Find the centre and radius of the circle with the equation
`x^2+6x+y^2-y+3=0` (2 marks)
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Find the centre and radius of the circle with the equation
`x^2+6x+y^2-y+3=0` (2 marks)
`text{Centre}\ (-3,1/2),\ text{Radius}\ = 5/2`
`x^2+6x+y^2-y+3` | `=0` | |
`x^2+6x+9+y^2-y+1/4-25/4` | `=0` | |
`(x+3)^2+(y-1/2)^2` | `=25/4` | |
`(x+3)^2+(y-1/2)^2` | `=(5/2)^2` |
`text{Centre}\ (-3,1/2),\ text{Radius}\ = 5/2`
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Find the centre and radius of the circle with the equation
`x^2+ y^2+8y= 0` (2 marks)
`text(Centre)\ (0,-4)`
`text(Radius = 4)`
`x^2+ y^2+8y` | `= 0` |
`x^2+ y^2+8y+16-16` | `= 0` |
`x^2+(y+4)^2` | `= 4^2` |
`:.\ text(Centre)\ (0,-4)`
`:.\ text(Radius = 4)`
Find the centre and radius of the circle with the equation
`x^2+10x + y^2-6y+33 = 0` (2 marks)
`text(Centre)\ (-5,3)`
`text(Radius = 1)`
`x^2+10x + y^2-6y+33` | `= 0` |
`x^2+10x + 25 + y^2-6y+9-1` | `= 0` |
`(x+5)^2 + (y-3)^2` | `= 1` |
`:.\ text(Centre)\ (-5,3)`
`:.\ text(Radius = 1)`
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a. `(x+3)^2 + (y+2)^2 = 3^2`
`:.\ text(Centre:)\ (-3,-2)\text(, Radius:)\ 3`
b.
a. | `x^2+6x+y^2+4y+4` | `=0` |
`x^2+6x+9+y^2+4y+4-9` | `=0` | |
`(x+3)^2+(y+2)^2` | `=3^2` |
`:.\ text(Centre:)\ (-3,-2)\text(, Radius:)\ 3`
b.
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a. `(x-1)^2 + (y+2)^2 = 4`
b.
a. `text{Circle with centre}\ (1,-2),\ r = 2:`
`(x-1)^2 + (y+2)^2 = 4`
b.
Write down the equation of the circle with centre `(0, -3)` and radius 4. (1 mark)
`x^2 + (y+3)^2 = 16`
`text{Circle with centre}\ (0, -3),\ r = 4:`
`x^2 + (y+3)^2 = 16`
Find the centre and radius of the circle with the equation
`x^2-12x + y^2 + 2y-12 = 0` (2 marks)
`text(Centre)\ (6, −1)`
`text(Radius = 7)`
`x^2-12x + y^2 + 2y-12` | `= 0` |
`(x-6)^2 + (y + 1)^2-36-1-12` | `= 0` |
`(x-6)^2 + (y + 1)^2` | `= 49` |
`:.\ text(Centre)\ (6, −1)`
`:.\ text(Radius = 7)`
A circle with centre `(a,-2)` and radius 5 units has equation
`x^2-6x + y^2 + 4y = b` where `a` and `b` are real constants.
The values of `a` and `b` are respectively
`B`
`x^2-6x + y^2 + 4y=b`
`text(Completing the squares:)`
`x^2-6x + 3^2-9 + y^2 + 4y + 2^2-4` | `= b` |
`(x-3)^2 + (y + 2)^2-13` | `= b` |
`(x-3)^2 + (y + 2)^2` | `= b + 13` |
`:. a=3`
`:. b+13=25\ \ =>\ \ b=12`
`=> B`
The diagram shows the curve with equation `y = x^2-7x + 10`. The curve intersects the `x`-axis at points `A and B`. The point `C` on the curve has the same `y`-coordinate as the `y`-intercept of the curve.
