Chapter
12
Quadratic
functions
Syllabus reference: 2.7, 4.3, 4.6
Contents:
A
B
C
D
E
F
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G
H
Quadratic functions
Graphs of quadratic
functions
Axes intercepts
Axis of symmetry
Vertex
Finding a quadratic from its
graph
Where functions meet
Quadratic modelling
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366
QUADRATIC FUNCTIONS
(Chapter 12)
OPENING PROBLEM
An athlete throws a javelin. Its height
above the ground after it has travelled a
horizontal distance of x metres is given
1 2
by H(x) = ¡ 60
x + x metres.
Things to think about:
a What would the graph of H against x look like?
b How high is the javelin after it has travelled
20 metres?
c What is the maximum height reached by the
javelin?
d How far did the javelin travel before it hit the ground?
The height function in the Opening Problem is an example of a quadratic function. In this
chapter we will look at the properties of quadratic functions, and how to draw their graphs.
We will use the quadratic equation solving techniques studied in Chapter 9 to solve problems
involving quadratic functions.
A
QUADRATIC FUNCTIONS
A quadratic function is a relationship between two variables which can be
written in the form y = ax2 + bx + c where x and y are the variables and
a, b and c represent constants, a 6= 0.
Using function notation, y = ax2 + bx + c can be written as f(x) = ax2 + bx + c.
FINDING y GIVEN x
For any value of x, the corresponding value of y can be found using substitution.
Example 1
Self Tutor
If y = 2x2 + 4x ¡ 5 find the value of y when: a x = 0
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b When x = 3,
y = 2(3)2 + 4(3) ¡ 5
= 2(9) + 12 ¡ 5
= 18 + 12 ¡ 5
= 25
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a When x = 0,
y = 2(0)2 + 4(0) ¡ 5
=0+0¡5
= ¡5
b x = 3.
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QUADRATIC FUNCTIONS
(Chapter 12)
367
We can test whether an ordered pair (x, y) satisfies a quadratic function by substituting the
x-coordinate into the function, and seeing whether the resulting value matches the y-coordinate.
Example 2
Self Tutor
State whether the following quadratic functions are satisfied by the given ordered pairs:
a y = 3x2 + 2x (2, 16)
b f (x) = ¡x2 ¡ 2x + 1 (¡3, 1)
a When x = 2,
y = 3(2)2 + 2(2)
= 12 + 4
= 16
So, when x = 2, y = 16
) (2, 16) does satisfy
y = 3x2 + 2x
b
f (¡3) = ¡(¡3)2 ¡ 2(¡3) + 1
= ¡9 + 6 + 1
= ¡2
So, f (¡3) 6= 1
) (¡3, 1) does not satisfy
f (x) = ¡x2 ¡ 2x + 1
When we substitute a value for y into a quadratic function, we are left with a quadratic
equation. Solving the quadratic equation gives us the values of x corresponding to that
y-value. Since the equation is quadratic, there may be 0, 1, or 2 values of x for any one
value of y.
Example 3
Self Tutor
If y = x2 ¡ 6x + 8 find the value(s) of x when: a y = 15 b y = ¡1
a If y = 15 then
x2 ¡ 6x + 8 = 15
) x2 ¡ 6x ¡ 7 = 0
) (x + 1)(x ¡ 7) = 0
) x = ¡1 or x = 7
So, there are 2 solutions.
b If y = ¡1 then
x2 ¡ 6x + 8 = ¡1
) x2 ¡ 6x + 9 = 0
) (x ¡ 3)2 = 0
) x=3
So, there is only one solution.
EXERCISE 12A
1 Which of the following are quadratic functions?
a y = 2x2 ¡ 4x + 10
b y = 15x ¡ 8
c y = ¡2x2
d y = 13 x2 + 6
e 3y + 2x2 ¡ 7 = 0
f y = 15x3 + 2x ¡ 16
2 For each of the following functions, find the value of y for the given value of x:
a y = x2 + 5x ¡ 14 when x = 2
b y = 2x2 + 9x when x = ¡5
c y = ¡2x2 + 3x ¡ 6 when x = 3
d y = 4x2 + 7x + 10 when x = ¡2
a If f (x) = x2 + 3x ¡ 7, find f (2) and f (¡1).
3
b If f (x) = 2x2 ¡ x + 1, find f (0) and f (¡3).
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c If g(x) = ¡3x2 ¡ 2x + 4, find g(3) and g(¡2).
