Linear Functions

Functions, 2ADV EQ-Bank 4

Solve the following simultaneous equations

$6 x-7 y=-6$

$4 x+2 y=16$ (2 marks)

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$x=\dfrac{5}{2}, \ y=3$

Show Worked Solution

$6x-7y=-6\ \ldots\ (1)$

$4x+2y=16\ \ldots\ (2)$

$\text{Mult \((1) \times 2\ $ and $\ (2) \times 3$:}\)

$12 x-14 y=-12\ \ldots\ \left(1^{\prime}\right)$

$12 x+6 y=48\ \ldots\ \left(2^{\prime}\right)$

$\text{Subtract}\ \ \left(2^{\prime}\right)-\left(1^{\prime}\right):$

$20 y=60 \ \Rightarrow \ y=3$

$\text{Substitute} \ \ y=3 \ \ \text{into (2):}$

$4 x+6=16 \ \Rightarrow \ x=\dfrac{5}{2}$

Functions, 2ADV EQ-Bank 3

Solve the following simultaneous equations:

$3 x-4 y=5$

$x+2 y=15$. (2 marks)

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$x=7, \ y=4$

Show Worked Solution

$3 x-4 y=5\ \ldots\ (1)$

$ x+2 y=15\ \ldots\ (2)$

$\text{Multiply (2)} \times 3:$

$3 x+6 y=45\ \ldots\ \left(2^{\prime}\right) $

$\text{Subtract} \ \left(2^{\prime}\right)-(1)$ :

$10 y=40 \ \Rightarrow \ y=4$

$\text{Substitute} \ \ y=4 \ \ \text{into (2):}$

$x+2 \times 4=15 \ \Rightarrow x=7$

Functions, 2ADV EQ-Bank 12

Show the equation of $AC$ is $3 x+4 y-16=0$. ( 2 marks)

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Show that $BC$ is not perpendicular to $AC$. (2 marks)

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a. $\text{Proof (See Worked Solutions)}$

b. $\text{Proof (See Worked Solutions)}$

Show Worked Solution

a. $A(0,4), C(4,1)$

$m_{AC}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{1-4}{4-0}=-\dfrac{3}{4}$

$\text{Equation of line with} \ \ m=-\dfrac{3}{4} \ \ \text{through}\ (0,4):$

$y-y_1$	$=m\left(x-x_1\right)$
$y-4$	$=-\dfrac{3}{4}(x-0)$
$y-4$	$=-\dfrac{3}{4} x$
$4 y-16$	$=-3 x$
$3 x+4 y-16$	$=0$

b. $B(3,0), C(4,1)$

$m_{BC}=\dfrac{1-0}{4-3}=1$

$m_{AC} \times m_{BC}=-\dfrac{4}{3} \times 1=-\dfrac{4}{3} \neq-1$

$\therefore AC \ \text{is not} \ \perp \text{to} \ BC.$

Functions, 2ADV 2011 HSC 3c

The diagram shows a line `l_1`, with equation `3x + 4y-12 = 0`, which intersects the `y`-axis at `B`.

A second line `l_2`, with equation `4x-3y = 0`, passes through the origin `O` and intersects `l_1` at `E`.

Show that the coordinates of `B` are `(0, 3)`. (1 mark)
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Show that `l_1` is perpendicular to `l_2`. (2 marks)
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a. `\text{See Worked Solution}`

b. `\text{See Worked Solution}`

Show Worked Solution

a. `B\ text(is)\ y text(-intercept of)\ l_1`

`text(When)\ x = 0:`

`(3 xx 0) + 4y-12=0\ \ =>\ \ y=3`

`:.\ B\ text(is)\ (0,3)`

b.	`l_1:\ \ 3x + 4y -12`	`= 0`
	`4y`	`= -3x + 12`
	`y`	`= -3/4x + 3`

`m(l_1)=-3/4`

`l_2:\ \ 4x-3y`	`= 0`
`3y`	`= 4x`
`y`	`= 4/3 x`

`m(l_2)=4/3`

`m (l_1) xx m (l_2)= -3/4 xx 4/3= -1`

`:.\ l_1\ text(and)\ l_2\ text(are perpendicular)`

Functions, 2ADV EQ-Bank 2

Find the equation of the line that passes through $(2,1)$ and $(-3,4)$. (2 marks)

