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Functions, 2ADV F2 EQ-Bank 2

A graph of the hyperbola  \(y=\dfrac{1}{x+p}+q\)  is shown, where \(p\) and \(q\) are constants.
 

Find the values of \(p\) and \(q\) and hence the graphs \(x\)-axis intercept.   (2 marks)

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\(\text {Vertical asymptote at } \ x=-1 \ \ \Rightarrow p=1\)

\(\text {Horizontal asymptote at } \ y=\dfrac{3}{2} \ \  \Rightarrow q=\dfrac{3}{2}\)

\(y=\dfrac{1}{x+1}+\dfrac{3}{2}\)
 

\(x\text{-intercept when} \ \ y=0:\)

  \(\dfrac{1}{x+1}+\dfrac{3}{2}\)  \(=0\)
  \(\dfrac{1}{x+1}\)  \(=-\dfrac{3}{2}\)
  \(3 x+3\)  \(=-2\)
  \(3x\)  \(=-5\)
  \(x\)  \(=-\dfrac{5}{3}\)

Show Worked Solution

\(\text {Vertical asymptote at } \ x=-1 \ \ \Rightarrow p=1\)

\(\text {Horizontal asymptote at } \ y=\dfrac{3}{2} \ \  \Rightarrow q=\dfrac{3}{2}\)

\(y=\dfrac{1}{x+1}+\dfrac{3}{2}\)
 

\(x\text{-intercept when} \ \ y=0:\)

  \(\dfrac{1}{x+1}+\dfrac{3}{2}\)  \(=0\)
  \(\dfrac{1}{x+1}\)  \(=-\dfrac{3}{2}\)
  \(3 x+3\)  \(=-2\)
  \(3x\)  \(=-5\)
  \(x\)  \(=-\dfrac{5}{3}\)

Functions, 2ADV F2 2022 SPEC2 3 MC

The graph of  `y=\frac{x^2+2x+c}{x^2-4}`, where `c \in R`, will always have

  1. two vertical asymptotes and one horizontal asymptote.
  2. a vertical asymptote with equation `x=-2` and one horizontal asymptote with equation `y=1`.
  3. one horizontal asymptote with equation `y=1` and only one vertical asymptote with equation `x=2`.
  4. a horizontal asymptote with equation `y=1` and at least one vertical asymptote.
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`D`

Show Worked Solution

`y=\frac{x^2+2x+c}{x^2-4}\ \ =>\ \ y=\frac{x^2+2x+c}{(x-2)(x+2)}`

`text{Vertical asymptotes:}\ x=2, \ x=-2`

`->\ text{However, if}\ c=0,\ \text{only 1 vertical asymptote at}\ \ x=2.`

`text{Horizontal asymptote:}\ y=1`

`=>D`


♦♦ Mean mark 38%.
41% of students chose `A` and did not consider when `c = 0`

Functions, 2ADV F2 2023 MET1 3

  1. Sketch the graph of  \(f(x)=2-\dfrac{3}{x-1}\) on the axes below, labelling all asymptotes with their equation and axial intercepts with their coordinates.   (3 marks)
     


  2. Find the values of \(x\) for which \(f(x)\leq1\).   (1 mark)
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a.   

b.    \(1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]\)

Show Worked Solution

a.    \(\text{Vertical asymptote when}\ \ x=1\)

\(y\text{-int:}\ y=2-(-3)\ \ \Rightarrow\ \ y=5\)

\(x\text{-int:}\ 2-\dfrac{3}{x-1}=0\ \ \Rightarrow\ \ x=\dfrac{5}{2} \)

\(\text{As}\ \ x \rightarrow \infty, \ \ y \rightarrow 2^{-}; \ \ x \rightarrow -\infty, \ \ y \rightarrow 2^{+} \)

b.    \(\text{From the graph:}\)

\(f(x)=1\ \text{when }x=4\)

\(x>1\ \text{to the right of the vertical asymptote}\)

\(\therefore\ f(x)\leq1\ \text{when}\ \ 1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]\)


♦♦ Mean mark (b) 38%.
MARKER’S COMMENT: Many students did not use their graph from part (a).

