analyze Behavior Around Asymptote

- For \( x > 4 \), as \( x \) approaches 4 from the right (\( x \to 4^+ \)), \( x - 4 \to 0^+ \), so \( \frac{1}{x - 4} \to +\infty \), and \( y \to +\infty + 1 = +\infty \).
- For \( x < 4 \), as \( x \) approaches 4 from the left (\( x \to 4^- \)), \( x - 4 \to 0^- \), so \( \frac{1}{x - 4} \to -\infty \), and \( y \to -\infty + 1 = -\infty \).
- Also, when \( x = 5 \), \( y = \frac{1}{5 - 4} + 1 = 2 \); when \( x = 3 \), \( y = \frac{1}{3 - 4} + 1 = 0 \).

Now, checking the graphs:
- The first graph has vertical asymptote at \( x = -4 \), so eliminate.
- The second graph: vertical asymptote at \( x = 4 \)? Wait, no, let's check the third graph (right middle) and the fourth? Wait, the third graph (right top) has vertical asymptote at \( x = 5 \)? No, wait the second row first graph? Wait, let's re - check. The function \( y=\frac{1}{x - 4}+1\) has vertical asymptote \( x = 4 \). Let's look at the graphs:

Looking at the graphs, the middle graph in the top row (second graph) has vertical asymptote at \( x = 4 \)? Wait, no, the third graph (right top) has vertical asymptote at \( x = 5 \)? Wait, no, let's check the x - axis. The graph in the middle of the top row (second graph) has a vertical asymptote at \( x = 4 \)? Wait, the x - axis labels: in the second graph (top middle), the vertical asymptote is at \( x = 4 \)? Wait, the open circle is at \( x = 4 \). Let's check the behavior:

For \( x>4 \), the graph should go to \( +\infty \), and for \( x < 4 \), go to \( -\infty \), with horizontal asymptote \( y = 1 \). The second graph (top middle) has horizontal asymptote \( y = 1 \), vertical asymptote \( x = 4 \), and when \( x>4 \), it goes up, \( x < 4 \) goes down. Wait, no, the third graph (right top) has vertical asymptote at \( x = 5 \)? No, the x - axis in the third graph (right top) has the open circle at \( x = 5 \)? Wait, no, the function is \( y=\frac{1}{x - 4}+1\), so vertical asymptote at \( x = 4 \). Let's check the graph in the middle of the top row (second graph):

In the second graph (top middle), the vertical asymptote is at \( x = 4 \) (the open circle is at \( x = 4 \)), horizontal asymptote \( y = 1 \), for \( x>4 \), the graph is increasing towards \( +\infty \), for \( x < 4 \), decreasing towards \( -\infty \), and when \( x = 5 \), \( y = 2 \) (which matches the graph's behavior as \( x = 5 \) is in \( x>4 \) and the graph is above \( y = 1 \)). Wait, no, let's check the third graph (right top):

Wait, maybe I made a mistake. Let's re - express the function as \( y - 1=\frac{1}{x - 4}\), which is a hyperbola, a transformation of \( y=\frac{1}{x}\) (shifted right 4 units and up 1 unit). The parent function \( y = \frac{1}{x}\) has branches in the first and third quadrants. After shifting right 4 and up 1, the branches are in the regions \( x>4,y > 1\) and \( x < 4,y < 1\).

Looking at the graphs, the graph in the middle of the top row (second graph) has:
- Horizontal asymptote \( y = 1 \)
- Vertical asymptote \( x = 4 \)
- For \( x>4 \), the graph is in the region \( y>1 \) (going up as \( x\to4^+ \))
- For \( x < 4 \), the graph is in the region \( y < 1 \) (going down as \( x\to4^- \))

Yes, this matches the function \( y=\frac{1}{x - 4}+1\).

Answer:

The Middle Graph in the Top Row (the second graph from the left in the top row)