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Determine whether $(7,-2)$ lies on the line. (1 mark)

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a. $y=\dfrac{4-1}{-3-2}=-\dfrac{3}{5}$

b. $\text {Substitute}\ (7,-2) \ \text{into equation:}$

$-2$	$=-\dfrac{3}{5} \times 7+\dfrac{11}{5}$
$-2$	$=-\dfrac{21}{5}+\dfrac{11}{5}$
$-2$	$=-2 \ \text{(correct)}$

$\therefore (7,-2) \text{ lies on line.}$

Show Worked Solution

a. $(2,1),(-3,4)$

$\text{Gradient}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{4-1}{-3-2}=-\dfrac{3}{5}$

$\text{Find equation with} \ \ m=-\dfrac{3}{5} \ \ \text{through}\ \ (2,1):$

$y-1$	$=-\dfrac{3}{5}(x-2)$
$y$	$=-\dfrac{3}{5} x+\dfrac{11}{5}$

b. $\text {Substitute}\ (7,-2)\ \text{into equation:}$

$-2$	$=-\dfrac{3}{5} \times 7+\dfrac{11}{5}$
$-2$	$=-\dfrac{21}{5}+\dfrac{11}{5}$
$-2$	$=-2 \ \text{(correct)}$

$\therefore (7,-2) \text{ lies on line.}$

Functions, 2ADV F1 2024 HSC 1 MC

Consider the function shown.

Which of the following could be the equation of this function?

$y=2 x+3$
$y=2 x-3$
$y=-2 x+3$
$y=-2 x-3$

Show Answers Only

$C$

Show Worked Solution

$\text {Gradient is negative (top left } \rightarrow \text { bottom right)}$

$y \text{-intercept = 3 (only positive option)}$

$\Rightarrow C$

Functions, 2ADV F1 EQ-Bank 25

Prove that the line between $(1,-1)$ and $(4,-3)$ is perpendicular to the line $3x-2y-4=0$ (2 marks)

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$\text {Perpendicular lines}\ \ \Rightarrow\ m_1 \times m_2 = -1$

$\text {Line 1 gradient:}$

$P_1 (1,-1), P_2(4,-3) $

$m_1=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-3+1}{4-1}=-\dfrac{2}{3}$

$\text {Line 2 gradient:}$

$3x-2y-4=0\ \ \Rightarrow \ y= \dfrac{3}{2}x-2\ \ \Rightarrow m_2=\dfrac{3}{2}$

$m_1 \times m_2 = -\dfrac{2}{3} \times \dfrac{3}{2} = -1$

$\therefore\ \text{Lines are perpendicular.}$

Show Worked Solution

$\text {Perpendicular lines}\ \ \Rightarrow\ m_1 \times m_2 = -1$

$\text {Line 1 gradient:}$

$P_1 (1,-1), P_2(4,-3) $

$m_1=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-3+1}{4-1}=-\dfrac{2}{3}$

$\text {Line 2 gradient:}$

$3x-2y-4=0\ \ \Rightarrow \ y= \dfrac{3}{2}x-2\ \ \Rightarrow m_2=\dfrac{3}{2}$

$m_1 \times m_2 = -\dfrac{2}{3} \times \dfrac{3}{2} = -1$

$\therefore\ \text{Lines are perpendicular.}$

Functions, 2ADV F1 2022 HSC 1 MC

Which of the following could be the graph of `y= -2 x+2`?

Show Answers Only

`A`

Show Worked Solution

`text{By elimination:}`

`y text{-intercept = 2 → Eliminate}\ B and C`

`text{Gradient is negative → Eliminate}\ D`

`=>A`

Functions, 2ADV F1 2020 HSC 11

There are two tanks on a property, Tank `A` and Tank `B`. Initially, Tank `A` holds 1000 litres of water and Tank B is empty.

Tank `A` begins to lose water at a constant rate of 20 litres per minute. The volume of water in Tank `A` is modelled by `V = 1000 - 20t` where `V` is the volume in litres and `t` is the time in minutes from when the tank begins to lose water. (1 mark)

On the grid below, draw the graph of this model and label it as Tank `A`.
Tank `B` remains empty until `t=15` when water is added to it at a constant rate of 30 litres per minute.