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368
QUADRATIC FUNCTIONS
(Chapter 12)
4 State whether the following quadratic functions are satisfied by the given ordered pairs:
a f (x) = 6x2 ¡ 10
b y = 2x2 ¡ 5x ¡ 3
(0, 4)
c y = ¡4x + 6x
(¡ 12 ,
e f (x) = 3x2 ¡ 11x + 20
(2, ¡10)
2
2
¡4)
(4, 9)
d y = ¡7x + 9x + 11
(¡1, ¡6)
f f (x) = ¡3x2 + x + 6
( 13 , 4)
5 For each of the following quadratic functions, find any values of x for the given value
of y:
a y = x2 + 6x + 10 when y = 1
b y = x2 + 5x + 8 when y = 2
c y = x2 ¡ 5x + 1 when y = ¡3
d y = 3x2 when y = ¡3:
6 Find the value(s) of x for which:
a f (x) = 3x2 ¡ 3x + 6 takes the value 6
b f (x) = x2 ¡ 2x ¡ 7 takes the value ¡4
c f (x) = ¡2x2 ¡ 13x + 3 takes the value ¡4
d f (x) = 2x2 ¡ 10x + 1 takes the value ¡11.
Example 4
Self Tutor
A stone is thrown into the air. Its height above the ground is given by the function
h(t) = ¡5t2 + 30t + 2 metres where t is the time in seconds from when the stone is
thrown.
a How high is the stone above the ground at time t = 3 seconds?
b From what height above the ground was the stone released?
c At what time is the stone 27 m above the ground?
a h(3) = ¡5(3)2 + 30(3) + 2
= ¡45 + 90 + 2
= 47
) the stone is 47 m above the ground.
b The stone was released when
t = 0 s.
) h(0) = ¡5(0)2 + 30(0) + 2 = 2
) the stone was released from 2 m
above ground level.
c When h(t) = 27,
¡5t2 + 30t + 2 = 27
) ¡5t2 + 30t ¡ 25 = 0
) t2 ¡ 6t + 5 = 0 fdividing each term by ¡5g
) (t ¡ 1)(t ¡ 5) = 0 ffactorisingg
) t = 1 or 5
) the stone is 27 m above the ground after 1 second and after 5 seconds.
7 An object is projected into the air with a velocity of 80 m s¡1 . Its height after t seconds
is given by the function h(t) = 80t ¡ 5t2 metres.
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a Calculate the height after: i 1 second ii 3 seconds
b Calculate the time(s) at which the height is: i 140 m
c Explain your answers to part b.
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ii 0 m.
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QUADRATIC FUNCTIONS
369
(Chapter 12)
8 A cake manufacturer finds that the profit from making
x cakes per day is given by the function
P (x) = ¡ 12 x2 + 36x ¡ 40 dollars.
a Calculate the profit if: i 0 cakes ii 20 cakes
are made per day.
b How many cakes per day are made if the profit is
$270?
B
GRAPHS OF QUADRATIC FUNCTIONS
Consider the quadratic function f (x) = x2 .
y
The table below shows the values of f (x) corresponding
to x-values from ¡3 to 3.
x
¡3
¡2
¡1
0
1
2
3
f (x)
9
4
1
0
1
4
9
8
4
These points can be used to draw the graph of
f (x) = x2 .
y = x2
x
The shape formed is a parabola. The graphs of
quadratic functions all have this same basic shape.
-2
2
vertex
Notice that the curve y = x2 :
² opens upwards
² is symmetric about the y-axis since, for example,
f (¡2) = (¡2)2 = 4 and
f (2) = 22 = 4, so f (¡2) = f (2):
The vertex is the point
where the graph is at its
maximum or minimum.
² has a vertex or turning point at (0, 0).
Example 5
Self Tutor
Draw the graph of y = x2 ¡ 2x ¡ 5 from a table of values.
Consider f (x) = x2 ¡ 2x ¡ 5:
Now, f (¡3) = (¡3)2 ¡ 2(¡3) ¡ 5
=9+6¡5
= 10
y
12
8
We can do the same for the
other values of x.
4
The table of values is:
-4
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x ¡3 ¡2 ¡1 0
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y 10 3 ¡2 ¡5 ¡6 ¡5 ¡2
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370
QUADRATIC FUNCTIONS
(Chapter 12)
EXERCISE 12B.1
1 Construct a table of values for x = ¡3, ¡2, ¡1, 0, 1, 2, 3 for each of the following
functions. Hence, draw the graph of each.
a y = x2 + 2x ¡ 2
b y = x2 ¡ 3
c y = x2 ¡ 2x
2
2
d f (x) = ¡x + x + 2
e y = x ¡ 4x + 4
f f (x) = ¡2x2 + 3x + 10
2 Use the graphing package or your graphics calculator to check your
graphs in 1.
