SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Graphs, SPEC2 2025 VCAA 3 MC

The graph of  \(y=\dfrac{x^2+a}{b x+c}\)  has an asymptote given by  \(y=-\dfrac{1}{2} x+\dfrac{1}{4}\)  and a \(y\)-intercept of –2 .

The values of \(a, b\) and \(c\) are

  1. \(a=2, \ b=-2, \ c=-1\)
  2. \(a=2, \ b=2, \ c=-1\)
  3. \(a=-2, \ b=-2, \ c=1\)
  4. \(a=-2, \ b=-2, \ c=-1\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text { Expand expression by CAS: }\)

\(\dfrac{x^2+a}{b x+c}=\dfrac{a b^2+c^2}{b^2(b x+c)}+\dfrac{x}{b}-\dfrac{c}{b^2}\)

\(\text{Equating coefficients:}\)

\(\dfrac{x}{b}-\dfrac{c}{b^2}=-\dfrac{1}{2} x-\dfrac{1}{4}\  \ldots\ (1)\)
 

\(\text {Since } y(0)=-2\)

\(\dfrac{a}{c}=-2\ \ldots\ (2)\)
 

\(\text{Solve }(1) \text { and }(2) \text { for } a, b, c:\)

\(a=2, \ b=-2, \ c=-1\)

\(\Rightarrow A\)

Graphs, SPEC1 2025 VCAA 9

Let  \(f: R \backslash\{-1,1\} \rightarrow R, f(x)=\dfrac{x^3+x^2-2 x}{1-x^2}\).

  1. Show that \(f(x)\) can be written in the form  \(f(x)=-x-1+\dfrac{1}{x+1}\), for  \(x \in R \backslash\{-1,1\}\).   (2 marks)

  2. Consider the function with rule
    1. \begin{align*}
      g(x)=\left\{\begin{array}{cl}
      \dfrac{x^3+x^2-2 x}{1-x^2}, & x \in R \backslash\{-1,1\} \\
      k, & x \in\{1\}
      \end{array}\right.
      \end{align*}
  3. Find the value of \(k\) such that the graph of \(g\) is continuous at  \(x=1\).   (1 mark)

  4. Sketch the graph of  \(y=f(x)\) on the axes below.
  5. Label the asymptotes with their equations.   (3 marks)

Show Answers Only
a.     \(f(x)\) \(=\dfrac{x^3+x^2-2 x}{1-x^2}\)
    \(=\dfrac{x\left(x^2+x-2\right)}{(1-x)(1+x)}\)
    \(=\dfrac{-x(x+2)(1-x)}{(1-x)(1+x)}\)
    \(=\dfrac{-x(x+2)}{x+1}\)
    \(=\dfrac{-x^2-2 x}{x+1}\)
    \(=\dfrac{-(x+1)^2+1}{x+1}\)
    \(=-x-1+\dfrac{1}{x+1}\)

 

b.    \(k=-\dfrac{3}{2}\)

c.    
       

Show Worked Solution
a.     \(f(x)\) \(=\dfrac{x^3+x^2-2 x}{1-x^2}\)
    \(=\dfrac{x\left(x^2+x-2\right)}{(1-x)(1+x)}\)
    \(=\dfrac{-x(x+2)(1-x)}{(1-x)(1+x)}\)
    \(=\dfrac{-x(x+2)}{x+1}\)
    \(=\dfrac{-x^2-2 x}{x+1}\)
    \(=\dfrac{-(x+1)^2+1}{x+1}\)
    \(=-x-1+\dfrac{1}{x+1}\)

 

b.    \(\text{Find \(k\) such that \(g(x)\) is continuous:}\)

\(k\) \(=\dfrac{k^3+k^2-2 k}{1-k^2}\)
\(k-k^3\) \(=k^3+k^2-2 k\)
\(1-k^2\) \(=k^2+k-2\)
\(0\) \(=2 k^2+k-3\)
\(0\) \(=(2 k+3)(k-1)\)