By drawing a line on the grid (above), or otherwise, find the value of `t` when the two tanks contain the same volume of water. (2 marks)
Using the graphs drawn, or otherwise, find the value of `t` (where `t > 0`) when the total volume of water in the two tanks is 1000 litres. (1 mark)

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`text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}`
`29 \ text{minutes}`
`45 \ text{minutes}`

Show Worked Solution

a. `text{T} text{ank} \ A \ text{will pass trough (0, 1000) and (50, 0)}`

b. `text{T} text{ank} \ B \ text{will pass through (15, 0) and (45, 900)}`

`text{By inspection, the two graphs intersect at} \ \ t = 29 \ text{minutes}`

c. `text{Strategy 1}`

`text{By inspection of the graph, consider} \ \ t = 45`

`text{T} text{ank A} = 100 \ text{L} , \ text{T} text{ank B} =900 \ text{L} `

`:.\ text(Total volume = 1000 L when t = 45)`

`text{Strategy 2}`

`text{Total Volume}`	`=text{T} text{ank A} + text{T} text{ank B}`
`1000`	`= 1000 – 20t + (t – 15) xx 30`
`1000`	`= 1000 – 20t + 30t – 450 `
`10t`	`= 450`
`t`	`= 45 \ text{minutes}`

Algebra, STD2 A2 2019 HSC 34

The relationship between British pounds `(p)` and Australian dollars `(d)` on a particular day is shown in the graph.

Write the direct variation equation relating British pounds to Australian dollars in the form `p = md`. Leave `m` as a fraction. (1 mark)

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The relationship between Japanese yen `(y)` and Australian dollars `(d)` on the same day is given by the equation `y = 76d`.

Convert 93 100 Japanese yen to British pounds. (2 marks)

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`p = 4/7 d`
`93\ 100\ text(Yen = 700 pounds)`

Show Worked Solution

a. `m = text(rise)/text(run) = 4/7`

♦ Mean mark 42%.

`p = 4/7 d`

b. `text(Yen to Australian dollars:)`

`y`	`=76d`
`93\ 100`	`= 76d`
`d`	`= (93\ 100)/76= 1225`

`text(Aust dollars to pounds:)`

`p= 4/7 xx 1225= 700\ text(pounds)`

`:. 93\ 100\ text(Yen = 700 pounds)`

Algebra, STD2 A2 2019 HSC 14 MC

Last Saturday, Luke had 165 followers on social media. Rhys had 537 followers. On average, Luke gains another 3 followers per day and Rhys loses 2 followers per day.

If `x` represents the number of days since last Saturday and `y` represents the number of followers, which pair of equations model this situation?

A.	`text(Luke:)\ \ y = 165x + 3` `text(Rhys:)\ \ y = 537x - 2`
B.	`text(Luke:)\ \ y = 165 + 3x` `text(Rhys:)\ \ y = 537 - 2x`
C.	`text(Luke:)\ \ y = 3x + 165` `text(Rhys:)\ \ y = 2x - 537`
D.	`text(Luke:)\ \ y = 3 + 165x` `text(Rhys:)\ \ y = 2 - 537x`

Show Answers Only

`B`

Show Worked Solution

`text(Luke starts with 165 and adds 3 per day:)`

`y = 165 + 3x`

`text(Rhys starts with 537 and loses 2 per day:)`

`y = 537 – 2x`

`=> B`

Functions, 2ADV F1 2019 HSC 2 MC

What values of `x` satisfy `4-3x <= 12`?

`x <= -16/3`
`x >= -16/3`
`x <= -8/3`
`x >= -8/3`

Show Answers Only

`D`

Show Worked Solution

`4-3x`	`<= 12`
`-3x`	`<= 8`
`x`	`>= -8/3`

`=> D`

Functions, 2ADV F1 SM-Bank 25

Damon owns a swim school and purchased a new pool pump for $3250.

He writes down the value of the pool pump by 8% of the original price each year.

