Question

how is this graph obtained from the graph $f(x) = \frac{1}{x}$ using transformations?
a reflected across the y - axis and shifted up 2 units.
b reflected across the x - axis and shifted left 2 units.
c reflected across the y - axis and shifted down 2 units.
d reflected across the x - axis and shifted right 2 units.
e reflected across the x - axis and shifted right $\frac{1}{2}$ units.

Explanation:

Step1: Analyze the parent function

The parent function is \( f(x)=\frac{1}{x} \), which has two branches: one in the first quadrant (where \( x>0,y>0 \)) and one in the third quadrant (where \( x<0,y<0 \)).

Step2: Analyze the transformation - Reflection

Looking at the given graph, one branch is in the first quadrant (similar to the parent function's first - quadrant branch) and the other is in the fourth quadrant (where \( x > 0,y<0 \)). The parent function's third - quadrant branch (\( x<0,y<0 \)) is transformed to the fourth - quadrant branch (\( x > 0,y<0 \))? No, wait. Wait, the parent function \( y = \frac{1}{x} \): when \( x>0 \), \( y>0 \); when \( x < 0 \), \( y<0 \). The given graph has one branch for \( x>0,y>0 \) (similar to the parent's \( x > 0 \) branch) and one branch for \( x>2,y<0 \). Wait, no, let's think about reflection over \( x \) - axis: the transformation for reflection over \( x \) - axis is \( y=-f(x) \). So if we have \( y = \frac{1}{x} \), reflecting over \( x \) - axis gives \( y=-\frac{1}{x} \), which has a branch in the fourth quadrant (\( x>0,y < 0 \)) and a branch in the second quadrant (\( x<0,y>0 \)). But our graph has a branch in the first quadrant (\( x>0,y>0 \)) and a branch in the fourth quadrant (\( x>2,y < 0 \)). Wait, maybe I made a mistake. Wait, let's check the horizontal shift.

Step3: Analyze the transformation - Horizontal shift

The vertical asymptote of the parent function \( y=\frac{1}{x} \) is \( x = 0 \). For the given graph, the vertical asymptote seems to be at \( x = 2 \). The formula for a horizontal shift of \( y = f(x) \) to the right by \( h \) units is \( y=f(x - h) \). If we shift \( y=\frac{1}{x} \) to the right by 2 units, we get \( y=\frac{1}{x - 2} \). But we also have a reflection. Let's consider the reflection over \( x \) - axis: if we first reflect \( y=\frac{1}{x} \) over \( x \) - axis, we get \( y =-\frac{1}{x} \), then shift it to the right by 2 units, we get \( y=-\frac{1}{x - 2} \).

Let's check the quadrants:
- For \( y =-\frac{1}{x-2} \), when \( x>2 \), \( x - 2>0 \), so \( y=-\frac{1}{x - 2}<0 \) (fourth quadrant branch), and when \( x<2 \), \( x - 2<0 \), so \( y=-\frac{1}{x - 2}>0 \) (first quadrant branch when \( x>2 \)? No, wait, when \( x<2 \), \( x - 2<0 \), so \( y =-\frac{1}{x - 2}=\frac{1}{2 - x}>0 \), so for \( x<2 \), \( y>0 \) (first quadrant when \( x<2,y>0 \))? Wait, no, the graph in the picture: one branch is for \( x<2,y>0 \) (since at \( x = 0 \), the \( y \) - value is positive, around \( y = 1 \) or so) and one branch for \( x>2,y<0 \).

Wait, let's check the options:
- Option D: Reflected across the \( x \) - axis and shifted right 2 units. Let's verify the transformation.
- First, reflect \( f(x)=\frac{1}{x} \) across the \( x \) - axis: the transformation is \( y=-f(x)=-\frac{1}{x} \).
- Then shift right 2 units: the transformation for a horizontal shift right by \( h \) units is \( y = - f(x - 2)=-\frac{1}{x - 2} \).
- For \( x<2 \), \( x - 2<0 \), so \( y=-\frac{1}{x - 2}=\frac{1}{2 - x}>0 \) (so the branch is in the region \( x<2,y>0 \), which matches the left - hand branch of the given graph).
- For \( x>2 \), \( x - 2>0 \), so \( y=-\frac{1}{x - 2}<0 \) (so the branch is in the region \( x>2,y<0 \), which matches the right - hand branch of the given graph).

Now let's check other options:
- Option A: Reflected across \( y \) - axis and shifted up 2 units. Reflecting \( y=\frac{1}{x} \) across \( y \) - axis gives \( y=\frac{1}{-x}=-\frac{1}{x} \), then shifting up 2 units gives \( y =-\frac{1}{x}+2 \). For \( x>0 \), when \( x \) is small, \( -\frac{1}{x} \) is a large negative number, \( y=-\frac{1}{x}+2 \) would be negative, but our graph has a positive \( y \) - value for small \( x \) (around \( x = 0 \)), so A is wrong.
- Option B: Reflected across \( x \) - axis and shifted left 2 units. Reflecting across \( x \) - axis gives \( y =-\frac{1}{x} \), shifting left 2 units gives \( y=-\frac{1}{x + 2} \). For \( x>-2 \), when \( x = 0 \), \( y=-\frac{1}{2}<0 \), but our graph has \( y>0 \) at \( x = 0 \), so B is wrong.
- Option C: Reflected across \( y \) - axis and shifted down 2 units. Reflecting across \( y \) - axis gives \( y=-\frac{1}{x} \), shifting down 2 units gives \( y=-\frac{1}{x}-2 \). At \( x = 0.5 \), \( y=-2 - 2=-4<0 \), but our graph has \( y>0 \) at \( x = 0.5 \), so C is wrong.
- Option E: Reflected across \( x \) - axis and shifted right \( \frac{1}{2} \) units. Shifting right \( \frac{1}{2} \) units gives \( y=-\frac{1}{x-\frac{1}{2}} \). The vertical asymptote would be at \( x=\frac{1}{2} \), but our graph's vertical asymptote is at \( x = 2 \), so E is wrong.

Answer:

D. Reflected across the x - axis and shifted right 2 units.