smc-6408-60-Combinations

Functions, 2ADV 2026 HSC 3 MC

The graph of $y=f(x)$ is shown for $0 \le x \le 4$.

A new graph $y=g(x)$ is obtained by transforming $y=f(x)$ as shown.

Which of the following expressions is equivalent to $g(x)$?

$f\left(-\dfrac{1}{2}x\right)$
$-f\left(\dfrac{1}{2}x\right)$
$f(-2x)$
$-f(2x)$

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$B$

Show Worked Solution

$\text{Domain $f(x)$:}\ \ 0 \le x \le 4\ \ \rightarrow\ \ \text{Domain $g(x)$:}\ \ 0 \le x \le 8$

$\text{Horizontal dilation (factor 2)}\ \Rightarrow\ f\left(\tfrac{1}{2}x\right)\ \ \text{(eliminate C and D)}$

$\text{At $\ x=0$:}$

$\text{Option A:}\ f(-0)=2 \neq g(0)\ \text{(eliminate A)}$

$\text{Option B:}\ -f(0)=-2 = g(0)\ \checkmark$

$\Rightarrow B$

Functions, 2ADV F2 2025 HSC 6 MC

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

Which of the following is the graph of $y=-f(-x)$ ?

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$C$

Show Worked Solution

$y=-f(x)\ \ \Rightarrow\ \ \text{Reflect \(f(x)$ in the $x$-axis.}\)

$y=-f(-x)\ \ \Rightarrow\ \ \text{Reflect \(-f(x)$ in the $y$-axis.}\)

$\text{In this combination of translations, the order is not important.}$

$\Rightarrow C$

Functions, 2ADV F2 2025 HSC 30

The parabola with equation $y=(x-1)(x-5)$ is translated both horizontally to the right and vertically up by $k$ units, where $k$ is positive.

The translated parabola passes through the point $(6,11)$.

Find the value of $k$. (3 marks)

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$k=6$

Show Worked Solution

$y=(x-1)(x-5)$

$\text{Translate \(k$ units to the right:}\)

$y \rightarrow y^{\prime}=(x-k-1)(x-k-5)$

$\text{Translate \(k$ units vertically up:}\)

$y^{\prime} \rightarrow y^{\prime \prime}=(x-k-1)(x-k-5)+k$

$y^{\prime \prime} \ \text{passes through}\ (6,11):$

$11=(6-k-1)(6-k-5)+k$

$11=(5-k)(1-k)+k$

$11=5-6 k+k^2+k$

$0=k^2-5 k-6$

$0=(k-6)(k+1)$

$\therefore k=6 \quad(k>0)$

♦ Mean mark 39%.

Functions, 2ADV F2 EQ-Bank 27

The curve $f(x)=x^2$ is transformed to $g(x)=3 f[2(x+2)]$

Write the equation of $g(x)$ (1 mark)
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$P(-3,9)$ lies on $f(x)=x^2$
Determine the corresponding co-ordinates of $P$ on the curve $g(x)$. (2 marks)

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a. $g(x)=12(x+2)^2$

b. $\left( -\dfrac{7}{2}, 27 \right) $

Show Worked Solution

a.	$g(x)$	$=3[2(x+2)]^2$
		$=3 \times 4(x+2)^2$
		$=12(x+2)^2$

b. $P(-3,9)\ \text{lies on}\ \ f(x)=x^2 $

$\text{Find corresponding point on}\ f(x)$

$\text{Mapping}\ x_f\ \text{to}\ x_g: $

$2(x_g +2)=x_f\ \ \Rightarrow\ \ x_g=\dfrac{1}{2} x_f-2 $

$x_g=\dfrac {1}{2} \times -3 -2=-\dfrac{7}{2} $

$\text{Mapping}\ y_f\ \text{to}\ y_g: $

$y_g=3 \times y_f = 3 \times 9=27$

$\therefore\ \text{Corresponding point}\ = \left( -\dfrac{7}{2}, 27 \right) $

Functions, 2ADV F2 2024 MET2 12 MC

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

Which of the following options best represents the graph of $y=f(2 x+1)$ ?

