Let \(g: R \backslash\{-3\} \rightarrow R, g(x)=\dfrac{1}{(x+3)^2}-2\). --- 0 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- a. b. \(\dfrac{10}{3}\ \text{sq units}\) 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.
\(-\displaystyle\int_{-2}^0 (x+3)^{-3}-2\,dx\)
\(=-\left[\dfrac{1}{-1}(x+3)^{-1}-2x\right]_{-2}^0\)
\(=-\left[\dfrac{-1}{x+3}-2x\right]_{-2}^0\)
\(=-\left[\dfrac{-1}{3}-\left(\dfrac{-1}{-2+3}-2(-2)\right)\right]\)
\(=\dfrac{10}{3}\ \text{sq units}\)
Graphs, MET1 2023 VCE SM-Bank 1
Let \(\displaystyle f:[-3,-2) \cup(-2, \infty) \rightarrow R, f(x)=1+\frac{1}{x+2}\). --- 0 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- a. b. \(x\in [-3, -2)\cap (-1, \infty)\) a. b. \(\text{From graph }f(x)\leq -2\ \text{ for}\ -3\leq x <2\ \text{ and when }\ x\geq -1\) \(\rightarrow x\in [-3, -2)\cap (-1, \infty)\)
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Graphs, MET1 2023 VCAA 3
- 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)
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- 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).
MARKER’S COMMENT: Many students did not use their graph from part (a).
Graphs, MET2 2020 VCAA 18 MC
Let `a \in(0, \infty)` and `b \in R`.
Consider the function `h:[-a, 0) \cup(0, a] \rightarrow R, h(x)=\frac{a}{x}+b`.
The range of `h` is
- `[b-1,b+1]`
- `(b-1,b+1)`
- `(-oo,b-1)uu(b+1,oo)`
- `(-oo,b-1]uu[b+1,oo)`
- `[b-1,oo)`
Graphs, MET2 2020 VCAA 5 MC
The graph of the function `f:D rarr R,f(x)=(3x+2)/(5-x)`, where `D` is the maximal domain, has asymptotes
- `x=-5,y=(3)/(2)`
- `x=-3,y=5`
- `x=(2)/(3),y=-3`
- `x=5,y=3`
- `x=5,y=-3`
Graphs, MET1 2021 VCAA 4
- 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)
- Find the values of `x` for which `1 - 2/(x - 2) >= 3`. (1 mark)
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a.
b. `x in [1, 2)`
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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)`
Graphs, MET2-NHT 2019 VCAA 4 MC
The graph of the function `ƒ : D → R, \ f(x) = (2x -3)/(4 + x)`, where `D` is the maximal domain, has asymptotes
- `x = –4, \ y = 2`
- `x = (3)/(2), \ y = –4`
- `x = –4, \ y = (3)/(2)`
- `x = (3)/(2), \ y = 2`
- `x = 2, \ y = 1`
Algebra, MET2 2018 VCAA 3 MC
Consider the function `f: [a, b) -> R,\ f(x) = 1/x`, where `a` and `b` are positive real numbers.
The range of `f` is
- `[1/a, 1/b)`
- `(1/a, 1/b]`
- `[1/b, 1/a)`
- `(1/b, 1/a]`
- `[a, b)`
Graphs, MET2 2018 VCAA 2 MC
The maximal domain of the function `f` is `R text(\{1})`.
A possible rule for `f` is
- `f(x) = (x^2 - 5)/(x - 1)`
- `f(x) = (x + 4)/(x - 5)`
- `f(x) = (x^2 + x + 4)/(x^2 + 1)`
- `f(x) = (5 - x^2)/(1 + x)`
- `f(x) = sqrt (x - 1)`
Graphs, MET2 2008 VCAA 8 MC
The graph of the function `f: D -> R,\ f(x) = (x - 3)/(2 - x),` where `D` is the maximal domain has asymptotes
- `x = 3,\ \ \ \ \ \ \ \ \ \ y = 2`
- `x = -2,\ \ \ \ \ y = 1`
- `x = 1,\ \ \ \ \ \ \ \ \ \ y = -1`
- `x = 2,\ \ \ \ \ \ \ \ \ \ y = -1`
- `x = 2,\ \ \ \ \ \ \ \ \ \ y = 1`
Calculus, MET1 2016 VCAA 3
Let `f: R text{\}{1} -> R` where `f(x) = 2 + 3/(x - 1)`.
- Sketch the graph of `f`. Label the axis intercepts with their coordinates and label any asymptotes with the appropriate equation. (3 marks)
- 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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-
- `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)` |
Graphs, MET1 2008 VCAA 2
On the axes below, sketch the graph of `f: R\ text(\)\ {−1} -> R`, `f(x) = 2 - 4/(x + 1)`.
Label all axis intercepts. Label each asymptote with its equation. (4 marks)
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Show Worked Solution
`text(Asymptotes:)`
`x = −1`
`y = 2`
Calculus, MET2 2012 VCAA 2
Let `f: R text(\{2}) -> R,\ f(x) = 1/(2x-4) + 3.`
- 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)
- i. Find `f^{′}(x)`. (1 mark)
- ii. State the range of `f ^{′}`. (1 mark)
- iii. Using the result of part ii. explain why `f` has no stationary points. (1 mark)
- 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)
- 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)
- 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
- `T([(x), (y)]) = [(a, 0), (0, 1)] [(x), (y)] + [(c), (d)]`, where `a`, `c` and `d` are non-zero real numbers.
- Find the values of `a, c` and `d`. (2 marks)
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-
-
- `f^{′}(x) = (−2)/((2x-4)^3)`
- `text(Range) = (−∞,0)`
- `text(See Worked Solutions)`
- `text(See Worked Solutions)`
- `text(Coordinates:) (1,5/2)\ text(or)\ (5,19/6)`
- `a = 2, c = −4, d = −3`
Show Worked Solution
a. `text(Asymptotes:)`
`x = 2`
`y = 3`
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`