GRAPHING
PACKAGE
GRAPHS OF THE FORM y__=_ ax2
INVESTIGATION 1
What to do:
1 Use the graphing package or your graphics calculator
to graph each pair of functions on the same set of axes.
a y = x2 and y = 2x2
b y = x2 and y = 4x2
c y = x2 and y = 12 x2
d y = x2 and y = ¡x2
e y = x2 and y = ¡2x2
f y = x2 and y = ¡ 12 x2
GRAPHING
PACKAGE
2 These functions are all members of the family y = ax2 .
What effect does a have on:
a the direction in which the graph opens
b the shape of the graph?
From the investigation we make the following observations:
If a > 0, y = ax2 opens upwards. It has the shape
If a < 0, y = ax2 opens downwards. It has the shape
If a < ¡1 or a > 1, y = ax2 is ‘thinner’ than y = x2 .
If ¡1 < a < 1, a 6= 0, y = ax2 is ‘wider’ than y = x2 .
Example 6
Self Tutor
Sketch y = x2 on a set of axes and hence sketch:
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b y = ¡3x2
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a y = 3x2
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QUADRATIC FUNCTIONS
a y = 3x2 is ‘thinner’ than y = x2 .
(Chapter 12)
371
y
y = x2
b y = ¡3x2 has the same shape as
y = 3x2 but opens downwards.
y = 3x2
x
y = -3x2
EXERCISE 12B.2
1 On separate sets of axes, sketch y = x2 and each of the following. In each case
comment on the direction in which the graph opens and the shape of the graph.
a y = 5x2
b y = ¡5x2
c y = 13 x2
d y = ¡ 13 x2
f y = 14 x2
e y = ¡4x2
2 Use your graphics calculator or graphing package to check your graphs
in 1.
C
GRAPHING
PACKAGE
AXES INTERCEPTS
The axes intercepts are an important property of quadratic functions. Knowing the axes
intercepts helps us to graph the function.
y
An x-intercept is a value of x where the graph
meets the x-axis. It is found by letting y = 0.
x-intercepts
A y-intercept is a value of y where the graph meets
the y-axis. It is found by letting x = 0.
x
y-intercept
INVESTIGATION 2
THE y-INTERCEPT OF A QUADRATIC
What to do:
1 Use the graphing package or your graphics calculator
to find the y-intercept of:
b y = 2x2 + x + 4
a y = x2 + 3x ¡ 5
d y = x2 ¡ 6x
c y = x2 ¡ 8
2
e y = 3x + 7x ¡ 11
f y = 2x2 ¡ 5x + 17
GRAPHING
PACKAGE
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2 Use your observations in 1 to copy and complete:
“The y-intercept of y = ax2 + bx + c is .... ”.
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372
QUADRATIC FUNCTIONS
(Chapter 12)
You should have found that the y-intercept of y = ax2 + bx + c is the constant term c.
We can see this since, letting x = 0, y = a(0)2 + b(0) + c
=0+0+c
=c
EXERCISE 12C.1
1 State the y-intercept for the following functions:
a y = x2 + 2x + 3
b y = 2x2 + 5x ¡ 1
c y = ¡x2 ¡ 3x ¡ 4
d f (x) = 3x2 ¡ 10x + 1
e y = 3x2 + 5
f y = 4x2 ¡ x
g y = 8 ¡ x ¡ 2x2
h f(x) = 2x ¡ x2 ¡ 5
i y = 6x2 + 2 ¡ 5x
2 Find the y-intercept for the following functions:
a y = (x + 1)(x + 3)
b y = (x ¡ 2)(x + 3)
c y = (x ¡ 7)2
d y = (2x + 5)(3 ¡ x)
e y = x(x ¡ 4)
f y = ¡(x + 4)(x ¡ 5)
THE x-INTERCEPTS
When a quadratic function is given in the form
y = a(x ¡ ®)(x ¡ ¯), we can find the x-intercepts by
letting y = 0 and using the Null Factor law:
0 = a(x ¡ ®)(x ¡ ¯)
) x ¡ ® = 0 or x ¡ ¯ = 0
) x = ® or
x=¯
The x-intercepts are
also called the zeros of
the function.
where a 6= 0
fNull Factor lawg
The x-intercepts of y = a(x ¡ ®)(x ¡ ¯) are ® and ¯ for a 6= 0.