Construct a function to represent the value of the pool pump after `t` years. (1 mark)

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Draw the graph of the function and state its domain and range. (2 marks)

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`text(Value) = 3259 – 260t`
`text(Domain)\ {t: 0 <= t <= 12.5}`
`text(Range)\ {y: 0 <= y <= 3250}`

Show Worked Solution

i. `text(Depreciation each year)`	`= 8text(%) xx 3250`
	`= $260`

`:.\ text(Value) = 3250 – 260t`

ii.

`text(Find)\ \ t\ \ text(when value = 0)`

`3250 – 260t`	`= 0`
`t`	`= 3250/260`
	`= 12.5\ text(years)`

`text(Domain)\ \ {t: 0 <= t <= 12.5}`

`text(Range)\ \ {y: 0 <= y <= 3250}`

Functions, 2ADV F1 SM-Bank 24

Ita publishes and sells calendars for $25 each. The cost of producing the calendars is $8 each plus a set up cost of $5950.

How many calendars does Ita need to sell to breakeven? (2 marks)

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`350`

Show Worked Solution

`text(Let)\ \ x =\ text(number of calendars sold)`

`text(C)text(ost) = 5950 + 8x`

`text(Sales revenue) = 25x`

`text(Breakeven occurs when:)`

`25x`	`= 5950 + 8x`
`17x`	`= 5950`
`:. x`	`= 350`

Functions, 2ADV F1 2018 HSC 11b

Solve `1 - 3x > 10`. (2 marks)

Show Answers Only

`x < -3`

Show Worked Solution

`1 – 3x`	`> 10`
`-3x`	`> 9`
`x`	`< -3`

Functions, 2ADV F1 2018 HSC 3 MC

What is the `x`-intercept of the line `x + 3y + 6 = 0`?

`(-6, 0)`
`(6, 0)`
`(0, -2)`
`(0, 2)`

Show Answers Only

`A`

Show Worked Solution

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

`x + 0 + 6`	`= 0`
`x`	`= -6`

`:. x text{-intercept is}\ (-6, 0)`

`=> A`

Functions, 2ADV F1 2017 HSC 1 MC

What is the gradient of the line `2x + 3y + 4 = 0`?

`-2/3`
`2/3`
`-3/2`
`3/2`

Show Answers Only

`A`

Show Worked Solution

`2x + 3y + 4`	`= 0`
`3y`	`= -2x-4`
`y`	`= -2/3 x-4/3`
`:.\ text(Gradient)`	`= -2/3`

`=> A`

Functions, 2ADV F1 2016 HSC 12a*

The diagram shows points `A(1, 0), B(2, 4)` and `C(6, 1).` The point `D` lies on `BC` such that `AD _|_ BC.`

Show that the equation of `BC` is `3x + 4y-22 = 0`. (2 marks)

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Determine the gradient of `AD`. (1 mark)

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i. `text(Proof)\ \ text{(See Worked Solutions)}`

ii. `m_(AD)=4/3`

Show Worked Solution

i. `B (2, 4),\ \ C (6, 1)`

`m_(BC) = (y_2-y_1)/(x_2-x_1) = (1-4)/(6-2) =-3/4`

`text(Equation of)\ \ BC,\ \ m=-3/4\ \ text(through)\ \ (2, 4):`

`y-y_1`	`= m(x-x_1)`
`y-4`	`=-3/4 (x-2)`
`4y-16`	`= -3x + 6`
`3x + 4y-22`	`= 0\ text(… as required.)`

ii. `\text{Perpendicular lines:}\ m_1 xx m_2 = -1`

`m_(BC) =-3/4\ \ =>\ \ m_(AD)=4/3\ (BC _|_ AD)`

Functions, 2ADV F1 2007 HSC 1f

Find the equation of the line that passes through the point `(1, 3)` and is perpendicular to `2x + y + 4 = 0`. (2 marks)

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`x-2y + 7 = 0`

Show Worked Solution

`2x + y + 4`	`= 0`
`y`	`= -2x-4`

`=>\ text(Gradient) = -2`

`:. text(⊥ gradient) = 1/2\ \ \ (m_1 m_2=-1)`

`text(Equation of line)\ \ m = 1/2, \ text(through)\ (1, 3):`

`y-y_1`	`= m (x-x_1)`
`y-3`	`= 1/2 (x-1)`
`y`	`= 1/2 x + 5/2`
`2y`	`= x + 5`
`:. x-2y + 5`	`= 0`

Functions, 2ADV F1 2007 HSC 1b

Solve `2x-5> -3` and graph the solution on a number line. (2 marks)

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`x > 1`

Show Worked Solution

`2x-5`	`> -3`
`2x`	`> 2`
`x`	`> 1`

Algebra, STD2 A2 2015 HSC 27c

Ariana’s parents have given her an interest‑free loan of $4800 to buy a car. She will pay them back by paying `$x` immediately and `$y` every month until she has repaid the loan in full.