Functions, 2ADV F2 SM-Bank 20

  1. Sketch the graph of  `y = 1 - 2/(x - 2)`  on the axes below. Label asymptotes with their equations and axis intercepts with their coordinates.  (3 marks)
     

     
  2. Find the values of  `x`  for which  `1 - 2/(x - 2) >= 3`.  (1 mark)

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a. 

b. `x in [1, 2)`

Show Worked Solution

a.   `text(Asymptotes:)`

`x=2`

`text(As)\ \ x→ +-oo, \ y→1\ \ =>\ text(Asymptote at)\ \ y=1`

`ytext(-intercept at)\ (0,2)`

`xtext(-intercept at)\ (4,0)`


 

b.   `text(By inspection of the graph:)`

`1-2/(x-2) >=3\ \ text(for)\ \ x in [1, 2)`

Functions, 2ADV F2 2021 HSC 19

Without using calculus, sketch the graph of  `y = 2 + 1/(x + 4)`, showing the asymptotes and the `x` and `y` intercepts.  (3 marks)

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Show Worked Solution

`text(Asymptotes:)\ x = -4`

`text(As)\ \ x -> ∞, y -> 2`

`ytext(-intercept occurs when)\ \ x = 0:`

`y = 2.25`

`xtext(-intercept occurs when)\ \ y = 0:`

`2 + 1/(x + 4) = 0 \ => \ x = -4.5`
 

Functions, 2ADV F2 EQ-Bank 9

Consider the function  `f(x) = 1/(4x - 1)`.

  1. Find the domain of  `f(x)`.  (1 mark)

  2. Sketch  `f(x)`, showing all asymptotes and intercepts?  (2 marks)

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  1. `{text(all real)\ x,  x!=1/4}`
  2.   
Show Worked Solution
i.    `4x – 1` `!= 0`
  `x` `!= 1/4`

 
`:.\ text(Domain:)\ {text(all real)\ x,  x!=1/4}`

 

ii.   `text(When)\ \ x = 0, \ y = −1`

`text(As)\ \ x -> ∞, \ y -> 0^+`

`text(As)\ \ x -> −∞, \ y -> 0^−`

Functions, 2ADV F2 SM-Bank 12

Sketch the graph of  `f(x) = (2x+1)/(x-1)`. Label the axis intercepts with their coordinates and label any asymptotes with the appropriate equation.  (4 marks)

  

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Show Worked Solution
`(2x+1)/(x-1)` `=(2x-2+3)/(x-1)`  
  `=(2(x-1)+3)/(x-1)`  
  `=2 + 3/(x-1)`  

COMMENT: Manipulation of the equation makes graphing much easier.

 
`text(Asymptotes:)\ \ x = 1,\ \ y = 2`

`text(As)\ \ x->oo,\ \ y->2(+)`

`text(As)\ \ x->-oo,\ \ y->2(-)`

`text(As)\ \ x->-1 (-),\ \ y->-oo`

`text(As)\ \ x->-1 (+),\ \ y->oo`

 

 

Functions, 2ADV F2 EQ-Bank 11

On the axes below, sketch the graph of   `f(x) = (2x-2)/(x + 1)`.

Label all axis intercepts. Label each asymptote with its equation. (4 marks)

 
met1-2008-vcaa-q2

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met1-2008-vcaa-q2-answer1

Show Worked Solution
`(2x-2)/(x + 1)` `=(2(x+1) – 4)/(x+1)`
  `=2- 4/((x+1)`

 
`text(Asymptotes:)\ \ x = −1,\ \ y = 2`

`text(As)\ \ x->oo,\ \ y->2(-)`

`text(As)\ \ x->-oo,\ \ y->2(+)`

`text(As)\ \ x->-1 (-),\ \ y->oo`

`text(As)\ \ x->-1 (+),\ \ y->-oo`

 

met1-2008-vcaa-q2-answer1

Functions, 2ADV F2 SM-Bank 10

Consider the function  `f(x) = x/(4 - x^2)`.

