smc-6922-10-Find Vertex

Algebra, STD2 EQ-Bank 38

The graph of the parabola $y=a(x+2)(x-6)$ for some value of $a$ is shown.

By first finding the value of $a$, find the coordinates of the vertex. (3 marks)

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$\text{Vertex:}\ (2,32)$

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$\text{Since graph passes through}\ (0,24):$

$24$	$=a(0+2)(0-6)$
$24$	$=-12a$
$a$	$=-2$

$\text{Vertex is halfway between \(x$-intercepts.}\)

$\Rightarrow \ x=\dfrac{-2+6}{2}=2$

$y$	$=-2(2+2)(2-6)$
	$=-2 \times 4 \times (-4)=32$

$\therefore\ \text{Vertex at}\ (2,32).$

Algebra, STD2 EQ-Bank 35

The graph of the parabola $y=k(x-2)(x-8)$ for some value of $k$ is shown.

By first finding the value of $k$, find the coordinates of the vertex. (3 marks)

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$\text{Vertex:}\ (5,-18)$

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$\text{Since graph passes through}\ (0,32):$

$32$	$=k(0-2)(0-8)$
$32$	$=16k$
$k$	$=2$

$\text{Vertex is halfway between \(x$-intercepts.}\)

$\Rightarrow \ x=\dfrac{2+8}{2}=5$

$y$	$=2(5-2)(5-8)$
	$=2 \times 3 \times (-3)$
	$=-18$

$\therefore\ \text{Vertex at}\ (5,-18).$

Algebra, STD2 EQ-Bank 36

The graph of the parabola $y=a(x-1)(x-9)$ for some value of $a$ is shown.

By first finding the value of $a$, find the coordinates of the vertex. (3 marks)

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$\text{Vertex:}\ (5,32)$

Show Worked Solution

$\text{Since graph passes through}\ (0,-18):$

$-18$	$=a(0-1)(0-9)$
$-18$	$=9a$
$a$	$=-2$

$\text{Vertex is halfway between \(x$-intercepts}\ \ \Rightarrow\ \ x=5\)

$y$	$=-2(5-1)(5-9)$
	$=-2 \times 4 \times (-4)=32$

$\therefore\ \text{Vertex at}\ (5,32).$

Algebra, STD2 A4 2025 HSC 20

The graph of a quadratic function represented by the equation $h=t^2-8 t+12$ is shown.

Find the values of $t$ and $h$ at the turning point of the graph. (2 marks)

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The graph shows $h=12$ when $t=0$.
What is the other value of $t$ for which $h=12$? (1 mark)

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a. $\text{Turning point at} \ \ (4,-4)$

b. $t=8$

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a. $\text{Axis of quadratic occurs when}\ \ t= \dfrac{2+6}{2} = 4$

$\text{At} \ \ t=4:$

$h=4^2-8 \times 4+12=-4$

$\therefore \ \text{Turning point at} \ \ (4,-4)$

b. $\text {When} \ \ h=12:$

$t^2-8t+12$	$=12$
$t(t-8)$	$=0$

$\therefore \ \text{Other value:} \ \ t=8$

Algebra, STD2 A4 2024 HSC 26

A sheet of metal is folded to make a gutter, as shown. The cross-section of the gutter is a rectangle of width $w$ cm and height $h$ cm.

The area, $A$ cm$^{2}$, of the cross-section can be modelled by the quadratic formula

$A=-0.5w^{2}+20w$

A graph of this model is shown.

Find the width and height of the rectangle which will give the greatest possible area of the cross-section. (3 marks)

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$w=20\ \text{cm}, h=10\ \text{cm}$

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$\text{Graph cuts}\ w\text{-axis at}\ \ w=0\ \ \text{and}\ \ w=40.$

$\text{By symmetry, maximum area occurs at}\ \ w=20.$

$A_{\text{max}}=-0.5 \times 20^2 + 20 \times 20 = 200\ \text{cm}^{2}$

$A_{\text{max}}$	$=w \times h$
$200$	$=20 \times h$
$h$	$=10\ \text{cm}$

$\therefore A_{\text{max}}\ \text{occurs when:}\ w=20\ \text{cm}, \ h=10\ \text{cm}$

♦ Mean mark 43%.

Algebra, STD2 A4 2023 HSC 20

On another planet, a ball is launched vertically into the air from the ground.

The height above the ground, `h` metres, can be modelled using the function `h=-6 t^2+24t`, where `t` is measured in seconds. The graph of the function is shown.

Based on the graph, what is the maximum height reached by the ball? (1 mark)

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Based on the graph, at what TWO times is the ball at `3/4` of its maximum height? (2 marks)

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a. `h_max = 24\ text{metres}`

b. `t=1 and 3\ text{seconds}`

Show Worked Solution

a. `h_max = 24\ text{metres}`

b. `3/4 xx h_max = 3/4xx24=18\ text{metres}`

`text{From graph, ball is at at 18 metres when:}`

`t=1 and 3\ text{seconds}`

Algebra, STD2 A4 2021 HSC 35

A publisher sells a book for $10. At this price, 5000 copies of the book will be sold and the revenue raised will be `5000 xx 10=$50\ 000`.

The publisher is considering increasing the price of the book. For every dollar the price of the book is increased, the publisher will sell 50 fewer copies of the book.

If the publisher charges `(10+x)` dollars for each book, a quadratic model for the revenue raised, `R`, from selling books is

`R=-50x^2+4500x + 50\ 000`

By first finding a suitable value of `x`, find the price the publisher should charge for each book to maximise the revenues raised from sales of the book. (2 marks)

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Find the value of the intercept of the parabola with the vertical axis. (1 mark)

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a. `$55`

b. `$50\ 000`

Show Worked Solution

a. `text{Highest revenue}\ (R)\ text(occurs halfway between)`

`x= –10 and x=100.`

`text{Midpoint}\ =(-10 + 100)/2 = 45`

♦♦♦ Mean mark part (a) 16%.

`:.\ text(Price of book for)\ R_text(max)`

`=45 + 10=$55`

♦♦♦ Mean mark part (b) 21%.

b. `ytext(-intercept → find)\ \ R\ \ text(when)\ \ x=0:`

`R= -50(0)^2 + 4500(0) + 50\ 000=$50\ 000`

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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a. `200 \ text(m)`

b. `x = 150 \ text(m and) \ y = 225 \ text(m)`

c. `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`

\(24\)	\(=a(0+2)(0-6)\)
\(24\)	\(=-12a\)
\(a\)	\(=-2\)

\(32\)	\(=k(0-2)(0-8)\)
\(32\)	\(=16k\)
\(k\)	\(=2\)

\(y\)	\(=2(5-2)(5-8)\)
	\(=2 \times 3 \times (-3)\)
	\(=-18\)

\(-18\)	\(=a(0-1)(0-9)\)
\(-18\)	\(=9a\)
\(a\)	\(=-2\)

\(t^2-8t+12\)	\(=12\)
\(t(t-8)\)	\(=0\)

\(y\)	\(=-2(2+2)(2-6)\)
	\(=-2 \times 4 \times (-4)=32\)

\(y\)	\(=-2(5-1)(5-9)\)
	\(=-2 \times 4 \times (-4)=32\)

\(A_{\text{max}}\)	\(=w \times h\)
\(200\)	\(=20 \times h\)
\(h\)	\(=10\ \text{cm}\)