After 18 months Ariana has paid back $1510, and after 36 months she has paid back $2770.

This information can be represented by the following equations.

`x + 18y = 1510`

`x + 36y = 2770`

Graph these equations below and use to solve simultaneously for the values of `x` and `y`. (2 marks)

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How many months will it take Ariana to repay the loan in full? (2 marks)

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`x = 250, \ y = 70`
`text(65 months)`

Show Worked Solution

`:.\ text(Solution is)\ \ x = 250, \ y = 70`

ii. `text(Let)\ \ A = text(the amount paid back after)\ n\ text(months)`

`A = 250 + 70n`

♦ Mean mark 44%.

`text(Find)\ n\ text(when)\ A = 4800`

`250 + 70n`	`= 4800`
`70n`	`= 4550`
`n`	`= 65`

`:.\ text(It will take Ariana 65 months to repay)`

`text(the loan in full.)`

Functions, 2ADV F1 2015 HSC 2 MC

What is the slope of the line with equation `2x-4y + 3 = 0`?

`-2`
`-1/2`
`1/2`
`2`

Show Answers Only

`C`

Show Worked Solution

`2x-4y + 3`	`= 0`
`4y`	`= 2x + 3`
`y`	`= 1/2 x + 3/4`

`=> C`

Functions, 2ADV F1 2006 HSC 1e

Solve `3-5x <= 2`. (2 marks)

Show Answers Only

`x >= 1/5`

Show Worked Solution

`3-5x`	`<= 2`
`-5x`	`<= -1`
`x`	`>= 1/5`

Algebra, STD2 A4 2005 HSC 28b

Sue and Mikey are planning a fund-raising dance. They can hire a hall for $400 and a band for $300. Refreshments will cost them $12 per person.

Write a formula for the cost ($C) of running the dance for `x` people. (1 mark)

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The graph shows planned income and costs when the ticket price is $20

Estimate the minimum number of people needed at the dance to cover the costs. (1 mark)

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How much profit will be made if 150 people attend the dance? (1 mark)

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Sue and Mikey plan to sell 200 tickets. They want to make a profit of $1500.

What should be the price of a ticket, assuming all 200 tickets will be sold? (3 marks)

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`700 + 12x`
`text(Approximately 90)`
`$500`
`$23`

Show Worked Solution

i.	`$C`	`= 400 + 300 + (12 xx x)`
		`= 700 + 12x`

ii. `text(Using the graph intersection)`

`text(Approximately 90 people are needed)`

`text(to cover the costs.)`

iii. `text(If 150 people attend)`

`text(Income)`	`= 150 xx $20`
	`= $3000`

`text(C)text(osts)`	`= 700 + (12 xx 150)`
	`= $2500`

`:.\ text(Profit)`	`= 3000 − 2500`
	`= $500`

iv. `text(C)text(osts when)\ x = 200:`

`C`	`= 700 + (12 xx 200)`
	`= $3100`

`text(Income required to make $1500 profit)`

`= 3100 + 1500`

`= $4600`

`:.\ text(Price per ticket)`	`= 4600/200`
	`= $23`

Algebra, STD2 A4 SM-Bank 27

Fiona and John are planning to hold a fund-raising event for cancer research. They can hire a function room for $650 and a band for $850. Drinks will cost them $25 per person.