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i. `y` | `= x^2-7x + 10` |
`= (x-2) (x-5)` |
`:.x = 2 or 5`
`:.\ \ x text(-coordinate of)\ \ A = 2`
`x text(-coordinate of)\ \ B = 5`
ii. `y\ text(intercept occurs when)\ \ x = 0`
`=>y text(-intercept) = 10`
`C\ text(occurs at intercept:)`
`y` | `= x^2-7x + 10` | `\ \ \ \ \ text{… (1)}` |
`y` | `= 10` | `\ \ \ \ \ text{… (2)}` |
`(1) = (2)`
`x^2-7x + 10` | `= 10` |
`x^2-7x` | `= 10` |
`x (x-7)` | `= 10` |
`x = 0 or 7`
`:.\ C\ \ text(is)\ \ (7, 10)`
The parabola `y = −2x^2 + 8x` and the line `y = 2x` intersect at the origin and at the point `A`.
Find the `x`-coordinate of the point `A`. (2 marks)
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`x=3`
Sketch the graph of `y=4/(x-3)`. (3 marks)
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`text{Vertical asymptote at}\ \ x=3`
`text{As}\ \ x->oo, \ y->0`
`text{Horizontal asymptote at}\ \ y=0`
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -1 & \ \ 0\ \ & \ \ 2\ \ & \ \ 4\ \ & \ \ 5\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & -1 & -\frac{4}{3} & -4 & 4 & 2\\
\hline
\end{array}
Sketch the graph of `y=2/(3-x)`. (3 marks)
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`text{Vertical asymptote at}\ \ x=3`
`text{As}\ \ x->oo, \ y->0`
`text{Horizontal asymptote at}\ \ y=0`
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -1 & \ \ 0\ \ & \ \ 2\ \ & \ \ 4\ \ & \ \ 5\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & \frac{1}{2} & \frac{2}{3} & 2 & -2 & -1\\
\hline
\end{array}
Sketch the graph of `y=3/(x+1)`. (2 marks)
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`text{Vertical asymptote at}\ \ x=-1`
`text{As}\ \ x->oo, \ y->0`
`text{Horizontal asymptote at}\ \ y=0`
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -3 & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & -\frac{3}{2} & -3 & ∞ & 3 & \frac{3}{2} \\
\hline
\end{array}
Sketch the graph of `y=1/(x-2)`. (2 marks)
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`text{Vertical asymptote at}\ \ x=2`
`text{As}\ \ x->oo, \ y->0`
`text{Horizontal asymptote at}\ \ y=0`
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ & \ \ 3\ \ & \ \ 4\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & -\frac{1}{2} & -1 & ∞ & 1 & \frac{1}{2} \\
\hline
\end{array}
Sketch the graph of `f(x) = (2x+1)/(x-1)`. Label the axis intercepts with their coordinates and label any asymptotes with the appropriate equation. (4 marks)
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`(2x+1)/(x-1)` | `=(2x-2+3)/(x-1)` | |
`=(2(x-1)+3)/(x-1)` | ||
`=2 + 3/(x-1)` |
`text(Asymptotes:)\ \ x = 1,\ \ y = 2`
`text(As)\ \ x->oo,\ \ y->2(+)`
`text(As)\ \ x->-oo,\ \ y->2(-)`
`text(As)\ \ x->-1 (-),\ \ y->-oo`
`text(As)\ \ x->-1 (+),\ \ y->oo`
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i. `y=2-1/x`
`text{Vertical asymptote at}\ \ x=0`
`text{As}\ x->oo, \ 1/x -> 0\ \ => 2-1/x -> 2`
`text{Horizontal asymptote at}\ \ y=2`
ii.
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & \frac{5}{2} & 3 & ∞ & 1 & \frac{3}{2} \\
\hline
\end{array}
Sketch the graph of `y=2/x+2`.
