Quadratics

Quadratics, SMB-015

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.

Find the `x`-coordinates of points `A and B.` (2 marks)

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Write down the coordinates of `C.` (2 marks)

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`A = 2,\ \ B = 5`
`(7, 10)`

Show Worked Solution

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)`

Quadratics, SMB-014

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`

Show Worked Solution

`x\text(-coordinate of)\ A:`

`y`	`= 2x\ \ \ \ \ …\ text{(i)}`
`y`	`= -2x^2 + 8x\ \ \ \ \ …\ text{(ii)}`

`text(Subst)\ y = 2x\ text{from (i) into (ii):}`

`-2x^2 + 8x`	`= 2x`
`-2x^2 + 6x`	`= 0`
`-2x (x-3)`	`= 0`

`:.\ x = 0\ text(or)\ 3`

`:.\ x\text(-coordinate of)\ A\ text(is 3)`

Quadratics, SMB-013

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)`

Show Worked Solution

`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)`

Quadratics, SMB-012

Factorise `y=x^2-8x+15` (1 mark)

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Find the vertex of the parabola with equation `y=x^2-8x+15` (2 marks)

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`y=(x-5)(x-3)`
`text{Vertex}\ = (4,-1)`

Show Worked Solution

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)`

Quadratics, SMB-011

Factorise `y=2x^2+5x-3` (1 mark)

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Find the vertex of the parabola with equation `y=2x^2+5x-3` (2 marks)

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`y=(2x-1)(x+3)`
`text{Vertex}\ = (-5/4,-49/8)`

Show Worked Solution

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)`

Quadratics, SMB-010

Factorise `y=6-x-x^2` (2 marks)

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Find the vertex of the parabola with equation `y=6-x-x^2` (2 marks)

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`y=(2-x)(x+3)`
`text{Vertex}\ = (-1/2,6 1/4)`

Show Worked Solution

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)`

Quadratics, SMB-009

By completing the square, find the coordinates of the vertex of the parabola with equation

`y=x^2-3x+1` (3 marks)

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`(3/2,-5/4)`

Show Worked Solution

`y`	`=x^2-3x+1`
	`=x^2-3x+9/4-5/4`
	`=(x-3/2)^2-5/4`

`:.\ text{Vertex}\ = (3/2,-5/4)`

Quadratics, SMB-008

By completing the square, find the coordinates of the vertex of the parabola with equation

`y=x^2+8x+9` (3 marks)

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`(-4,-7)`

Show Worked Solution

`y`	`=x^2+8x+9`
	`=x^2+8x+16-7`
	`=(x+4)^2-7`

`:.\ text{Vertex}\ = (-4,-7)`

Quadratics, SMB-007

By completing the square, find the coordinates of the vertex of the parabola with equation

`y=x^2-6x-4` (3 marks)

Show Answers Only

`(3,-13)`

Show Worked Solution

`y`	`=x^2-6x-4`
	`=x^2-6x+9-13`
	`=(x-3)^2-13`

`:.\ text{Vertex}\ = (3,-13)`

Quadratics, SMB-006

Complete the table of values for the equation `y=2-x^2/2`. (1 mark)

\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}

Sketch the graph `y=2-x^2/2` (2 marks)
For what range of `x`-values is the parabola concave up? (1 mark)

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\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.}`

Show Worked Solution

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.}`

Quadratics, SMB-005

Complete the table of values for the equation `y=8-2x^2`. (1 mark)

\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}

Sketch the graph `y=8-2x^2` (2 marks)
For what range of `x`-values is the parabola concave down? (1 mark)

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ii.

iii. `text{The parabola is concave down for all values of}\ x.`

Show Worked Solution

ii.

iii. `text{The parabola is concave down for all values of}\ x.`

Quadratics, SMB-004

Complete the table of values for the equation `y=4-x^2`. (1 mark)

\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}

Sketch the graph `y=4-x^2` (2 marks)
What are the coordinates of the vertex of the parabola? (1 mark)

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Show Worked Solution

\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} & 0 & 3 & 4 & 3 & 0 \\
\hline
\end{array}

ii.

iii. `text{Vertex at (0,4)}`

Quadratics, SMB-003

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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\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} & 5 & -1 & -3 & -1 & 5 \\
\hline
\end{array}

Show Worked Solution

Quadratics, SMB-002

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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\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} & 7 & 4 & 3 & 4 & 7 \\
\hline
\end{array}

Show Worked Solution

Quadratics, SMB-001

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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Show Worked Solution

\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 & -1 & 0 & -1 & 2 \\
\hline
\end{array}

Algebra, STD1 A3 2019 HSC 9 MC

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.

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`D`

Show Worked Solution

`text(Depth will decrease slowly at first and accelerate.)`

`=> D`

Algebra, STD2 A4 2022 HSC 9 MC

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?

4 seconds
6 seconds
8 seconds
10 seconds

Show Answers Only

`A`

Show Worked Solution

`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`

Algebra, STD2 A4 2020 HSC 19

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.

If the area of the paddock is `30 \ 000\ text(m)^2`, what is the largest possible value of `x`? (1 mark)

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Find the values of `x` and `y` so that the area of the paddock is as large as possible. (2 marks)

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Using your value from part (b), find the largest possible area of the paddock. (1 mark)

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`200 \ text(m)`
`x = 150 \ text(m and) \ y = 225 \ text(m)`
`33 \ 750 \ text(m)^2`

Show Worked Solution

a. `text(From the graph, an area of)\ 30\ 000\ text(m)^2`

♦ Mean mark part (a) 39%.

`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`

♦♦ Mean mark part (b) 34%.

`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)`

♦ Mean mark part (c) 40%.

c.	`A_(max)`	`= xy`
		`= 150 xx 225`
		`= 33 \ 750 \ text(m)^2`

Algebra, STD2 A4 2018 HSC 4 MC

Which graph best represents the equation `y = x^2-2`?

A.		B.
C.		D.

Show Answers Only

`A`

Show Worked Solution

`y = x^2-2`

`ytext(-intercept) = -2\ \ \ (text(when) = 0)`

`text(Quadratic is positive with vertex at)\ \ y = -2`

`=>A`

Algebra, STD2 A4 2014 HSC 3 MC

The diagram shows the graph of an equation.

Which of the following equations does the graph best represent?

`y = 3/x + 1`
`y = 3^x + 1`
`y = 3x^2 + 1`
`y = 3x^3 + 1`

Show Answers Only

`C`

Show Worked Solution

`text(Graph is a parabola that passes through)\ (0, 1).`

`=> C`

Algebra, STD2 A4 2012 HSC 30b

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.

What is the maximum height the ball reaches above the ground? (1 mark)

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There are two occasions when the golf ball is at a height of 35 metres.

What horizontal distance does the ball travel in the period between these two occasions? (1 mark)

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What is the height of the ball above the ground when it still has to travel a horizontal distance of 50 metres to hit the ground at point `B`? (1 mark)

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Only part of the graph applies to this model.

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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`40 text(m)`
`140 text(m)`
`text(17.5 m)`
`d < 0\ text(and)\ d>300`

Show Worked Solution

i. `text(Max height) = 40 text(m)`

COMMENT: With a mean mark of 92% in (i), a classic example of low hanging fruit in later questions.

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`

MARKER’S COMMENT: Responses for (iii) in the range `17<=\ h\ <=18` were deemed acceptable estimates read off the graph.

`=>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).`

♦♦♦ Mean mark (iv) 12%
MARKER’S COMMENT: Many students did not refer to the domain `d>300` as unsuitable to the model.

`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.)`