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Exponentials, SMB-005

Sketch the graph of  `y=3^x+2`, clearly labelling any asymptotes.  (3 marks)

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`y=3^x+2`

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1  & \ \ 0\ \  & \ \ 1\ \  & \ \ 2\ \  \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 2 \frac{1}{9} & 2 \frac{1}{3} & 3 & 5 & 11\\
\hline
\end{array}

Exponentials, SMB-004

Sketch the graph of  `y=3(2^(-x))-1`  (3 marks)

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\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -1 & \ \ 0\ \  & \ \ 1\ \  & \ \ 2\ \  & \ \ 3\ \  \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 5 & 2 & \frac{1}{2} & -\frac{1}{4} & -\frac{5}{8}\\
\hline
\end{array}

Exponential, SMB-003

By completing the table of values, sketch the graph of  `y=2^(-x)`  (3 marks)

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

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\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \  & \ \ 1\ \  & \ \ 2\ \  \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 4 & 2 & 1 & \frac{1}{2} & \frac{1}{4}\\
\hline
\end{array}

Exponentials, SMB-002

  1. Complete the table of values below given  `y=2^x-1`  (2 marks)

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

  1. Draw the graph of  `y=2^x-1`  on the Cartesian plane below.  (2 marks)


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i.   

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \  & \ \ 1\ \  & \ \ 2\ \  \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & -\frac{3}{4} & -\frac{1}{2} & 0 & 1 & 3\\
\hline
\end{array}

ii. 

Show Worked Solution

i.   

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \  & \ \ 1\ \  & \ \ 2\ \  \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & -\frac{3}{4} & -\frac{1}{2} & 0 & 1 & 3\\
\hline
\end{array}

ii.   

 

Exponentials, SMB-001

In 2004 there were 13.5 million registered motor vehicles in Australia. The number of registered motor vehicles is increasing at a rate of 2.3% per year.

Find an expression that represents the number (in millions) of registered motor vehicles `(V)`, if `y` represents the number of years after 2004?  (2 marks)

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`V = 13.5 xx (1.023)^y`

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`text(In 2004, 13.5 million)`

`Vtext{(1 year later)} = 13.5 xx (1.023)`

`Vtext{(2 years later)}= 13.5 xx (1.023) xx (1.023)= 13.5 xx (1.023)^2`

`:. V(y\ text{years later}) = 13.5 xx (1.023)^y`

Algebra, STD2 A4 2021 HSC 24

A population,  `P`,  is to be modelled using the function  `P = 2000 (1.2)^t`,  where `t` is the time in years.

  1. What is the initial population?  (1 mark)

  2. Find the population after 5 years.  (1 mark)

  3. On the axes below, draw the graph of the population against time, showing the points at  `t = 0`  and at  `t = 5`.  (2 marks)
      

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  1. `2000`
  2. `4977`
  3. `text{See Worked Solutions}`
Show Worked Solution
a.    `text{Initial population occurs when}\ \  t = 0.`
`P` `= 2000 (1.2)^0`  
  `= 2000`  

 

b.    `text{Find} \ P \ text{when} \ \ t = 5: `

`P` `= 2000 (1.2)^5`  
  `= 4976.64`  
  `= 4977 \ text{(nearest whole)}`  

 

♦ Mean mark (c) 48%.

c. 

Algebra, STD2 A4 2021 HSC 10 MC

Which of the following best represents the graph of  `y = 10 (0.8)^x`?
 

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`A`

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`\text{By elimination:}`

♦ Mean mark 41%.

`\text{When} \ x = 0 \ , \ y = 10(0.8) ^0 = 10`

`-> \ text{Eliminate B and D}`

`text(As)\ \ x→oo, \ y→0`

`-> \ text{Eliminate C}`

`=> A`

Algebra, STD1 A3 2020 HSC 19

Each year the number of fish in a pond is three times that of the year before.

  1. The table shows the number of fish in the pond for four years.
    \begin{array} {|l|c|c|c|c|}
    \hline
    \rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \  & \ \ \ 2021\ \ \  & \ \ \ 2022\ \ \  & \ \ \ 2023\ \ \ \\
    \hline
    \rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & & & 2700\\
    \hline
    \end{array}

    Complete the table above showing the number of fish in 2021 and 2022.   (2 marks)
     

  2. Plot the points from the  table in part (a) on the grid.   (2 marks)
     
  3. Which model is more suitable for this dataset: linear or exponential? Briefly explain your answer.   (2 marks)

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a.   

\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \  & \ \ \ 2021\ \ \  & \ \ \ 2022\ \ \  & \ \ \ 2023\ \ \ \\
\hline
\rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & 300 & 900 & 2700\\
\hline
\end{array}

b.   
       

c.     The more suitable model is exponential.

A linear dataset would graph a straight line which is not the case here.

An exponential curve can be used to graph populations that grow at an increasing rate, such as this example.

Show Worked Solution

a.        

\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \  & \ \ \ 2021\ \ \  & \ \ \ 2022\ \ \  & \ \ \ 2023\ \ \ \\
\hline
\rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & 300 & 900 & 2700\\
\hline
\end{array}

b.  

c.     The more suitable model is exponential.

A linear dataset would graph a straight line which is not the case here.

An exponential curve can be used to graph populations that grow at an increasing rate, such as this example.

♦ Mean mark (c) 31%.

Algebra, STD2 A4 2014 HSC 3 MC

The diagram shows the graph of an equation.
  

 Which of the following equations does the graph best represent?

  1. `y = 3/x + 1`
  2. `y = 3^x + 1`
  3. `y = 3x^2 + 1`
  4. `y = 3x^3 + 1`
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`C`

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`text(Graph is a parabola that passes through)\ (0, 1).`

`=>  C`

Algebra, STD2 A4 2011 HSC 6 MC

 

Which of the following graphs best represents the equation `y = a^x`, where `a` is a positive number greater than 1?
  

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`D`

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♦ Mean mark 48%

`text(At) \ x=0,\ y=1`

`text(As)\ x uarr, \ \ y ↑\ text(exponentially)`

`=>D`