Clearly mark all asymptotes. (3 marks)
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\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & 1 & 0 & ∞ & 4 & 3 \\
\hline
\end{array}
Sketch the graph of `y=-2/x`. (2 marks)
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\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ x\ \ \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex}\ \ \ y\ \ \rule[-1ex]{0pt}{0pt} & -\frac{3}{2} & -3 & ∞ & 3 & \frac{3}{2} \\
\hline
\end{array}
Factorise the parabola described by the equation `y=-x^2-x+12` and find its vertex. (3 marks)
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`y=(3-x)(x+4)`
`text{Vertex}\ = (-1/2,12 1/4)`
`y` | `=-x^2-x+12` | |
`=-(x^2+x-12)` | ||
`=-(x+4)(x-3)` | ||
`=(3-x)(x+4)` |
`text{Solutions at}\ \ x=3, -4`
`text{Line of symmetry at mid-point of solutions.}`
`x=(3+(-4))/2=-1/2`
`text{Substitute}\ \ x=-1/2\ \ text{into}\ \ y=-x^2-x+12`
`y=-1/4+1/2+12=12 1/4`
`:.\ text{Vertex}\ = (-1/2,12 1/4)`
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i. `y` | `=x^2-8x+15` | |
`=(x-5)(x-3)` |
ii. `text{Solutions at}\ \ x=3,5.`
`text{Line of symmetry at mid-point of solutions.}`
`x=(3+5)/2=4`
`text{Substitute}\ \ x=4\ \ text{into}\ \ y=x^2-8x+15`
`y=4^2-8xx4+15=-1`
`:.\ text{Vertex}\ = (4,-1)`
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i. `y` | `=2x^2+5x-3` | |
`=(2x-1)(x+3)` |
ii. `text{Solutions at}\ \ x=1/2,-3.`
`text{Line of symmetry at mid-point of solutions.}`
`x=(1/2+(-3))/2=-5/4`
`text{Substitute}\ \ x=-5/4\ \ text{into}\ \ y=2x^2+5x-3`
`y=2xx(-5/4)^2-5xx5/4-3=-49/8`
`:.\ text{Vertex}\ = (-5/4,-49/8)`
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i. `y` | `=6-x-x^2` | |
`=-(x^2+x-6)` | ||
`=-(x-2)(x+2)` | ||
`=(2-x)(x+3)` |
ii. `text{Solutions at}\ \ x=2,-3.`
`text{Line of symmetry at mid-point of solutions.}`
`x=(2+(-3))/2=-1/2`
`text{Substitute}\ \ x=-1/2\ \ text{into}\ \ y=6-x-x^2`
`y=6+1/2-1/4=6 1/4`
`:.\ text{Vertex}\ = (-1/2,6 1/4)`
By completing the square, find the coordinates of the vertex of the parabola with equation
`y=x^2-3x+1` (3 marks)
`(3/2,-5/4)`
`y` | `=x^2-3x+1` | |
`=x^2-3x+9/4-5/4` | ||
`=(x-3/2)^2-5/4` |
`:.\ text{Vertex}\ = (3/2,-5/4)`
By completing the square, find the coordinates of the vertex of the parabola with equation
`y=x^2+8x+9` (3 marks)
`(-4,-7)`
`y` | `=x^2+8x+9` | |
`=x^2+8x+16-7` | ||
`=(x+4)^2-7` |
`:.\ text{Vertex}\ = (-4,-7)`
By completing the square, find the coordinates of the vertex of the parabola with equation
`y=x^2-6x-4` (3 marks)
`(3,-13)`
`y` | `=x^2-6x-4` | |
`=x^2-6x+9-13` | ||
`=(x-3)^2-13` |
`:.\ text{Vertex}\ = (3,-13)`
\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2 \ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & & 2 & & 0 \\
\hline
\end{array}
i.
\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2 \ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 0 & \frac{3}{2} & 2 & \frac{3}{2} & 0 \\
\hline
\end{array}
ii.
iii. `text{The parabola is concave down for all values of}\ x.`
`=>\ text{There are no values of}\ x\ text{where the graph is concave up.}`
i.
\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2 \ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 0 & \frac{3}{2} & 2 & \frac{3}{2} & 0 \\
\hline
\end{array}
ii.
iii. `text{The parabola is concave down for all values of}\ x.`
`=>\ text{There are no values of}\ x\ text{where the graph is concave up.}`
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ & \ \ 3 \ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & & 6 & & & & -10 \\
\hline
\end{array}
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i.