Example 7
Self Tutor
Find the x-intercepts of:
a y = 3(x ¡ 4)(x + 2)
b y = (x ¡ 5)2
a We let y = 0
) 3(x ¡ 4)(x + 2) = 0
) x = 4 or x = ¡2
So, the x-intercepts are 4 and ¡2.
b We let y = 0
) (x ¡ 5)2 = 0
) x=5
So, the only x-intercept is 5.
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The x-intercepts are
easy to find when the
quadratic is in
factorised form.
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QUADRATIC FUNCTIONS
(Chapter 12)
373
EXERCISE 12C.2
1 Find the x-intercepts of the following functions:
a y = (x ¡ 2)(x ¡ 5)
b y = (x ¡ 3)(x + 4)
c y = 2(x + 6)(x + 3)
d y = ¡(x ¡ 7)(x + 1)
e y = x(x ¡ 8)
f y = ¡3(x + 5)(x ¡ 5)
g y = (x + 4)2
h y = 7(x ¡ 2)2
i y = ¡4(x + 1)2
FINDING x-INTERCEPTS
For quadratic functions written in the expanded form y = ax2 + bx + c, finding the
x-intercepts is not so easy.
The x-intercepts are the zeros of the quadratic function, and we find them by solving the
quadratic equation ax2 + bx + c = 0.
In Chapter 9 we saw that the quadratic equation can have two, one, or zero solutions. This
means that quadratic functions can have two, one, or zero x-intercepts.
INVESTIGATION 3
THE x-INTERCEPTS OF A QUADRATIC
In this investigation you can use the graphing package or
your graphics calculator to graph the quadratic functions.
GRAPHING
PACKAGE
What to do:
1 Use technology to graph each quadratic function. Sketch each graph in your notebook
and write down the x-intercepts.
a y = x2 ¡ x ¡ 12
b y = x2 ¡ 6x + 9
c y = 3x2 + 6x + 5
d y = ¡x2 + 5x ¡ 6
e y = ¡2x2 ¡ 4x ¡ 2
f y = ¡x2 + 4x ¡ 10
2 How many zeros has the quadratic function which:
a cuts the x-axis twice
c lies entirely below the x-axis?
b touches the x-axis
3 Use technology to solve graphically:
a x2 + 4x + 2 = 0
d x2 + 2x + 5 = 0
b x2 + 6x ¡ 2 = 0
e ¡x2 ¡ 4x ¡ 6 = 0
c 2x2 ¡ 3x ¡ 7 = 0
f 4x2 ¡ 4x + 1 = 0
When you are given a quadratic function in expanded form, you should first look to see if it
can be factorised. If it can, then you should do this to find the x-intercepts.
However, we saw in Chapter 9 that many quadratic equations are not easily factorised, even
those which have two solutions. To find the zeros or x-intercepts in these cases we can use
technology to assist us.
Example 8
Self Tutor
Find the x-intercepts of these quadratic functions:
a y = x2 ¡ 2x ¡ 15
b y = x2 + 2x + 1
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Check your answers using technology.
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QUADRATIC FUNCTIONS
(Chapter 12)
b We let y = 0
) x2 + 2x + 1 = 0
) (x + 1)2 = 0
) x = ¡1
So, the only x-intercept is ¡1.
a We let y = 0
) x2 ¡ 2x ¡ 15 = 0
) (x + 3)(x ¡ 5) = 0
) x = ¡3 or 5
So, the x-intercepts are ¡3 and 5.
EXERCISE 12C.3
If you cannot factorise the
function, use technology
to graph the function and
find any x-intercepts.
1 Find the x-intercepts of the following functions:
a y = x2 ¡ x ¡ 6
b y = x2 ¡ 16
c y = x2 + 5
d y = 3x ¡ x2
e y = x2 ¡ 12x + 36
f y = x2 + x ¡ 7
g y = ¡x2 ¡ 4x + 21
h y = 2x2 ¡ 20x + 50
i y = 2x2 ¡ 7x ¡ 15
j y = ¡2x2 + x ¡ 5
k y = ¡6x2 + x + 5
l y = 3x2 + x ¡ 1
2 Find the axes intercepts of the following functions:
a y = x2 + x ¡ 2
b y = (x + 3)2
2
c y = (x + 5)(x ¡ 2)
2
d y =x +x+4
e y = 3x ¡ 3x ¡ 36
f y = ¡x2 ¡ 8x ¡ 16
g y = x2 ¡ 7x
h y = ¡2x2 + 3x + 7
i y = 2x2 ¡ 18
j y = ¡x2 + 2x ¡ 9
k y = 4x2 ¡ 4x ¡ 3
l y = 6x2 ¡ 11x ¡ 10
GRAPHS FROM AXES INTERCEPTS
If we know the axes intercepts of a quadratic function, we can use them to draw its graph.