  1. Identify the domain of  `f(x)`.   (1 mark)

  2. Sketch the graph  `y = f(x)`, showing all intercepts and asymptotes.   (3 marks)

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  1. `text(Domain:)\ \ text(all)\ x,\ x != +- 2`
  2.  
     
Show Worked Solution

i.   `4-x^2 !=0`

`=> x!= 2 or -2`

`:.\ text(Domain:)\ \ text(all)\ x,\ x != +- 2`

 

ii. `text(Asymptotes at)\ \ x = +- 2`

`text(Passes through)\ (0,0)`

COMMENT: Note that  `x->2^(–)`  is a short way of writing as `x` approaches 2 from the negative (or left-hand) side. This notation can save time when required.
`text(As)` `\ \ x -> 2^(-),` `\ y -> oo`
  `\ \ x -> 2 ^(+),` `\ y -> – oo`
`text(As)` `\ \ x -> -2^ (-),` `\ y -> oo`
  `\ \ x -> -2^ (+),` `\ y -> -oo`
`text(As)` `\ \ x -> oo,` `\ y -> 0`
  `\ \ x -> – oo,` `\ y -> 0`

 

  EXT1 2013 11d

 

Functions, 2ADV’ F2 2007 HSC 3b

  1. Find the vertical and horizontal asymptotes of the hyperbola

     

    `qquad y = (x − 2)/(x − 4)`  and hence sketch the graph of  `y = (x − 2)/(x − 4)`.  (3 marks)

  2. Hence, or otherwise, find the values of  `x`  for which  `(x − 2)/(x − 4) ≤ 3`.  (2 marks)

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  1. `text(See Worked Solutions.)`
  2. `x < 4\ text(and)\ x ≥ 5`
Show Worked Solution

i.    `y = (x − 2)/(x − 4)`

`text(Vertical asymptote at)\ \ x = 4`

`lim_(x → ∞) (x − 2)/(x − 4)` `= lim_(x → ∞) (1 − 2/x)/(1 − 4/x)=1`

`ytext(–intercept)\ = 1/2`

`xtext(–intercept)\ = 2`

 

Geometry and Calculus, EXT1 2007 HSC 3b Answer

 

ii.  `text(Find)\ \ x\ \ text(so that)\ \ (x − 2)/(x − 4) ≤ 3`

`(x − 2)/(x − 4)` `= 3`
`x − 2` `= 3x − 12`
`2x` `= 10`
`x` `= 5`

 
`=>(5, 3)\ \ text(is the intersection of)\ \ y = 3\ and\ y = (x − 2)/(x − 4)`
 

`:. (x − 2)/(x − 4) ≤ 3\ \ text(when)\ \ x < 4\ \ text(and)\ \ x ≥ 5.`

Functions, 2ADV’ F2 2012 HSC 13b

  1. Find the horizontal asymptote of the graph  `y=(2x^2)/(x^2 + 9)`.   (1 mark)

  2. Without the use of calculus, sketch the graph  `y=(2x^2)/(x^2 + 9)`, showing the asymptote found in part (i).    (2 marks)

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  1. `text(Horizontal asymptote at)\ y = 2`
  2.  
    Geometry and Calculus, EXT1 2012 HSC 13b Answer
Show Worked Solution
i.    `y` `= (2x^2)/(x^2 +9)`
    `= 2/(1 + 9/(x^2))`

 

`text(As)\ \ x -> oo,\ y ->2`

`text(As)\ \ x -> – oo,\ y -> 2`

`:.\ text(Horizontal asymptote at)\ y = 2`

 

ii.    `text(At)\ \ x = 0,\ y = 0`

`f(x) = (2x^2)/(x^2 + 9) >= 0\ text(for all)\ x`

`f(–x) = (2(–x)^2)/((–x)^2 + 9) = (2x^2)/(x^2 + 9) = f(x)`

`text(S)text(ince)\ \ f(x) = f(–x) \ \ =>\ text(EVEN function)`

Geometry and Calculus, EXT1 2012 HSC 13b Answer