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$A$

Show Worked Solution

$\text{By elimination:}$

$\text{Graph has been dilated by a factor of}\ \dfrac{1}{2}\ \text{from}\ y\text{axis.}$

→ $\text{Eliminate C and D.}$

$\text{Graph is then translated}\ \dfrac{1}{2}\ \text{unit to the left.}$

$\text{Consider the turning point}\ (2, 1)\ \text{after translation:}$

$\left(2, 1\right)\ \rightarrow \ \left(2\times \dfrac{1}{2}-\dfrac{1}{2}, 1\right)=\left(\dfrac{1}{2}, 1\right)$

$\therefore\ \text{Option A is the only possible solution.}$

$\Rightarrow A$

♦ Mean mark 47%.

Functions, 2ADV F2 2024 HSC 7 MC

The diagram shows the graph $y = f(x)$.

Which of the following best represents the graph $y = f(2x-1)$?

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$C$

Show Worked Solution

$\text{At}\ \ x=0:$

$f(2x-1)=f(-1)\ \ \Rightarrow\ \ \text{Eliminate}\ A\ \text{and}\ B.$

$\text{Consider the transformations of}\ f(x) \rightarrow\ f(2x-1) $

$\rightarrow\ \text{Shift}\ f(x)\ \text{1 unit to the right.}$

$\rightarrow\ \text{Dilate}\ f(x-1)\ \text{by a factor of}\ \dfrac{1}{2}\ \text{from the}\ y\text{-axis.}$

$\Rightarrow C$

♦ Mean mark 47%.

Functions, 2ADV F2 2024 HSC 4 MC

The parabola $y=(x-3)^2-2$ is reflected about the $y$-axis. This is then reflected about the $x$-axis.

What is the equation of the resulting parabola?

$y=(x+3)^2+2$
$y=(x-3)^2+2$
$y=-(x+3)^2+2$
$y=-(x-3)^2+2$

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$ C $

Show Worked Solution

$y=(x-3)^2-2$

$\text{Reflect in the}\ y\text{-axis}\ (f(-x)):$

$y=(-x-3)^2-2$

$\text{Reflect in the}\ x\text{-axis}\ (-f(-x)):$

$y$	$=-\left[(-x-3)^2-2\right]$
	$=-(x+3)^2+2$

$ \Rightarrow C $

♦ Mean mark 54%.

Functions, 2ADV F2 2023 HSC 27

The graph of $y=f(x)$, where $f(x)=a|x-b|+c$, passes through the points $(3,-5), (6,7)$ and $(9,-5)$ as shown in the diagram.

Find the values of $a, b$ and $c$. (3 marks)

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The line $y=m x$ cuts the graph of $y=f(x)$ in two distinct places.
Find all possible values of $m$. (2 marks)

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a. $\ a=-4$ , $\ b=6$ , $\ c=7$

b. $ \text{2 solutions when}\ \ -4<m<7/6 $

Show Worked Solution

a. $\text{Consider the transformation of}\ \ y=-|x|$

$\text{Translate 6 units to the right}$

$y=-|x|\ \ \rightarrow\ \ y=-|x-6| $

$\therefore b=6$

$\text{Translate 7 units vertically up}$

$y=-|x-6|\ \ \rightarrow\ \ y=-|x-6|+7 $

$\therefore c=7$

$f(x)=a|x-6|+7\ \ \text{passes through}\ (3, -5):$

$-5$	$=a\|3-6\|+7$
$-5$	$=3a+7$
$3a$	$=-12$
$\therefore a$	$=-4$

b. $y=mx\ \ \text{passes through (0, 0)}$

$ \text{One solution when}\ \ y=mx\ \ \text{passes through (0, 0) and (6, 7)}$

$m=\dfrac{7-0}{6-0}=\dfrac{7}{6}$

$\text{As graph gets flatter and turns negative ⇒ 2 solutions}$

$\text{2 solutions continue until}\ \ y=mx\ \ \text{is parallel to}$

$\text{the line joining (6, 7) to}\ (9,-5),\ \text{where}: $

$m=\dfrac{7-(-5)}{6-9}=-\dfrac{12}{3}=-4 $

$\therefore \ \text{2 solutions when}\ \ -4<m< \dfrac{7}{6} $

♦♦♦ Mean mark (b) 23%.

Functions, 2ADV F2 2022 HSC 19

The graph of the function `f(x)=x^2` is translated `m` units to the right, dilated vertically by a scale factor of `k` and then translated 5 units down. The equation of the transformed function is `g(x)=3 x^2-12 x+7`.