Example 9
Self Tutor
Sketch the graphs of the following functions by considering:
i the value of a
ii the y-intercept
2
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b y = ¡2(x + 1)(x ¡ 2)
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a y = x ¡ 2x ¡ 3
iii the x-intercepts.
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QUADRATIC FUNCTIONS
a y = x2 ¡ 2x ¡ 3
i Since a = 1 which is > 0,
the parabola has shape
375
(Chapter 12)
b y = ¡2(x + 1)(x ¡ 2)
i Since a = ¡2 which is < 0,
the parabola has shape
ii The y-intercept occurs when
x = 0, so the y-intercept is ¡3.
iii x-intercepts occur when y = 0
) x2 ¡ 2x ¡ 3 = 0
) (x ¡ 3)(x + 1) = 0
) x = 3 or x = ¡1
) the x-intercepts are 3 and ¡1.
ii The y-intercept occurs when x = 0
) y = ¡2(0 + 1)(0 ¡ 2)
= ¡2 £ 1 £ ¡2
=4
) the y-intercept is 4
iii x-intercepts occur when y = 0
) ¡2(x + 1)(x ¡ 2) = 0
) x = ¡1 or x = 2
) the x-intercepts are ¡1 and 2.
Sketch:
Sketch:
y
y
4
x
-1
-1
x
2
3
-3
Example 10
Self Tutor
Sketch the graph of f (x) = 2(x ¡ 3)2 by considering:
a the value of a
b the y-intercept
c the x-intercepts.
f (x) = 2(x ¡ 3)2
a Since a = 2 which is > 0, the parabola has shape
b The y-intercept occurs when x = 0.
Now f(0) = 2(0 ¡ 3)2 = 18
) the y-intercept is 18.
c x-intercepts occur when f (x) = 0
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) 2(x ¡ 3)2 = 0
) x=3
) the x-intercept is 3.
There is only one x-intercept, which means
the graph touches the x-axis.
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IB_STSL-2ed
376
QUADRATIC FUNCTIONS
(Chapter 12)
If a quadratic function
has only one x-intercept
then its graph must
touch the x-axis.
EXERCISE 12C.4
1 Sketch the graph of the quadratic function with:
a x-intercepts ¡1 and 1, and y-intercept ¡1
b x-intercepts ¡3 and 1, and y-intercept 2
c x-intercepts 2 and 5, and y-intercept ¡4
d x-intercept 2 and y-intercept 4.
2 Sketch the graphs of the following by considering:
i the value of a
ii the y-intercept
2
iii the x-intercepts.
a y = x ¡ 4x + 4
b f (x) = (x ¡ 1)(x + 3) c y = 2(x + 2)2
d y = ¡(x ¡ 2)(x + 1)
e y = ¡3(x + 1)2
f f (x) = ¡3(x ¡ 4)(x ¡ 1)
2
g y = 2(x + 3)(x + 1)
i y = ¡2x2 ¡ 3x + 5
h f (x) = 2x + 3x + 2
D
AXIS OF SYMMETRY
y
The graph of any quadratic function is symmetric
about a vertical line called the axis of symmetry.
Since the axis of symmetry is vertical, its equation
will have the form x = k.
If a quadratic function has two x-intercepts, then the
axis of symmetry lies halfway between them.
x
axis of symmetry
Example 11
Self Tutor
Find the equation of the axis of symmetry for the following:
y
a
y
b
2
x
1
5
x
a The x-intercepts are 1 and 5, and 3
is halfway between 1 and 5.
So, the axis of symmetry has
equation x = 3.
b The only x-intercept is 2, so the axis
of symmetry has equation x = 2.
y
y
2
x
1
3
5
x
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\376IB_STSL-2_12.CDR Monday, 8 February 2010 12:37:43 PM PETER
IB_STSL-2ed
QUADRATIC FUNCTIONS
377
(Chapter 12)
EXERCISE 12D.1
1 For each of the following, find the equation of the axis of symmetry:
a
b
c
y
y
y
x
x
-5
6
x
3
-1
-2
d
e
y
f
y
x
-7
-1
y
x
3
2
-3
x
-4
2 Find the equation of the axis of symmetry for the following functions:
a y = (x ¡ 2)(x ¡ 6)
b y = x(x + 4)
c y = ¡(x + 3)(x ¡ 5)
d y = (x ¡ 3)(x ¡ 8)
e y = 2(x ¡ 5)2
f y = ¡3(x + 2)2
AXIS OF SYMMETRY OF y = ax2 + bx + c
We are often given quadratic functions in expanded form
such as y = 2x2 + 9x + 4, where we cannot easily identify
the x-intercepts.