 

\(\therefore k=-\dfrac{3}{2}\ \ (k \neq 1)\)

♦ Mean mark (b) 46%.

c.    \(\text{Intercepts where}\ \ -x-1+\dfrac{1}{x+1}=0:\)

\((x+1)^2=1 \ \ \Rightarrow \ \ x^2+2 x=0 \ \ \Rightarrow \ \ x=0 \ \ \text{or} \  -2\)

\(\text{Asymptotes:} \ \ y=-1-x, \ x=-1\)

\(\text{Hole at} \ \left(1,-\dfrac{3}{2}\right)\)

♦♦ Mean mark (c) 37%.

Functions, SPEC2 2024 VCAA 2 MC

Consider the function \(f\) with rule \(f(x)=\left\{\begin{array}{cl}\dfrac{x^2+3 x-10}{x-2} & , x \in R \backslash\{2\} \\ 7 & , x=2\end{array}\right.\)

Which of the following statements is correct?

  1. The function \(f\) is continuous.
  2. The graph of  \(y=f(x)\)  has a vertical asymptote.
  3. The graph of  \(y=f(x)\)  has a horizontal asymptote.
  4. The graph of  \(y=f(x)\)  has a point of discontinuity.
Show Answers Only

\(A\)

Show Worked Solution

\(\dfrac{x^2+3 x-10}{x-2} = \dfrac{(x+5)(x-2)}{x-2} \)

\(\text{As}\ x\rightarrow 2,\ \dfrac{(x+5)(x-2)}{x-2}\rightarrow 7 \)

\(\text{Piecewise function is continuous at}\ \ x=2.\)

\(\Rightarrow A\)

♦ Mean mark 48%.

Calculus, SPEC2 2022 VCAA 1

Consider the family of functions \(f\) with rule  \(f(x)=\dfrac{x^2}{x-k}\), where \(k \in R \backslash\{0\}\).

  1. Write down the equations of the two asymptotes of the graph of \(f\) when \(k=1\).   (2 marks)

  2. Sketch the graph of  \(y=f(x)\)  for  \(k=1\)  on the set of axes below. Clearly label any turning points with their coordinates and label any asymptotes with their equations.   (3 marks)
     

  1.  i. Find, in terms of \(k\), the equations of the asymptotes of the graph of  \(f(x)=\dfrac{x^2}{x-k}\).   (1 mark)

  2. ii. Find the distance between the two turning points of the graph of  \(f(x)=\dfrac{x^2}{x-k}\) in terms of \(k\).   (2 marks)

  3. Now consider the functions \(h\) and \(g\), where  \(h(x)=x+3\)  and  \(g(x)=\abs{\dfrac{x^2}{x-1}}\).
  4. The region bounded by the curves of \(h\) and \(g\) is rotated about the \(x\)-axis.
    1. Write down the definite integral that can be used to find the volume of the resulting solid.   (2 marks)

    2. Hence, find the volume of this solid. Give your answer correct to two decimal places.   (1 mark)

Show Answers Only

a.  \(\text {Asymptotes: } x=1,\  y=x+1\)

b.   
       

c.i.   \(\text {Asymptotes: } x=k,\  y=x+k\)

c.ii.  \(\text {Distance }=2 \sqrt{5}|k|\)

d.i.  \(\displaystyle V=\pi \int_{\frac{-\sqrt{7}-1}{2}}^{\frac{\sqrt{7}-1}{2}}(x+3)^2-\left(\frac{x^2}{x-1}\right)^2 dx\)

d.ii.  \(V=51.42\ \text{u}^3 \)

Show Worked Solution

a.    \(\text {When } k=1 :\)

\(f(x)=\dfrac{x^2}{x-1}=\dfrac{(x+1)(x-1)+1}{(x-1)}=x+1+\dfrac{1}{x-1}\)

\(\text {Asymptotes: } x=1,\  y=x+1\)
 

b.    
       

