Question

y = \frac{1}{x - 4} + 3
identify the asymptotes.
there is a vertical asymptote at x = 4.
there is a horizontal asymptote at y = 3.
choose the correct graph below.
\bigcirc a.\quad\includegraphicsheight=1cm{grapha}
\bigcirc b.\quad\includegraphicsheight=1cm{graphb}
\bigcirc c.\quad\includegraphicsheight=1cm{graphc}
\bigcirc d.\quad\includegraphicsheight=1cm{graphd}

Explanation:

Step1: Analyze Vertical Asymptote

The function \( y = \frac{1}{x - 4}+3 \) has a vertical asymptote at \( x = 4 \), so the graph should have a vertical dashed line at \( x = 4 \).

Step2: Analyze Horizontal Asymptote

The horizontal asymptote is \( y = 3 \), so the graph should approach the horizontal line \( y = 3 \) as \( x \to \pm\infty \).

Step3: Analyze Function Behavior

The parent function \( y=\frac{1}{x} \) is shifted right 4 units and up 3 units. For \( x>4 \), as \( x \) increases, \( y \) approaches 3 from above; for \( x<4 \), as \( x \) decreases, \( y \) approaches 3 from below. Now check the graphs:

• Graph A: Vertical asymptote at \( x = 4 \), horizontal at \( y = 3 \), and behavior matches (above 3 for \( x>4 \), below for \( x<4 \))? Wait, no, let's re - check. Wait, the function \( y=\frac{1}{x - 4}+3 \), when \( x>4 \), \( x - 4>0 \), so \( \frac{1}{x - 4}>0 \), so \( y=3+\text{positive} \), so \( y>3 \) for \( x>4 \); when \( x<4 \), \( x - 4<0 \), so \( \frac{1}{x - 4}<0 \), so \( y = 3+\text{negative} \), so \( y<3 \) for \( x<4 \). Now check the graphs:

Looking at the options, Graph D: Wait, no, let's check the vertical asymptote \( x = 4 \) (the line \( x = 4 \) is the vertical asymptote). The horizontal asymptote \( y = 3 \). Also, the function \( y=\frac{1}{x-4}+3\) has the same shape as \( y = \frac{1}{x} \) shifted. Let's check the quadrants (relative to asymptotes). For \( x>4 \), \( y>3 \); for \( x<4 \), \( y<3 \). Now looking at the graphs:
Graph D: Wait, no, let's re - evaluate. Wait, the correct graph should have vertical asymptote \( x = 4 \), horizontal \( y = 3 \), and for \( x>4 \), the graph is above \( y = 3 \), for \( x<4 \), below \( y = 3 \). Let's check the options:

• Option D: The graph has vertical asymptote \( x = 4 \), horizontal \( y = 3 \), and for \( x>4 \), the curve is above \( y = 3 \), for \( x<4 \), below \( y = 3 \), which matches the function \( y=\frac{1}{x - 4}+3 \). Wait, maybe I made a mistake earlier. Wait, let's check the original function's transformation. The parent function \( y=\frac{1}{x} \) is in the first and third quadrants. When we shift right 4 and up 3, the two branches are in the regions: for \( x>4 \), \( y>3 \) (like first quadrant relative to asymptotes) and for \( x<4 \), \( y<3 \) (like third quadrant relative to asymptotes). So the correct graph should be D? Wait, no, let's check the options again. Wait, maybe the correct graph is D? Wait, no, let's look at the graphs:

Wait, the function \( y=\frac{1}{x - 4}+3\), let's take a test point. When \( x = 5 \), \( y=\frac{1}{5 - 4}+3=1 + 3=4 \). When \( x = 3 \), \( y=\frac{1}{3 - 4}+3=-1 + 3 = 2 \). So at \( x = 5 \), \( y = 4 \) (above \( y = 3 \)), at \( x = 3 \), \( y = 2 \) (below \( y = 3 \)). Now check the graphs:

• Graph D: At \( x = 5 \) (right of \( x = 4 \)), the y - value is above 3, at \( x = 3 \) (left of \( x = 4 \)), the y - value is below 3, and has vertical asymptote \( x = 4 \) and horizontal \( y = 3 \). So the correct graph is D.

Answer:

D