Other quadratic functions do not have any x-intercepts, such
as the one illustrated alongside.
y
x
So, we need a method for finding the axis of symmetry of
a function without using x-intercepts.
f(x) = ax2 + bx + c
y
d
A
d
h
h_-_d
Consider the quadratic function f(x) = ax2 + bx + c
alongside, with axis of symmetry x = h.
Suppose A and B are two points on f (x), d units
either side of the axis of symmetry.
B
Thus, A and B have x-coordinates h ¡ d and h + d
respectively.
Since the function is symmetric about x = h, A and
B will have the same y-coordinate.
x
h_+_d
x=h
So, f (h ¡ d) = f (h + d)
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) a(h ¡ d) + b(h ¡ d) + c = a(h + d)2 + b(h + d) + c
5
2
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\377IB_STSL-2_12.CDR Monday, 8 February 2010 12:37:46 PM PETER
IB_STSL-2ed
378
QUADRATIC FUNCTIONS
(Chapter 12)
) a(h2 ¡ 2hd + d2 ) + bh ¡ bd = a(h2 + 2hd + d2 ) + bh + bd
2
) ¡4ahd = 2bd
¡b
) h=
2a
¡b
So, the axis of symmetry has equation x =
:
2a
The equation of the axis of symmetry of y = ax2 + bx + c is x =
Example 12
¡b
.
2a
Self Tutor
Find the equation of the axis of symmetry of y = 3x2 + 4x ¡ 5:
y = 3x2 + 4x ¡ 5 has a = 3, b = 4, c = ¡5.
¡b
¡4
=
= ¡ 23
2a
2£3
Now
So, the axis of symmetry has equation x = ¡ 23 .
EXERCISE 12D.2
1 Determine the equation of the axis of symmetry of:
a y = x2 + 6x + 2
b y = x2 ¡ 8x ¡ 1
c f (x) = 2x2 + 5x ¡ 3
d y = ¡x2 + 3x ¡ 7
e y = 2x2 ¡ 5
f y = ¡5x2 + 7x
g f (x) = x2 ¡ 6x + 9
h y = 10x ¡ 3x2
i y = 18 x2 + x ¡ 1
E
VERTEX
The vertex or turning point of a parabola is the point at which the function has a
maximum value for a < 0
, or a minimum value for a > 0
The vertex of a quadratic function always lies on the
axis of symmetry.
.
y
So, the axis of symmetry gives us the x-coordinate of
the vertex.
The y-coordinate can be found by substituting this
value of x into the function.
x
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vertex
axis of symmetry
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\378IB_STSL-2_12.CDR Monday, 8 February 2010 12:37:49 PM PETER
IB_STSL-2ed
QUADRATIC FUNCTIONS
Example 13
(Chapter 12)
379
Self Tutor
The vertex is either a
maximum or minimum
turning point.
Determine the coordinates of the vertex of
f (x) = x2 + 6x + 4.
f (x) = x2 + 6x + 4 has a = 1, b = 6, c = 4.
Now
¡b
¡6
=
= ¡3
2a
2£1
So, the axis of symmetry has equation x = ¡3.
f (¡3) = (¡3)2 + 6(¡3) + 4
= 9 ¡ 18 + 4
= ¡5
So, the vertex is (¡3, ¡5):
EXERCISE 12E
1 Find the vertex of each of the following quadratic functions:
a y = x2 ¡ 4x + 7
b y = x2 + 2x + 5
c f (x) = ¡x2 + 6x ¡ 1
d y = x2 + 3
e f (x) = 2x2 + 12x
f y = ¡3x2 + 6x ¡ 4
g y = x2 ¡ x ¡ 1
h y = ¡2x2 + 3x ¡ 2
i y = ¡ 14 x2 + 3x ¡ 2
2 For each of the functions in 1, state whether the vertex is a maximum or a minimum
turning point.
