smc-1008-70-Combinations

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:}\)

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

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

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

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

\(11=5-6 k+k^2+k\)

\(0=k^2-5 k-6\)

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

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

Functions, 2ADV F2 EQ-Bank 2

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 EQ-Bank 3 MC

The graph \(y=\dfrac{2}{x-2}\) undergoes the following transformations:

translated 3 units to the left
dilated vertically by a factor of 2

Determine which of the following is the new function.

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

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\(\Rightarrow D\)

Show Worked Solution

\(\text{Translate 3 units to the left:}\)

\(y=\dfrac{2}{x-2} \ \Rightarrow \ y^{′}=\dfrac{2}{(x+3)-2}=\dfrac{2}{x+1}\)

\(\text{Dilate vertically by a factor of 2:}\)

\(y^{′}=\dfrac{2}{x+1} \ \Rightarrow \ \dfrac{y^{″}}{2}=\dfrac{2}{x+1}\)

\(\Rightarrow D\)

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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\(C\)

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\(\text{At}\ \ x=0:\)

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

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

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

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

\(\Rightarrow C\)

♦ 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 SM-Bank 16

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)`

Show Worked Solution

`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 13

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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`y = 4k(x – 2)^2 + 4c`
`k = 2, c = −10`

Show Worked Solution

i. `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`

ii.	`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 14

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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`text(See Worked Solutions)`

Show Worked Solution

i. `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`

ii.

Functions, 2ADV F2 EQ-Bank 1

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 F2 SM-Bank 7 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 SM-Bank 6 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)`?

A. `f(x) = sqrt (5 - 4x)`

B. `f(x) = - sqrt (x - 5)`

C. `f(x) = sqrt (x + 5)`

D. `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 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)`
`:. f(x)`	`= −sqrt(4x – 5)`

`=> D`

Functions, 2ADV F2 SM-Bank 5 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 SM-Bank 4 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

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

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

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

D. `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 SM-Bank 1

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

ii. `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: Students need a clear graphical understanding of what they are finding to solve this very challenging, Band 6 question.

`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\ |\ \ text(are)\ \ 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`

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