 

c.i. \(f(x)=\dfrac{x^2}{x-k}=\dfrac{(x+k)(x-k)+k^2}{x-k}=x+k+\dfrac{k^2}{x-k}\)

\(\text {Using part a.}\)

\(\text {Asymptotes: } x=k,\  y=x+k\)
 

c.ii.  \(f^{\prime}(x)=1-\left(\dfrac{k}{x-k}\right)^2\)

\(\text {TP’s when } f^{\prime}(x)=0 \text { (by CAS):}\)

\(\Rightarrow(2 k, 4 k),(0,0)\)

\(\text {Distance }\displaystyle=\sqrt{(2 k-0)^2+(4 k-0)^2}=\sqrt{20 k^2}=2 \sqrt{5}|k|\)
 

d.i  \(\text {Solve for intersection of graphs (by CAS):}\)

\(\displaystyle x+3=\left|\frac{x^2}{x-1}\right|\)

\(\displaystyle \Rightarrow x=\frac{3}{2}, x=\frac{-1 \pm \sqrt{7}}{2}\)

\(\displaystyle V=\pi \int_{\frac{-\sqrt{7}-1}{2}}^{\frac{\sqrt{7}-1}{2}}(x+3)^2-\left(\frac{x^2}{x-1}\right)^2 dx\)
 

d.ii. \(V=51.42\ \text{u}^3 \text{ (by CAS) }\)

♦♦ Mean mark (d)(ii) 37%.

Graphs, SPEC2 2022 VCAA 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. two horizontal asymptotes and one vertical asymptote.
  3. a vertical asymptote with equation `x=-2` and one horizontal asymptote with equation `y=1`.
  4. one horizontal asymptote with equation `y=1` and only one vertical asymptote with equation `x=2`.
  5. a horizontal asymptote with equation `y=1` and at least one vertical asymptote.
Show Answers Only

`E`

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`

`text{Using CAS to graph for different values of}\ c:`

`=>E`


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

Functions, SPEC2 2023 VCAA 2 MC

The graph of  \(y=\dfrac{x^3}{a x^2+b x+c}\)  has asymptotes given by  \(y=2 x+1\)  and  \(x=1\). The values of \(a, b\) and \(c\) are, respectively

  1. \(2,-4,2\)
  2. \( \dfrac{1}{2},-\dfrac{1}{4},-\dfrac{1}{4} \)
  3. \( \dfrac{1}{2}, \dfrac{1}{4},-\dfrac{3}{4} \)
  4. \( \dfrac{1}{2},-\dfrac{1}{4},-\dfrac{3}{4} \)
  5. \(2,-4,-8\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Using partial fractions (by calc):}\)

\(\text{Remainder}\ =  \dfrac{x}{a}-\dfrac{b}{a^2} =  2x+1\)

\(a=\dfrac{1}{2},\ \ b=-\dfrac{1}{4} \)

\(x=1\ \ \text{is a solution of}\ \ ax^2+bx+c=0\)

\(c=-\dfrac{1}{4} \)

\(\Rightarrow B\)

Functions, SPEC1 2023 VCAA 1

Consider the function \(f\) with rule  \(f(x)=\dfrac{x^2+x-6}{x-1}\).

  1. Show that the rule for the function \(f\) can be written as  \(f(x)=x+2-\dfrac{4}{x-1}\).   (1 mark)
  1. Sketch the graph of \(f\) on the axes below, labelling any asymptotes with their equations.   (3 marks)

   

Show Answers Only
a.     \(f(x)\) \(=\dfrac{x^2+x-6}{x-1}\)
    \(=\dfrac{(x-1)(x+2)-4}{x-1}\)
    \(=x+2-\dfrac{4}{x-1} \)

 
b.    \(\text{Asymptotes at:}\ x=1, \ y=x+2 \)

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -1 & 0 & 1 & 2 \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 0 & 3 & 6 & \infty & 0 \\
\hline
\end{array}

 