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ & \ \ 3 \ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & -10 & 0 & 6 & 8 & 6 & 0 & -10 \\
\hline
\end{array}
ii.
iii. `text{The parabola is concave down for all values of}\ x.`
i.
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ & \ \ 3 \ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & -10 & 0 & 6 & 8 & 6 & 0 & -10 \\
\hline
\end{array}
ii.
iii. `text{The parabola is concave down for all values of}\ x.`
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & & & 3 & \\
\hline
\end{array}
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By completing the table of values, sketch the graph of `y=2x^2-3`. (3 marks)
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & & & -1 & \\
\hline
\end{array}
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By completing the table of values, sketch the graph of `y=x^2+3`. (3 marks)
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & & & 4 & \\
\hline
\end{array}
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By completing the table of values, sketch the graph of `y=x^2-2`. (3 marks)
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & -1 & & & \\
\hline
\end{array}
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Sketch the graph of `y=3^x+2`, clearly labelling any asymptotes. (3 marks)
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`y=3^x+2`
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 2 \frac{1}{9} & 2 \frac{1}{3} & 3 & 5 & 11\\
\hline
\end{array}
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ & \ \ 3\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 5 & 2 & \frac{1}{2} & -\frac{1}{4} & -\frac{5}{8}\\
\hline
\end{array}
By completing the table of values, sketch the graph of `y=2^(-x)` (3 marks)
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ x\ \ \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} \ \ y\ \ \rule[-1ex]{0pt}{0pt} & & & 1 & & \\
\hline
\end{array}
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\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 4 & 2 & 1 & \frac{1}{2} & \frac{1}{4}\\
\hline
\end{array}
\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & \ \ 0\ \ & \ \ 1\ \ & \ \ 2\ \ \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & & -\frac{1}{2} & & & 3\\
\hline
\end{array}
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In 2004 there were 13.5 million registered motor vehicles in Australia. The number of registered motor vehicles is increasing at a rate of 2.3% per year.
Find an expression that represents the number (in millions) of registered motor vehicles `(V)`, if `y` represents the number of years after 2004? (2 marks)
`V = 13.5 xx (1.023)^y`
`text(In 2004, 13.5 million)`
`Vtext{(1 year later)} = 13.5 xx (1.023)`
`Vtext{(2 years later)}= 13.5 xx (1.023) xx (1.023)= 13.5 xx (1.023)^2`
`:. V(y\ text{years later}) = 13.5 xx (1.023)^y`
The container shown is initially full of water.
Water leaks out of the bottom of the container at a constant rate.
Which graph best shows the depth of water in the container as time varies?
A. | B. | ||
C. | D. |
`D`
`text(Depth will decrease slowly at first and accelerate.)`
`=> D`
A student believes that the time it takes for an ice cube to melt (`M` minutes) varies inversely with the room temperature `(T^@ text{C})`. The student observes that at a room temperature of `15^@text{C}` it takes 12 minutes for an ice cube to melt.
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\begin{array} {|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ T\ \ \rule[-1ex]{0pt}{0pt} & \ \ \ 5\ \ \ & \ \ 15\ \ \ & \ \ \ 30\ \ \ \\
\hline
\rule{0pt}{2.5ex} \ \ M\ \ \rule[-1ex]{0pt}{0pt} & & & \\
\hline
\end{array}
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a. | `M` | `prop 1/T` |
`M` | `=k/T` | |
`12` | `=k/15` | |
`k` | `=15 xx 12` | |
`=180` |
`:.M=180/T`
b.
\begin{array} {|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ T\ \ \rule[-1ex]{0pt}{0pt} & \ \ \ 5\ \ \ & \ \ 15\ \ \ & \ \ 30\ \ \\
\hline
\rule{0pt}{2.5ex} \ \ M\ \ \rule[-1ex]{0pt}{0pt} & 36 & 12 & 6 \\
\hline
\end{array}
a. | `M` | `prop 1/T` |
`M` | `=k/T` | |
`12` | `=k/15` | |
`k` | `=15 xx 12` | |
`=180` |
`:.M=180/T`
b.