Find the values of `m` and `k`. (3 marks)

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`m=2, \ \ k=3`

Show Worked Solution

`text{Horizontal translation}\ m\ text{units to the right:}`

`x^2\ → \ (x-m)^2`

`text{Dilated vertically by scale factor}\ k:`

`(x-m)^2\ →\ k(x-m)^2`

`text{Vertical translation 5 units down:}`

`k(x-m)^2\ →\ k(x-m)^2-5`

`y`	`=k(x-m)^2-5`
	`=k(x^2-2mx+m^2)-5`
	`=kx^2-2kmx+(km^2-5)`

`:.k=3`

`-2km`	`=-12`
`:.m`	`=2`

♦ Mean mark 51%.

Functions, 2ADV F2 EQ-Bank 20

Let `f(x) = x^2-4`

Let the graph of `g(x)` be a transformation of the graph of `f(x)` where the transformations have been applied in the following order:
• dilation by a factor of `1/2` from the vertical axis (parallel to the horizontal axis)
• translation by two units to the right (in the direction of the positive horizontal axis

Find `g(x)` and the coordinates of the horizontal axis intercepts of the graph of `g(x)`. (3 marks)

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`(1,0) and (3,0)`

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`text(1st transformation)`

`text(Dilation by a factor of)\ 1/2\ text(from the)\ ytext(-axis:)`

`x^2 – 4 \ => \ (x/(1/2))^2-4 = 4x^2-4`

`text(2nd transformation)`

`text(Translation by 2 units to the right:)`

`4x^2-4 \ => \ g(x) = 4(x-2)^2-4`

`xtext(-axis intercept of)\ g(x):`

`4(x-2)^2-4`	`=0`
`(x-2)^2`	`=1`
`x-2`	`=+-1`

`x-2=1 \ => \ x=3`

`x-2=-1 \ => \ x=1`

`:.\ text(Horizontal axis intercepts occur at)\ (1,0) and (3,0).`

Functions, 2ADV F2 EQ-Bank 26

The curve `y = kx^2 + c` is subject to the following transformations

- Translated 2 units in the positive `x`-direction
- Dilated in the positive `y`-direction by a factor of 4
- Reflected in the `y`-axis

The final equation of the curve is `y = 8x^2 + 32x-8`.

Find the equation of the graph after the dilation. (1 mark)

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Find the values of `k` and `c`. (2 marks)

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a. `y = 4k(x-2)^2 + 4c`

b. `k = 2, c = -10`

Show Worked Solution

a. `y = kx^2 + c`

`text(Translate 2 units in positive)\ xtext(-direction.)`

`y = kx^2 + c \ => \ y = k(x-2)^2 + c`

`text(Dilate in the positive)\ ytext(-direction by a factor of 4.)`

`y = k(x-2)^2 + c \ => \ y = 4k(x-2)^2 + 4c`

b.	`y`	`= 4k(x^2-4x + 4) + 4c`
		`= 4kx^2-16kx + 16k + 4c`

`text(Reflect in the)\ ytext(-axis.)`

COMMENT: Using “swap” terminology for reflections in the y-axis is simpler and more intelligible for students in our view.

`=>\ text(Swap:)\ \ x →\ -x`

`y`	`= 4k(-x)^2-16k(-x) + 16k + 4c`
	`= 4kx^2 + 16kx + 16k + 4c`

`text(Equating co-efficients:)`

`4k=8\ \ =>\ \ k=2`

`16k + 4c`	`= -8`
`4c`	`= -40`
`:. c`	`=-10`

Functions, 2ADV F2 EQ-Bank 19

List a set of transformations that, when applied in order, would transform `y = x^2` to the graph with equation `y = 1-6x-x^2`. (3 marks)

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`text(T1: Translate 3 units in negative)\ xtext(-direction)`