Show Worked Solution
a.     \(f(x)\) \(=\dfrac{x^2+x-6}{x-1}\)
    \(=\dfrac{(x-1)(x+2)-4}{x-1}\)
    \(=x+2-\dfrac{4}{x-1} \)

 
b.    \(\text{Asymptotes at:}\ x=1, \ y=x+2 \)

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -1 & 0 & 1 & 2 \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 0 & 3 & 6 & \infty & 0 \\
\hline
\end{array}

 

Graphs, SPEC2 2019 VCAA 2 MC

The asymptote(s) of the graph of  `f(x) = (x^2 + 1)/(2x - 8)`  has equation(s)

  1. `x = 4`
  2. `x = 4 and y = x/2`
  3. `x = 4 and y = x/2 + 2`
  4. `x = 8 and y = x/2`
  5. `x = 8 and y = 2x + 2`
Show Answers Only

`C`

Show Worked Solution
`(x^2 + 1)/(2x – 8)` `= ((x – 4)(x + 4) + 17)/(2(x – 4))`
  `= (x + 4)/2 + 17/(2x – 8)`
  `= x/2 + 2 + 17/(2x – 8)`

 
`:.\ text(Asymptotes at)\ \ x = 4 and y = x/2 + 2`

`=>C`

Graphs, SPEC2 2011 VCAA 1 MC

The number of straight line asymptotes of the graph of  `y = (2x^3 + x^2 - 1)/(x^2 - x - 2)`  is

A.   0

B.   1

C.   2

D.   3

E.   4

Show Answers Only

`D`

Show Worked Solution
`y` `= (2x^3 + x^2 – 1)/(x^2 – x – 2)`
  `= 2x + 3 + (7x + 5)/((x – 2)(x + 1))`

 
`:.\ text(Straight line asymptotes occur at:)`

♦♦ Mean mark 34%.

`y = 2x + 3, \ x = 2\ \ text(and)\ \ x = −1`

`=> D`

Graphs, SPEC2 2012 VCAA 3 MC

The graph of  `y = f(x)`  is shown below.
 

All the axes below have the same scale as the axes in the diagram above.

The graph of  `y = 1/(f(x))`  is best represented by

A.    B.   
C.    D.   
E.       

Show Answers Only

`E`

Show Worked Solution

`y(-1) = 0`

`-> 1/(y(-1)):\ text(asymptote)\ x = -1`

`text(Eliminate A and B.)`

`text(Local max at)\ \ x = 1`

`-> 1/(y(1)):\ text(local min)`
 

`=> E`

Graphs, SPEC2 2012 VCAA 1 MC

The graph with equation  `y = 1/(2x^2 - x - 6)` has asymptotes given by

A.  `x = -3/2,\ x = 2 and y = 1`

B.   `x = -3/2 and x = 2`  only

C.   `x = 3/2,\ x = -2 and y = 0`

D.   `x = -3/2,\ x = 2 and y = 0`

E.   `x = 3/2 and x = -2`  only

Show Answers Only

`D`

Show Worked Solution
`y` `=1/(2x^2 – x – 6)`  
  `=1/((2x + 3)(x – 2))`  

 

`:.\ text(Asymptotes:)`   `x = 2`
  `x = -3/2`
  `y = 0`

 
`=> D`

Algebra, SPEC2 2013 VCAA 3 MC

The graph of  `y = 1/(ax^2 + bx + c)`  has asymptotes at  `x = − 5`, `x = 3`  and  `y = 0`.

Given that the graph has one stationary point with a `y`-coordinate of  `−1/8`, it follows that

A.   `a = 1`, `b = 2`, `c = −15`

B.   `a = 1/2`, `b = −1`, `c = −15/2`

C.   `a = −1/2`, `b = −1`, `c = 15`

D.   `a = −1`, `b = −2`, `c = −15`

E.   `a = 1/2`, `b = 1`, `c = −15/2`

Show Answers Only

`E`

Show Worked Solution
`y` `= 1/(ax^2 + bx + c)`
  `= 1/(a(x + 5)(x – 3))`
  `= 1/(ax^2 + 2ax – 15a)`

 
`ax^2 + bx + c = ax^2 + 2ax – 15a`

♦ Mean mark 47%.

