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Calculus, MET1 2024 VCAA 3

Let  \(g: R \backslash\{-3\} \rightarrow R, \ g(x)=\dfrac{1}{(x+3)^2}-2\).

  1. On the axes below, sketch the graph of  \(y=g(x)\),  labelling all asymptotes with their equations and axis intercepts with their coordinates.   (3 marks)


  2. Determine the area of the region bounded by the line  \(x=-2\),  the \(x\)-axis, the \(y\)-axis and the graph of \(y=g(x)\).   (2 marks)

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a.

b.    \(\dfrac{10}{3}\ \text{sq units}\)

Show Worked Solution

a.    \(y\text{-intercept:}\ x=0\)

\(y=\dfrac{1}{(0+3)^2}-2=-\dfrac{17}{9}\)

\(x\text{-intercepts:}\ y=0\)

\(\dfrac{1}{(x+3)^2}-2\) \(=0\)
\((x+3)^2\) \(=\dfrac{1}{2}\)
\(x+3\) \(=\pm\dfrac{1}{\sqrt{2}}\)
\(x\) \(=-3\pm\dfrac{1}{\sqrt{2}}\)

b.   \(\text{Area is below}\ x\text{-axis:}\)

  \(\text{Area}\) \(=-\displaystyle\int_{-2}^0 (x+3)^{-3}-2\,dx\)
    \(=-\left[\dfrac{1}{-1}(x+3)^{-1}-2x\right]_{-2}^0\)
    \(=-\left[\dfrac{-1}{x+3}-2x\right]_{-2}^0\)
    \(=-\left[\dfrac{-1}{3}-\left(\dfrac{-1}{-2+3}-2(-2)\right)\right]\)
    \(=-\Big[\dfrac{-1}{3}-(-1+4)\Big] \)
    \(=\dfrac{10}{3}\ \text{u}^{2}\)
♦ Mean mark (b) 40%.

Graphs, MET1 EQ-Bank 1

Let  \(\displaystyle f:[-3,-2) \cup(-2, \infty) \rightarrow R, f(x)=1+\frac{1}{x+2}\).

  1. On the axes below, sketch the graph of \(f\). Label any asymptotes with their equations, and endpoints and axial intercepts with their coordinates.   (3 marks)

  1. Find the values of \(x\) for which \(f(x) \leq 2\).   (2 marks)

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a.
   

b.   \(x\in [-3, -2)\cap (-1, \infty)\)

Show Worked Solution

a.
     

b.    \(\text{From graph }f(x)\leq -2\ \text{ for}\ -3\leq x <2\ \text{ and when }\ x\geq -1\) 

\(\rightarrow x\in [-3, -2)\cap (-1, \infty)\)

Graphs, MET1 2023 VCAA 3

  1. Sketch the graph of  \(f(x)=2-\dfrac{3}{x-1}\) on the axes below, labelling all asymptotes with their equation and axial intercepts with their coordinates.   (3 marks)
     


  2. Find the values of \(x\) for which \(f(x)\leq1\).   (1 mark)
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a.   

b.    \(1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]\)

Show Worked Solution

a.    \(\text{Vertical asymptote when}\ \ x=1\)

\(y\text{-int:}\ y=2-(-3)\ \ \Rightarrow\ \ y=5\)

\(x\text{-int:}\ 2-\dfrac{3}{x-1}=0\ \ \Rightarrow\ \ x=\dfrac{5}{2} \)

\(\text{As}\ \ x \rightarrow \infty, \ \ y \rightarrow 2^{-}; \ \ x \rightarrow -\infty, \ \ y \rightarrow 2^{+} \)

b.    \(\text{From the graph:}\)

\(f(x)=1\ \text{when }x=4\)

\(x>1\ \text{to the right of the vertical asymptote}\)

\(\therefore\ f(x)\leq1\ \text{when}\ \ 1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]\)


♦♦ Mean mark (b) 38%.
MARKER’S COMMENT: Many students did not use their graph from part (a).

Graphs, MET1 2021 VCAA 4

  1. Sketch the graph of  `y = 1-2/(x-2)`  on the axes below. Label asymptotes with their equations and axis intercepts with their coordinates.   (3 marks)

     

     

  2. Find the values of  `x`  for which  `1-2/(x-2) >= 3`.   (1 mark)

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a. 

b. `x in [1, 2)`

Show Worked Solution

a.   `text(Asymptotes:)`

`x=2`

`text(As)\ \ x→ +-oo, \ y→1\ \ =>\ text(Asymptote at)\ \ y=1`

`ytext(-intercept at)\ (0,2)`

`xtext(-intercept at)\ (4,0)`

♦ Mean mark part (b) 32%.

b.   `text(By inspection of the graph:)`

`1-2/(x-2) >=3\ \ text(for)\ \ x in [1, 2)`

Calculus, MET1 2016 VCAA 3

Let  `f: R text{\}{1} -> R`  where  `f(x) = 2 + 3/(x - 1)`.