\begin{array} {|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ T\ \ \rule[-1ex]{0pt}{0pt} & \ \ \ 5\ \ \ & \ \ 15\ \ \ & \ \ 30\ \ \\
\hline
\rule{0pt}{2.5ex} \ \ M\ \ \rule[-1ex]{0pt}{0pt} & 36 & 12 & 6 \\
\hline
\end{array}
An object is projected vertically into the air. Its height, `h` metres, above the ground after `t` seconds is given by `h=-5 t^2+80 t`.
For how long is the object at a height of 300 metres or more above the ground?
`A`
`text{Object reaches 300 m when}\ \ t=6\ text{seconds.}`
`text{Object drops back below 300 m when}\ \ t=10\ text{seconds.}`
`text{Time at 300 m or above}\ = 10-6=4\ text{seconds}`
`=>A`
Without using calculus, sketch the graph of `y = 2 + 1/(x + 4)`, showing the asymptotes and the `x` and `y` intercepts. (3 marks)
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A population, `P`, is to be modelled using the function `P = 2000 (1.2)^t`, where `t` is the time in years.
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Each year the number of fish in a pond is three times that of the year before.
Complete the table above showing the number of fish in 2021 and 2022. (2 marks)
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a.
\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \ & \ \ \ 2021\ \ \ & \ \ \ 2022\ \ \ & \ \ \ 2023\ \ \ \\
\hline
\rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & 300 & 900 & 2700\\
\hline
\end{array}
c. The more suitable model is exponential.
A linear dataset would graph a straight line which is not the case here.
An exponential curve can be used to graph populations that grow at an increasing rate, such as this example.
a.
\begin{array} {|l|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Year}\rule[-1ex]{0pt}{0pt} & \ \ \ 2020\ \ \ & \ \ \ 2021\ \ \ & \ \ \ 2022\ \ \ & \ \ \ 2023\ \ \ \\
\hline
\rule{0pt}{2.5ex}\textit{Number of fish}\rule[-1ex]{0pt}{0pt} & 100 & 300 & 900 & 2700\\
\hline
\end{array}
b.
c. The more suitable model is exponential.
A linear dataset would graph a straight line which is not the case here.
An exponential curve can be used to graph populations that grow at an increasing rate, such as this example.
The circle of `x^2-6x + y^2 + 4y-3 = 0` is reflected in the `x`-axis.
Sketch the reflected circle, showing the coordinates of the centre and the radius. (3 marks)
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`x^2-6x + y^2 + 4y-3` | `= 0` |
`x^2-6x + 9 + y^2 + 4y + 4-16` | `= 0` |
`(x-3)^2 + (y + 2)^2` | `= 16` |
`=>\ text{Original circle has centre (3, − 2), radius = 4}`
`text(Reflect in)\ xtext(-axis):`
`text{Centre (3, − 2) → (3, 2)}`
A fence is to be built around the outside of a rectangular paddock. An internal fence is also to be built.
The side lengths of the paddock are `x` metres and `y` metres, as shown in the diagram.
A total of 900 metres of fencing is to be used. Therefore `3x + 2y = 900`.
The area, `A`, in square metres, of the rectangular paddock is given by `A =450x - 1.5x^2`.
The graph of this equation is shown.
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a. `text(From the graph, an area of)\ 30\ 000\ text(m)^2`
`text(can have an)\ x text(-value of)\ \ x=100 or 200\ text(m.)`
`:. x_text(max) = 200 text(m)`
b. `A_text(max) \ text(occurs when) \ \ x = 150`
`text(Substitute)\ \ x=150\ \ text(into)\ \ 3x + 2y = 900:`
`3 xx 150 + 2y` | `= 900` |
`2y` | `= 450` |
`y` | `= 225` |
`therefore \ text(Maximum area when) \ \ x = 150 \ text(m and) \ \ y = 225 \ text(m)`
c. | `A_(max)` | `= xy` |
`= 150 xx 225` | ||
`= 33 \ 750 \ text(m)^2` |
Which graph best represents the equation `y = x^2-2`?