 
`text(Vertex:)\ \ ((− 5 + 3)/2, – 1/8) = (– 1,– 1/8)`

`- 1/8` `= 1/(a – 2a – 15a)`
`- 1/8` `= -1/(16a)`
`a` `= 1/2`

 

`:. a = 1/2, \ b = 2a = 1, \ c = − 15a = − 15/2`

`=> E`

Graphs, SPEC2 2016 VCAA 3 MC

The straight-line asymptote(s) of the graph of the function with rule  `f(x) = (x^3 - ax)/x^2`, where  `a`  is a non-zero real constant, is given by

A.  `x = 0`  only.

B.  `x = 0`  and  `y = 0`  only.

C.  `x = 0`  and  `y = x`  only.

D.  `x = 0, \ x = sqrt a`  and  `x = -sqrt a`  only.

E.  `x = 0`  and  `y = a`  only. 

Show Answers Only

`C`

Show Worked Solution

`f(x) = x – a/x,\ x != o`

`lim_(x -> oo) f(x) = x`

`:.\ text(Asymptotes):\ \ x = 0,\ \ y = x`

 
`=>  C`

Graphs, SPEC2-NHT 2017 VCAA 1 MC

The number of asymptotes of the graph of the function with rule  `f(x) = (x^3 - 7x + 5)/(x^2 + 3x - 4)`  is

A.   0

B.   1

C.   2

D.   3

E.   4

Show Answers Only

`D`

Show Worked Solution

`text(By CAS:)`

`(x^3 – 7x + 5)/(x^2 + 3x – 4)`

`=x-3 +31/(5(x+4)) – 1/(5(x-1))`

 
`:. text(Asymptotes at:)\ \ x=1,\ \ x=-4,\ \ y=x-3`

`=>   D`

Graphs, SPEC2 2014 VCAA 3 MC

The features of the graph of the function with rule  `f(x) = (x^2 - 4x + 3)/(x^2 - x - 6)` include

A.   asymptotes at  `x = 1`  and  `x = −2`

B.   asymptotes at  `x = 3`  and  `x = −2`

C.   an asymptote at  `x = 1`  and a point of discontinuity at  `x = 3`

D.   an asymptote at  `x = −2`  and a point of discontinuity at  `x = 3`

E.   an asymptote at  `x = 3`  and a point of discontinuity at  `x = −2`

Show Answers Only

`D`

Show Worked Solution
`f(x)` `= (x^2 – 4x + 3)/(x^2 – x – 6)`
  `=(x^2 -x-6 -3x+9)/(x^2 – x – 6)`
  `= 1 – (3(x – 3))/((x + 2)(x – 3))`
  `=1 – 3/(x + 2)\ text(where)\ x != 3`

 
`=> D`

Graphs, SPEC1 2018 VCAA 5

Sketch the graph of  `f(x) = (x + 1)/(x^2 - 4)`  on the axes provided below, labelling any asymptotes with their equations and any intercepts with their coordinates.  (4 marks)

Show Answers Only

Show Worked Solution

`text(Using partial fractions:)`

`(x + 1)/(x^2 – 4)= A/(x – 2) + B/(x + 2)`

`A(x + 2) + B (x – 2) = x + 1`

 
`text(When)\ \ x = 2,`

`4A` `= 3 => \ A=3/4`

 
`text(When)\ \ x = -2`

`-4B` `= -1 =>\ B=1/4`

 
`f(x)= 1/(4(x + 2)) + 3/(4(x – 2))`
 

`text(As)\ \ x->oo, \ f(x)->0^+`

`text(As)\ \ x->- oo, \ f(x)->0^-`

`f(0)= – 1/4`
 

`text(Find)\ x\ text(when)\ \ f(x)=0:`

`x + 1` `= 0`
`x` `= -1`