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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%.

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%.

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}

 

Calculus, SPEC2 2021 VCAA 1

Let  `f(x) = ((2x-3)(x + 5))/((x-1)(x + 2))`.

  1. Express `f(x)` in the form  `A + (Bx + C)/((x-1)(x + 2))`, where `A`, `B` an `C` are real constants.   (1 mark)

  2. State the equation of the asymptotes of the graph of `f`.   (2 marks)

  3. Sketch the graph of `f` on the set of axes below. Label the asymptotes with their equations, and label the maximum turning point and the point of inflection with their coordinates, correct to two decimal places. Label the intercepts with the coordinate axes.   (3 marks)

     

     

  4. Let  `g_k(x) = ((2x-3)(x + 5))/((x-k)(x + 2))`, where `k` is a real constant.
  5.  i. For what values of `k` will the graph of `g_k`, have two asymptotes?   (2 marks)

  6. ii. Given that the graph of `g_k` has more than two asymptotes, for what values of `k` will the graph of `g_k` have no stationary points?   (2 marks)

Show Answers Only
  1. `f(x) = 2 + (5x-11)/((x-1)(x + 2))`
  2. `text(Horizontal asymptote:)\ y = 2`

  3.  
  4.  i. `k = 3/2 => g_k(x) = 2 + 6/(x + 2)`
  5. ii. `k , -5\ \ text(or)\ \ k > 3/2`
Show Worked Solution

a.   `text{By CAS  (prop Frac}\ f(x)):`

`f(x) = 2 + (5x-11)/((x-1)(x + 2))`
 

b.   `text(Vertical asymptotes:)\ x = 1, x = –2`

`text(As)\ \ x -> ∞, y -> 2`

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

♦ Mean mark part (c) 48%.

 
c.   

 

d.i.   `text(Two asymptotes only when:)`

♦♦ Mean mark part (d)(i) 24%.

`k = -2 \ => \ g_k(x) = 2-(23 + x)/((x + 2)^2)`

`k = -5 \ => \ g_k(x) = 2-7/(x + 2)`

`k = 3/2 \ => \ g_k(x) = 2 + 6/(x + 2)`

 

d.ii.   `text(By CAS, solve)\ \ d/(dx)(g_k(x)) = 0\ \ text(for)\ \ x:`

♦♦♦ Mean mark part (d)(ii) 16%.

`x = (-4k + 15 ± sqrt(-21(2k^2 + 7k-15)))/(2k + 3)`
 

`text(No solutions occur when:)`

`k = -3/2\ \ text(or)`

`2k^2 + 7k-15 < 0`

`=> k < -5\ \ text(or)\ \ k > 3/2`

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`