A. | B. | ||
C. | D. |
`A`
`y = x^2-2`
`ytext(-intercept) = -2\ \ \ (text(when) = 0)`
`text(Quadratic is positive with vertex at)\ \ y = -2`
`=>A`
The cost of hiring an open space for a music festival is $120 000. The cost will be shared equally by the people attending the festival, so that `C` (in dollars) is the cost per person when `n` people attend the festival.
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i.
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}
ii. |
iii. `C = (120\ 000)/n`
`n\ text(must be a whole number)`
iv. `text(Limitations can include:)`
`•\ n\ text(must be a whole number)`
`•\ C > 0`
v. `text(If)\ C = 94:`
`94` | `= (120\ 000)/n` |
`94n` | `= 120\ 000` |
`n` | `= (120\ 000)/94` |
`= 1276.595…` |
`:.\ text(C)text(ost cannot be $94 per person,)`
`text(because)\ n\ text(isn’t a whole number.)`
i.
\begin{array} {|l|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\text{Number of people} (n) \rule[-1ex]{0pt}{0pt} & \ 500\ & \ 1000 \ & 1500 \ & 2000 \ & 2500\ & 3000 \ \\
\hline
\rule{0pt}{2.5ex}\text{Cost per person} (C)\rule[-1ex]{0pt}{0pt} & 240 & 120 & 80 & 60 & 48\ & 40 \ \\
\hline
\end{array}
ii. |
iii. `C = (120\ 000)/n`
iv. `text(Limitations can include:)`
`•\ n\ text(must be a whole number)`
`•\ C > 0`
v. `text(If)\ C = 94`
`=> 94` | `= (120\ 000)/n` |
`94n` | `= 120\ 000` |
`n` | `= (120\ 000)/94` |
`= 1276.595…` |
`:.\ text(C)text(ost cannot be $94 per person,)`
`text(because)\ n\ text(isn’t a whole number.)`
Write down the equation of the circle with centre `(-1, 2)` and radius 5. (1 mark)
`text{Circle with centre (-1,2)},\ r = 5`
`(x + 1)^2 + (y-2)^2 = 25`
`text{Circle with centre}\ (-1, 2),\ r = 5`
`(x + 1)^2 + (y-2)^2 = 25`
A golf ball is hit from point `A` to point `B`, which is on the ground as shown. Point `A` is 30 metres above the ground and the horizontal distance from point `A` to point `B` is 300 m.
The path of the golf ball is modelled using the equation
`h = 30 + 0.2d-0.001d^2`
where
`h` is the height of the golf ball above the ground in metres, and
`d` is the horizontal distance of the golf ball from point `A` in metres.
The graph of this equation is drawn below.
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What horizontal distance does the ball travel in the period between these two occasions? (1 mark)
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Find all values of `d` that are not suitable to use with this model, and explain why these values are not suitable. (2 marks)
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i. `text(Max height) = 40 text(m)`
ii. `text(From graph)`
`h = 35\ text(when)\ x = 30\ text(and)\ x = 170`
`:.\ text(Horizontal distance)` | `= 170-30` |
`= 140\ text(m)` |
iii. `text(Ball hits ground at)\ x = 300`
`=>text(Need to find)\ y\ text(when)\ x = 250`
`text(From graph,)\ y = 17.5 text(m)\ text(when)\ x = 250`
`:.\ text(Height of ball is 17.5 m at a horizontal)`
`text(distance of 50m before)\ B.`
iv. `text(Values of)\ d\ text(not suitable).`
`text(If)\ d < 0 text(, it assumes the ball is hit away)`
`text(from point)\ B text(. This is not the case in our)`
`text(example.)`
`text(If)\ d > 300 text(,)\ h\ text(becomes negative which is)`
`text(not possible given the ball cannot go)`
`text(below ground level.)`