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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.
 

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

  2. The line  \(y=m x\)  cuts the graph of  \(y=f(x)\)  in two distinct places.
  3. 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 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 2013 HSC 15c

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

  2. 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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  1.  
    2UA 2013 HSC 15c Answer
  2. `text(When)\ m = -2/3,\ m >= 2\ text(or)\ m<-2`
Show Worked Solution

i. 

♦ 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!

2UA 2013 HSC 15c Answer

 

ii.

   2UA 2013 HSC 15c1 Answer

 

`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`