Example 14
Self Tutor
Consider the quadratic function y = ¡x2 ¡ 4x + 5.
a Find the axes intercepts.
b Find the equation of the axis of symmetry.
c Find the coordinates of the vertex.
d Sketch the function, showing all important features.
b a = ¡1, b = ¡4, c = 5
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4
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2a
¡2
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So, the axis of symmetry is x = ¡2.
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a When x = 0, y = 5
So, the y-intercept is 5.
When y = 0,
¡x2 ¡ 4x + 5 = 0
) x2 + 4x ¡ 5 = 0
) (x + 5)(x ¡ 1) = 0
) x = ¡5 or 1
So, the x-intercepts are ¡5 and 1.
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\379IB_STSL-2_12.CDR Monday, 8 February 2010 12:37:51 PM PETER
IB_STSL-2ed
380
QUADRATIC FUNCTIONS
(Chapter 12)
c When x = ¡2,
y = ¡(¡2)2 ¡ 4(¡2) + 5
= ¡4 + 8 + 5
=9
So, the vertex is (¡2, 9).
d
vertex (-2, 9)
y
5
-5
1
x
x = -2
y = -x2 - 4x + 5
3 For each of the following functions:
i find the axes intercepts
ii find the equation of the axis of symmetry
iii find the coordinates of the vertex
iv sketch the function.
a y = x2 ¡ 4x + 3
b y = ¡x2 + 4x + 12
c f (x) = x2 + 2x
d y = (x + 5)(x ¡ 3)
e y = x2 ¡ 10x + 25
f f (x) = ¡4x2 ¡ 8x ¡ 3
g y = x2 ¡ 8x
h y = 2x2 ¡ 2x ¡ 12
i f : x 7! ¡(x + 4)2
j y = 9x2 ¡ 1
k y = 3x2 ¡ 4x + 1
l f : x 7! ¡ 12 x2 ¡ x + 12
F
FINDING A QUADRATIC
FROM ITS GRAPH
So far in this chapter we have learned how to draw a graph given an equation. Sometimes
we wish to reverse this process.
If we are given sufficient information on or about a graph we can determine the quadratic
equation that has that graph.
Example 15
Self Tutor
Find the equation of the quadratic with graph:
y
a
y
b
8
3
3
-1
x
2
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b Since it touches the x-axis at 2,
y = a(x ¡ 2)2 , a > 0.
But when x = 0, y = 8
) 8 = a(¡2)2
) a=2
So, y = 2(x ¡ 2)2 :
a Since the x-intercepts are ¡1 and 3,
y = a(x + 1)(x ¡ 3), a < 0.
But when x = 0, y = 3
) 3 = a(1)(¡3)
) a = ¡1
So, y = ¡(x + 1)(x ¡ 3):
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\380IB_STSL-2_12.CDR Monday, 8 February 2010 12:37:54 PM PETER
IB_STSL-2ed
QUADRATIC FUNCTIONS
Example 16
381
(Chapter 12)
Self Tutor
y
16
Find the equation of the quadratic
with graph:
-2
x
x=1
The axis of symmetry is x = 1, so the other x-intercept is 4
) y = a(x + 2)(x ¡ 4)
But when x = 0, y = 16
) 16 = a(2)(¡4)
) a = ¡2
) the quadratic is y = ¡2(x + 2)(x ¡ 4).
Example 17
Self Tutor
Find, in the form y = ax2 + bx + c, the equation of the quadratic whose graph cuts
the x-axis at 4 and ¡3 and passes through the point (2, ¡20).
Since the x-intercepts are 4 and ¡3, the equation is
y = a(x ¡ 4)(x + 3) where a 6= 0.
But when x = 2, y = ¡20
)
)
¡20 = a(2 ¡ 4)(2 + 3)
¡20 = a(¡2)(5)
) a=2
) the equation is y = 2(x ¡ 4)(x + 3) or y = 2x2 ¡ 2x ¡ 24.
EXERCISE 12F
1 Find the equation of the quadratic with graph:
y
a
y
b
8
4
1
d
3
x
2
x
2
e
y
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3
x
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x
-3
3
x
3
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1
-1
1
f
y
3
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c
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\381IB_STSL-2_12.CDR Monday, 8 February 2010 12:37:57 PM PETER
IB_STSL-2ed
382
QUADRATIC FUNCTIONS
(Chapter 12)
2 Find the quadratic with graph:
a
b
y
c
y
y
4
12
x
-4
-12
x
2
x=3
d
x
x = -3
x = -1
e
y
f
y
y
V(0, 9)
3
x
-3
x
1
x
V(2, -4)
V(1, -4)
3 Find, in the form y = ax2 + bx + c, the equation of the quadratic whose graph:
a cuts the x-axis at 5 and 1, and passes through (2, ¡9)
b cuts the x-axis at 2 and ¡ 12 , and passes through (3, ¡14)
c touches the x-axis at 3 and passes through (¡2, ¡25)
d touches the x-axis at ¡2 and passes through (¡1, 4)
e cuts the x-axis at 3, passes through (5, 12), and has axis of symmetry x = 2
f cuts the x-axis at 5, passes through (2, 5), and has axis of symmetry x = 1.