`text(T2: Translate 10 units in negative)\ ytext(-direction)`

`text(T3: Reflect in the)\ xtext(-axis)`

Show Worked Solution

`y = x^2`

`text(Transformation 1:)`

`text(Translate 3 units in negative)\ xtext(-direction)`

`y = (x + 3)^2`

`y = x^2 + 6x + 9`

`text(Transformation 2:)`

`text(Translate 10 units in negative)\ ytext(-direction)`

`y = x^2 + 6x-1`

`text(Transformation 3:)`

`text(Reflect in the)\ xtext(-axis)`

`y`	`= -(x^2 + 6x-1)`
	`= 1-6x-x^2`

Functions, 2ADV F2 EQ-Bank 2 MC

Which diagram best shows the graph

`y = 1 - 2(x + 1)^2`

A.		B.
C.		D.

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`A`

Show Worked Solution

`text(Transforming)\ \ y = x^2 :`

`text(Translate 1 unit left)\ \ => \ y = (x + 1)^2`

`text(Dilate from)\ xtext(-axis by a factor of 2)\ => \ y = 2(x + 1)^2`

`text(Reflect in)\ xtext(-axis)\ \ => \ y= −2(x + 1)^2`

`text(Translate 1 unit up)\ \ => \ y = 1 – 2(x + 1)^2`

`:.\ text(Transformations describe graph)\ A.`

`=>\ A`

Functions, 2ADV F2 EQ-Bank 16

`y = -(x + 2)^4/3` has been produced by three successive transformations: a translation, a dilation and then a reflection.

Describe each transformation and state the equation of the graph after each transformation. (2 marks)

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Sketch the graph. (1 mark)

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a. `text(See Worked Solutions)`

Show Worked Solution

a. `text(Transformation 1:)`

`text(Translate)\ \ y = x^4\ \ 2\ text(units to the left.)`

`y = x^4 \ => \ y = (x + 2)^4`

`text(Transformation 2:)`

`text(Dilate)\ \ y = (x + 2)^4\ \ text(by a factor of)\ 1/3\ text(from the)\ xtext(-axis)`

`y = (x + 2)^4 \ => \ y = ((x + 2)^4)/3`

`text(Transformation 3:)`

`text(Reflect)\ \ y = ((x + 2)^4)/3\ \ text(in the)\ xtext(-axis).`

`y = ((x + 2)^4)/3 \ => \ y = -(x + 2)^4/3`

Functions, 2ADV F2 EQ-Bank 18

The function `f(x) = |x|` is transformed and the equation of the new function is `y = kf(x + b) + c`.

The graph of the new function is shown below.

What are the values of `k`, `b` and `c`. (2 marks)

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`k = −1/3, b = 3, c = 2`

Show Worked Solution

`y = |x|`

`text(Translate 3 units left) \ => \ y = |x + 3|`

`text(Reflect in the)\ xtext(-axis) \ => \ y = −|x + 3|`

`text(Dilate by)\ 1/3\ text(from the)\ x text(-axis)`

`=>\ text(Multiply by)\ 1/3 \ => \ y = −1/3|x + 3|`

`text(Translate 2 units up) \ => \ y = −1/3 |x + 3| + 2`

`:. k = −1/3, b = 3, c = 2`

Functions, 2ADV F1 2019 HSC 13e

Sketch the graph of `y = |x-1|` for `-4 <= x <= 4`. (1 mark)

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Using the sketch from part i, or otherwise, solve `|x-1| = 2x + 4`. (2 marks)

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`text(See Worked Solutions)`
`(-1, 2)`

Show Worked Solution

ii. `text(By inspection, intersection when)\ x = -1`

`text(Test:)`

`\|-1-1\|`	`= -2 + 4`
`2`	`= 2`

`:.\ text(Intersection at)\ (-1, 2)`

Functions, 2ADV F2 EQ-Bank 9 MC

The point `A (3, 2)` lies on the graph of the function `f(x)`. A transformation maps the graph of `f(x)` to the graph of `g(x)`,

where `g(x) = 1/2 f(x - 1)`. The same transformation maps the point `A` to the point `P`.

The coordinates of the point `P` are

A. `(2, 1)`

B. `(2, 4)`

C. `(4, 1)`

D. `(4, 2)`

Show Answers Only

`C`

Show Worked Solution

`g(x) = 1/2 f(x – 1),\ A(3, 2)`

`text(Dilate by a factor of)\ 1/2\ text(from)\ x text(-axis:)`

`A(3, 2) -> A′(3, 1)`

`text(Translate 1 unit to right:)`

`A′(3, 1) -> P(4, 1)`

`=> C`

Functions, 2ADV F2 EQ-Bank 5 MC

The graph of a function `f(x)` is obtained from the graph of the function `g(x) = sqrt (2x-5)` by a reflection in the `x`-axis followed by a dilation from the `y`-axis by a factor of `1/2`.

Which one of the following is the function `f(x)`?

`f(x) = sqrt (5-4x)`
`f(x) =-sqrt (x-5)`
`f(x) = sqrt (x + 5)`
`f(x) = −sqrt (4x-5)`

Show Answers Only

`D`

Show Worked Solution

`text(Let)\ \ y=sqrt(2x-5)`

`text(1st transformation:)`

`y =-sqrt(2x-5)`

COMMENT: Using “swap” terminology for dilations from the y-axis is simpler and often more intelligible for students in our view.

