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Functions, 2ADV 2026 HSC 22

  1. Show that  \(\dfrac{x+2}{x+1}=1+\dfrac{1}{x+1}\).   (1 mark)
  2. Describe the transformations required to obtain the graph  \(y=\dfrac{x+2}{x+1}\) from the graph of  \(y=\dfrac{1}{x}\).   (2 marks)
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a.    \(\text{See worked solutions}\)

b.    \(\text{Translate 1 unit to the left, then 1 unit up.}\)

Show Worked Solution
a.     \(\text{LHS}\) \(=\dfrac{x+2}{x+1}\)
    \(=\dfrac{(x+1)+1}{x+1}\)
    \(=1+\dfrac{1}{x+1}=\ \text{RHS}\)

 
b.    \(y=\dfrac{1}{x}\)

\(\text{Translate 1 unit to the left:}\)

\(y=\dfrac{1}{x}\ \rightarrow\ y=\dfrac{1}{x+1}\)
 

\(\text{Translate 1 unit up:}\)

\(y=\dfrac{1}{x+1}\ \rightarrow\ y=1+\dfrac{1}{x+1}=\dfrac{x+2}{x+1}\)

Functions, 2ADV F2 EQ-Bank 7 MC

The graph  \(y=\dfrac{2}{x-2}\)  undergoes the following transformations:

  • translated 3 units to the left
  • dilated vertically by a factor of 2

Determine which of the following is the new function.

  1.  \(2 y=\dfrac{2}{x+1}\)
  2.  \(\dfrac{y}{2}=\dfrac{2}{x-5}\)
  3.  \(2 y=\dfrac{2}{x-5}\)
  4.  \(\dfrac{y}{2}=\dfrac{2}{x+1}\)
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\(\Rightarrow D\)

Show Worked Solution

\(\text{Translate 3 units to the left:}\)

\(y=\dfrac{2}{x-2} \ \Rightarrow \ y^{′}=\dfrac{2}{(x+3)-2}=\dfrac{2}{x+1}\)
 

\(\text{Dilate vertically by a factor of 2:}\)

\(y^{′}=\dfrac{2}{x+1} \ \Rightarrow \ \dfrac{y^{″}}{2}=\dfrac{2}{x+1}\)

\(\Rightarrow D\)