Write a formula for the cost ($C) of holding the charity event for `x` people. (1 mark)

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The graph below shows the planned income and costs if they charge $50 per ticket. Estimate the number of guests they need to break even. (1 mark)

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How much profit will Fiona and John make if 80 people attend their event? (1 mark)

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Show Answers Only

`$C = 1500 + 25x`
`60`
`$500`

Show Worked Solution

i.	`text(Fixed C) text(osts)`	`= 650 + 850`
		`= $1500`

`text(Variable C) text(osts) = $25x`

`:.\ $C = 1500 + 25x`

ii.	`text(From the graph)`
	`text(C) text(osts = Income when)\ x = 60`
	`text{(i.e. where graphs intersect)}`

iii. `text(When)\ \ x = 80:`

`text(Income)`	`= 80 xx 50`
	`= $4000`

`$C`	`= 1500 + 25 xx 80`
	`= $3500`

`:.\ text(Profit)`	`= 4000 – 3500`
	`= $500`

Algebra, STD2 A2 2007 HSC 27b

A clubhouse uses four long-life light globes for five hours every night of the year. The purchase price of each light globe is $6.00 and they each cost `$d` per hour to run.

Write an equation for the total cost (`$c`) of purchasing and running these four light globes for one year in terms of `d`. (2 marks)

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Find the value of `d` (correct to three decimal places) if the total cost of running these four light globes for one year is $250. (1 mark)

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If the use of the light globes increases to ten hours per night every night of the year, does the total cost double? Justify your answer with appropriate calculations. (1 mark)

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The manufacturer’s specifications state that the expected life of the light globes is normally distributed with a standard deviation of 170 hours.

What is the mean life, in hours, of these light globes if 97.5% will last up to 5000 hours? (1 mark)

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`$c = 24 + 7300d`
`0.031\ $ text(/hr)\ text{(3 d.p.)}`
`text(Proof)\ \ text{(See Worked Solutions)}`
`text(4660 hours.)`

Show Worked Solution

a. `text(Purchase price) = 4 xx 6 = $24`

`text(Running cost)`	`= text(# Hours) xx text(Cost per hour)`
	`= 4 xx 5 xx 365 xx d= 7300d`

`:.\ $c = 24 + 7300d`

b. `text(Given)\ \ $c = $250`

`250`	`= 24 + 7300d`
`7300d`	`= 226`
`d`	`= 226/7300= 0.03095…`
	`= 0.031\ $ text(/hr)\ text{(3 d.p.)}`

c. `text(If)\ d\ text(doubles to 0.062)\ \ $text(/hr)`

`$c= 24 + 7300 xx 0.062= $476.60`

`text(S) text(ince $476.60 is less than)\ 2 xx $250\ ($500),`

`text(the total cost increases to less than double)`

`text(the original cost.)`

d. `sigma = 170`

`z\ text(-score of 5000 hours) = 2`

`z`	`= (x – mu)/sigma`
`2`	`= (5000 – mu)/170`
`340`	`= 5000 – mu`
`mu`	`= 4660`

`:.\ text(The mean life of these globes is 4660 hours.)`

Functions, 2ADV F1 2014 HSC 5 MC

Which equation represents the line perpendicular to `2x-3y = 8`, passing through the point `(2, 0)`?

`3x + 2y = 4`
`3x + 2y = 6`
`3x-2y = -4`
`3x-2y = 6`

Show Answers Only

`B`

Show Worked Solution

`2x-3y`	`= 8`
`3y`	`= 2x-8`
`y`	`= 2/3x-8/3`

`m= 2/3`

`:.\ m_text(perp)= -3/2\ \ \ (m_1 m_2=-1\text( for)_|_text{lines)}`

`text(Equation of line)\ \ m = -3/2\ \ text(through)\ \ (2,0):`

`y-y_1`	`= m (x-x_1)`
`y-0`	`= -3/2 (x-2)`
`y`	`= -3/2x + 3`
`2y`	`= -3x + 6`
`3x + 2y`	`= 6`

`=> B`

Functions, 2ADV F1 2009 HSC 5a

In the diagram, the points `A` and `C` lie on the `y`-axis and the point `B` lies on the `x`-axis. The line `AB` has equation `y = sqrt3x-3`. The line `BC` is perpendicular to `AB`.