`text(2nd transformation:)`

`text(Dilate from)\ y text(-axis by a factor of)\ 1/2`

`=>\ text(Swap)\ \ x → 2x`

`y=-sqrt(2(2x)-5)=- sqrt(4x-5)`

`=> D`

Functions, 2ADV F2 EQ-Bank 3 MC

The point `P\ text{(4, -3)}` lies on the graph of a function `f(x)`. The graph of `f(x)` is translated four units vertically up and then reflected in the `y`-axis.

The coordinates of the final image of `P` are

`(-4, 1)`
`(-4, 3)`
`(0, -3)`
`(4, -6)`

Show Answers Only

`A`

Show Worked Solution

`text(1st transformation:)`

`P(4,−3)\ ->\ (4,1)`

`text(2nd transformation:)`

`(4,1)\ ->\ (-4,1)`

`=> A`

Functions, 2ADV F2 EQ-Bank 6 MC

The graph of the function `f(x) = 3x^(5/2)` is reflected in the `x`-axis and then translated 3 units to the right and 4 units down.

The equation of the new graph is

`y = 3(x-3)^(5/2) + 4`
`y = -3 (x-3)^(5/2)-4`
`y = -3 (x + 3)^(5/2)-1`
`y = -3 (x-4)^(5/2) + 3`

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ y= 3x^(5/2)`

`text(Reflect in the)\ x text(-axis:)`

`y=-3x^(5/2)`

`text(Translate 3 units to the right:)`

`y=- 3(x-3)^(5/2)`

`text(Translate 4 units down:)`

`y=- 3(x-3)^(5/2)-4`

`=> B`

Functions, 2ADV F2 EQ-Bank 12

Draw the graph `y = ln x`. (1 mark)

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Explain how the above graph can be transformed to produce the graph
`y = 3ln(x + 2)`
and sketch the graph, clearly identifying all intercepts. (3 marks)

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Show Worked Solution

b. `text(Transforming)\ \ y = ln x => \ y = ln(x + 2)`

`y = ln x\ \ =>\ text(shift 2 units to left.)`

`text(Transforming)\ \ y = ln(x + 2)\ \ text(to)\ \ y = 3ln(x + 2)`

`=>\ text(increase each)\ y\ text(value by a factor of 3)`

Functions, 2ADV F2 2013 HSC 15c

Sketch the graph `y = |2x-3|`. (1 mark)

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Using the graph from part (i), or otherwise, find all values of `m` for which the equation `|2x-3| = mx + 1` has exactly one solution. (2 marks)

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`text(When)\ m = -2/3,\ m >= 2\ text(or)\ m<-2`

Show Worked Solution

♦ Mean mark 49%
MARKER’S COMMENT: Many students drew diagrams that were “too small”, didn’t use rulers or didn’t use a consistent scale on the axes!

ii.

`text(Line of intersection)\ \ y=mx + 1\ \ text(passes through)\ (0,1)`

♦♦ Mean mark 25%.
COMMENT: A clear graphical understanding of what the solution is crucial here.

`text(If it also passes through)\ (1.5, 0) => text(1 solution)`

`m=(y_2-y_1)/(x_2-x_1)= (1-0)/(0-3/2)=-2/3`

`text(Gradients of)\ \ y=|2x-3| => 2\ \text(or)\ -2`

`text(Considering a line through)\ (0,1):`

`text(If)\ \ m >= 2\ text(, only intersects once.)`

`text(Similarly,)`

`text(If)\ \ m<-2 text(, only intersects once.)`

`:.\ text(Only one solution when)\ \ m = -2/3,\ \ m >= 2\ \ text(or)\ \ m<-2`

a.	\(g(x)\)	\(=3[2(x+2)]^2\)
		\(=3 \times 4(x+2)^2\)
		\(=12(x+2)^2\)

\(3\)	\(=k f(-3)+2\)
\(3\)	\(=k+2\)
\(k\)	\(=1\)

\(-5\)	\(=a\|3-6\|+7\)
\(-5\)	\(=3a+7\)
\(3a\)	\(=-12\)
\(\therefore a\)	\(=-4\)

\(y\)	\(=-\left[(-x-3)^2-2\right]\)
	\(=-(x+3)^2+2\)