  1. Sketch the graph of  `f`. Label the axis intercepts with their coordinates and label any asymptotes with the appropriate equation.   (3 marks)
     

     

  2. Find the area enclosed by the graph of  `f`, the lines  `x = 2`  and  `x = 4`, and the `x`-axis.   (2 marks)

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  1.  
  2. `4 + 3log_e(3)\ text(units)`
Show Worked Solution
a.   

 

b.    `text(Area)` `= int_2^4 2 + 3(x – 1)^(−1)\ dx`
    `= [2x + 3 log_e(x – 1)]_2^4`
    `= (8 + 3log_e(3)) – (4 + 3log_e(1))`
    `= 4 + 3log_e(3)\ \ text(u²)`

Calculus, MET2 2012 VCAA 2

Let  `f: R text(\{2}) -> R,\ f(x) = 1/(2x-4) + 3.`

  1. Sketch the graph of  `y = f(x)` on the set of axes below. Label the axes intercepts with their coordinates and label each of the asymptotes with its equation.   (3 marks)

           VCAA 2012 2a
     

  2.   i. Find `f^{′}(x)`.   (1 mark)

  3.  ii. State the range of  `f ^{′}`.   (1 mark)

  4. iii. Using the result of part ii. explain why `f` has no stationary points.   (1 mark)

  5. If  `(p, q)`  is any point on the graph of  `y = f(x)`, show that the equation of the tangent to  `y = f(x)`  at this point can be written as  `(2p-4)^2 (y-3) = -2x + 4p-4.`   (2 marks)

  6. Find the coordinates of the points on the graph of  `y = f(x)`  such that the tangents to the graph at these points intersect at  `(-1, 7/2).`   (4 marks)

  7. A transformation  `T: R^2 -> R^2`  that maps the graph of  `f` to the graph of the function  `g: R text(\{0}) -> R,\ g(x) = 1/x`  has rule
  8.      `T([(x), (y)]) = [(a, 0), (0, 1)] [(x), (y)] + [(c), (d)]`, where `a`, `c` and `d` are non-zero real numbers.
  9. Find the values of `a, c` and `d`.   (2 marks)

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  1. met2-2012-vcaa-sec2-answer
  2.   i. `f^{′}(x) = (−2)/((2x-4)^3)`
     ii. `text(Range) = (−∞,0)`
    iii. `text(See Worked Solutions)`
  3. `text(See Worked Solutions)`
  4. `text(Coordinates:) (1,5/2)\ text(or)\ (5,19/6)`
  5. `a = 2, c = −4, d = −3`
Show Worked Solution

a.   `text(Asymptotes:)`

`x = 2`

`y = 3`

met2-2012-vcaa-sec2-answer

 

b.i.   `f^{′}(x) = (−2)/((2x-4)^2)`

 

b.ii.   `text(Range) = (−∞,0), or  R^-`

MARKER’S COMMENT: Incorrect notation in part (b)(ii) was common, including `{-oo,0}, -R, (0,-oo)`.

 

b.iii.    `text(As)\ \ ` `f^{′}(x) < 0quadtext(for)quadx ∈ R text(\{2})`
    `f^{′}(x) != 0`

 
`:. f\ text(has no stationary points.)`
 

c.   `text(Point of tangency) = P(p,1/(2p-4) + 3)`

♦♦ Mean mark 29%.
`m_text(tang)` `= f^{′}(p)`
  `= (-2)/((2p-4)^2)`

 
`text(Equation of tangent using:)`

`y-y_1` `= m(x-x_1)`
`y-(1/(2p-4) + 3)` `= (-2)/((2p-4)^2)(x-p)`
`y-3` `= (-2(x-p))/((2p-4)^2) + (2p-4)/((2p-4)^2)`
`(2p-4)^2(y-3)` `=-2x + 2p + 2p-4`
`:. (2p-4)^2(y-3)` `=-2x + 4p-4\ \ text(… as required)`

 

d.   `text(Substitute)\ \ (−1,7/2)\ text{into tangent (part c),}`

♦♦♦ Mean mark 19%.

`text(Solve)\ \ (2p-4)^2(7/2-3) = −2(-1) + 4p-4\ \ text(for)\ p:`

`:. p = 1,\ text(or)\ 5`

`text(Substitute)\ \ p = 1\ text(and)\ p = 5\ text(into)\ \ P(p,1/(2p-4) + 3)`

`:. text(Coordinates:)\ (1,5/2)\ text(or)\ (5,19/6)`
 

e.   `text(Determine transformations that that take)\ f -> g:`

`text(Dilate the graph of)\ \ f(x) = 1/(2x-4) + 3\ \ text(by a)`

`text(factor of 2 from the)\ \ ytext(-axis).`

`y = 1/(2(x/2)-4) + 3= 1/(x-4) + 3`

`text(Translate the graph 4 units to the left and 3)`

`text(units down to obtain)\ \ g(x).`
 

`text(Using the transformation matrix,)`

`x^{′}` `=ax+c`
`y^{′}` `=y+d`

 
`f -> g:\ \ 1/(2x-4) -> 1/(x^{′})`

`x^{′}=2x-4`

`=> a=2,\ \ c=-4`
 

`f -> g:\ \ y -> y^{′} + 3`

`y^{′}=y -3`

`=>\ \ d=-3`