Find the equation of the line `BC`. (2 marks)

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Find the area of the triangle `ABC`. (2 marks)

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`y=-1/sqrt3 x +1`
`text(Area)\ Delta ABC = 2 sqrt 3\ text(u²)`

Show Worked Solution

i. `text(Gradient of)\ \ AB = sqrt 3`

`:. m_(BC) = -1/(sqrt3)\ \ (BC _|_ AB)`

`text(Finding)\ B,`

`0`	`= sqrt3 x-3`
`sqrt 3 x`	`= 3`
`x`	`= 3/sqrt3 xx sqrt3/sqrt3= sqrt3`

`:. B (sqrt3, 0)`

`text(Equation of)\ \ BC\ \ text(has)\ \ m =-1/sqrt3\ \ text(through)\ \ (sqrt3, 0):`

`y\-y_1`	`= m (x\-x_1)`
`y\-0`	`=-1/sqrt3 (x\-sqrt3)`
`y`	`=-1/sqrt3 x +1`

ii. `AB\ \ text(cuts)\ y text(-axis when)\ \ x = 0, \ \ y=-3`

`=> A (0,–3)`

`BC\ \ text(cuts)\ y text(-axis when)\ \ x = 0, \ \ y=1`

`=> C (0,1)`

`:. AC`	`= 4`
`OB`	`= sqrt 3`

`text(Area)\ \ Delta ABC`	`= 1/2 xx AC xx OB`
	`= 1/2 xx 4 xx sqrt 3`
	`= 2 sqrt 3\ text(u²)`

Functions, 2ADV F1 2009 HSC 1a

Sketch the graph of `y-2x = 3`, showing the intercepts on both axes. (2 marks)

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Show Worked Solution

`y-2x=3\ \ =>\ \ y=2x+3`

`ytext{-intercept}\ = 3`

`text{Find}\ x\ text{when}\ y=0:`

`0-2x=3\ \ =>\ \ x=-3/2`

Functions, 2ADV F1 2011 HSC 1e

Solve `2 -3x <= 8`. (2 marks)

Show Answers Only

`x>=-2`

Show Worked Solution

`2 -3x`	`<= 8`
`-3x`	`<= 6`
`x`	`>= – 6/3`
`x`	`>= -2`

Algebra, STD2 A2 2010 HSC 27c

The graph shows tax payable against taxable income, in thousands of dollars.

2010 27c

Use the graph to find the tax payable on a taxable income of $21 000. (1 mark)

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Use suitable points from the graph to show that the gradient of the section of the graph marked `A` is `1/3`. (1 mark)

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How much of each dollar earned between $21 000 and $39 000 is payable in tax? (1 mark)

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Write an equation that could be used to calculate the tax payable, `T`, in terms of the taxable income, `I`, for taxable incomes between $21 000 and $39 000. (2 marks)

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`$3000\ \ \ text{(from graph)}`
`1/3`
`33 1/3\ text(cents per dollar earned)`
`text(Tax payable on)\ I = 1/3 I\-4000`

Show Worked Solution

`text(Income on)\ $21\ 000=$3000\ \ \ text{(from graph)}`

b. `text(Using the points)\ (21,3)\ text(and)\ (39,9)`

♦♦ Mean mark 25%

`text(Gradient at)\ A`	`= (y_2\-y_1)/(x_2\ -x_1)`
	`= (9000-3000)/(39\ 000-21\ 000)`
	`= 6000/(18\ 000)`
	`= 1/3\ \ \ \ \ text(… as required)`

c. `text(The gradient represents the tax applicable to each dollar)`

♦♦♦ Mean mark 12%!
MARKER’S COMMENT: Interpreting gradients is an examiner favourite, so make sure you are confident in this area.

`text(Tax)`	` = 1/3\ text(of each dollar earned)`
	` = 33 1/3\ text(cents per dollar earned)`

d. `text( Tax payable up to $21 000 = $3000)`

`text(Tax payable on income between $21 000 and $39 000)`

♦♦♦ Mean mark 15%.
STRATEGY: The earlier parts of this question direct students to the most efficient way to solve this question. Make sure earlier parts of a question are front and centre of your mind when devising strategy.

` = 1/3 (I\-21\ 000)`

`:.\ text(Tax payable on)\ \ I`	`= 3000 + 1/3 (I\-21\ 000)`
	`= 3000 + 1/3 I\-7000`
	`= 1/3 I\-4000`

Algebra, STD2 A4 2011 HSC 20 MC

A function centre hosts events for up to 500 people. The cost `C`, in dollars, for the centre
to host an event, where `x` people attend, is given by:

`C = 10\ 000 + 50x`

The centre charges $100 per person. Its income `I`, in dollars, is given by:

`I = 100x`

How much greater is the income of the function centre when 500 people attend an event, than its income at the breakeven point?

`$15\ 000`
`$20\ 000`
`$30\ 000`
`$40\ 000`

Show Answers Only

`C`

Show Worked Solution

♦ Mean mark 50%
COMMENT: Students can read the income levels directly off the graph to save time and then check with the equations given.

`text(When)\ x=500,\ I=100xx500=$50\ 000`

`text(Breakeven when)\ \ x=200\ \ \ text{(from graph)}`

`text(When)\ \ x=200,\ I=100xx200=$20\ 000`

`text(Difference)`	`=50\ 000-20\ 000`
	`=$30\ 000`

`=> C`

Algebra, STD2 A2 2009 HSC 24d

A factory makes boots and sandals. In any week

• the total number of pairs of boots and sandals that are made is 200
• the maximum number of pairs of boots made is 120
• the maximum number of pairs of sandals made is 150.

The factory manager has drawn a graph to show the numbers of pairs of boots (`x`) and sandals (`y`) that can be made.

Find the equation of the line `AD`. (1 mark)

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Explain why this line is only relevant between `B` and `C` for this factory. (1 mark)

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The profit per week, `$P`, can be found by using the equation `P = 24x + 15y`.

Compare the profits at `B` and `C`. (2 marks)

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`x + y = 200`
`text(S)text(ince the max amount of boots = 120)`

`=> x\ text(cannot)\ >120`

`text(S)text(ince the max amount of sandals = 150`

`=> y\ text(cannot)\ >150`

`:.\ text(The line)\ AD\ text(is only possible between)\ B\ text(and)\ C.`
`text(The profits at)\ C\ text(are $630 more than at)\ B.`

Show Worked Solution

a. `text{We are told the number of boots}\ (x),`

♦♦♦ Mean mark part (i) 14%.
Using `y=mx+b` is a less efficient but equally valid method, using `m=–1` and `b=200` (`y`-intercept).

`text{and shoes}\ (y),\ text(made in any week = 200)`

`=>text(Equation of)\ AD\ text(is)\ \ x + y = 200`

b. `text(S)text(ince the max amount of boots = 120)`

♦ Mean mark 49%

`=> x\ text(cannot)\ >120`

`text(S)text(ince the max amount of sandals = 150`

`=> y\ text(cannot)\ >150`

`:.\ text(The line)\ AD\ text(is only possible between)\ B\ text(and)\ C.`

c. `text(At)\ B,\ \ x = 50,\ y = 150`

♦ Mean mark 40%.

`=>$P (text(at)\ B)`	`= 24 xx 50 + 15 xx 150`
	`= 1200 + 2250`
	` = $3450`

`text(At)\ C,\ \ x = 120 text(,)\ y = 80`

`=> $P (text(at)\ C)`	`= 24 xx 120 + 15 xx 80`
	`= 2880 + 1200`
	`= $4080`

`:.\ text(The profits at)\ C\ text(are $630 more than at)\ B.`

\(y-y_1\)	\(=m\left(x-x_1\right)\)
\(y-4\)	\(=-\dfrac{3}{4}(x-0)\)
\(y-4\)	\(=-\dfrac{3}{4} x\)
\(4 y-16\)	\(=-3 x\)
\(3 x+4 y-16\)	\(=0\)

\(-2\)	\(=-\dfrac{3}{5} \times 7+\dfrac{11}{5}\)
\(-2\)	\(=-\dfrac{21}{5}+\dfrac{11}{5}\)
\(-2\)	\(=-2 \ \text{(correct)}\)

\(y-1\)	\(=-\dfrac{3}{5}(x-2)\)
\(y\)	\(=-\dfrac{3}{5} x+\dfrac{11}{5}\)

\(-2\)	\(=-\dfrac{3}{5} \times 7+\dfrac{11}{5}\)
\(-2\)	\(=-\dfrac{21}{5}+\dfrac{11}{5}\)
\(-2\)	\(=-2 \ \text{(correct)}\)