G
WHERE FUNCTIONS MEET
Consider the graphs of a quadratic function and a linear function on the same set of axes.
We could have:
cutting
2 points of intersection
touching
1 point of intersection
missing
no points of intersection
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If the graphs meet, the coordinates of the points of intersection of the graph of the two
functions can be found by solving the two equations simultaneously.
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\382IB_STSL-2_12.CDR Thursday, 25 February 2010 12:46:07 PM PETER
IB_STSL-2ed
QUADRATIC FUNCTIONS
(Chapter 12)
Example 18
383
Self Tutor
Find the coordinates of the points of intersection of the graphs with equations
y = x2 ¡ x ¡ 18 and y = x ¡ 3:
y = x2 ¡ x ¡ 18 meets y = x ¡ 3 where
x2 ¡ x ¡ 18 = x ¡ 3
) x2 ¡ 2x ¡ 15 = 0
fRHS = 0g
) (x ¡ 5)(x + 3) = 0
ffactorisingg
) x = 5 or ¡3
Substituting into y = x ¡ 3, when x = 5, y = 2 and when x = ¡3, y = ¡6.
) the graphs meet at (5, 2) and (¡3, ¡6).
EXERCISE 12G
1 Find the coordinates of the point(s) of intersection of the graphs with equations:
a y = x2 + x ¡ 1 and y = 2x ¡ 1
b y = ¡x2 + 2x + 3 and y = 2x ¡ 1
c y = x2 ¡ 2x + 8 and y = x + 6
d y = ¡x2 + 3x + 9 and y = 2x ¡ 3
e y = x2 ¡ 4x + 3 and y = 2x ¡ 6
f y = ¡x2 + 4x ¡ 7 and y = 5x ¡ 4
2 Use a graphing package or a graphics calculator to find the coordinates of the points
of intersection, to 3 significant figures, of the graphs with equations:
a y = x2 ¡ 3x + 7 and y = x + 5
b y = x2 ¡ 5x + 2 and y = x ¡ 7
c y = ¡x2 ¡ 2x + 4 and y = x + 8
GRAPHING
PACKAGE
d y = ¡x2 + 4x ¡ 2 and y = 5x ¡ 6
e y = x2 + 5x ¡ 4 and y = ¡ 12 x ¡ 3
f y = ¡x2 + 7x + 1 and y = ¡3x + 2
g y = 3x2 ¡ x ¡ 2 and y = 2x ¡
H
11
4
QUADRATIC MODELLING
There are many situations in the real world where the relationship between two variables is
a quadratic function.
The graph of such relationships will have shape
have a minimum or maximum value.
or
and the function will
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For the quadratic function y = ax2 + bx + c, we have already seen that the vertex or turning
point lies on the axis of symmetry.
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\383IB_STSL-2_12.CDR Monday, 8 February 2010 12:38:02 PM PETER
IB_STSL-2ed
384
QUADRATIC FUNCTIONS
² If a > 0,
the minimum
value of y occurs
b
at x = ¡ .
2a
(Chapter 12)
x=¡
² If a < 0,
the maximum
value of y occurs
b
at x = ¡ .
2a
b
2a
min.
max.
x=¡
b
2a
The process of finding the maximum or minimum value of a function is called optimisation.
Example 19
Self Tutor
When a baseballer hits the ball straight up, its height above the ground t seconds after
being hit is H(t) = 30t ¡ 5t2 metres, t > 0.
a How long does it take for the ball to reach its maximum height?
b What is the maximum height reached by the ball?
c How long does it take for the ball to hit the ground?
a H(t) = 30t ¡ 5t2 has a = ¡5 which is < 0, so its shape is
.
¡b
¡30
=
=3
2a
2 £ (¡5)
So, the maximum height is reached after 3 seconds.
The maximum height occurs when t =
b H(3) = 30(3) ¡ 5(3)2
= 90 ¡ 45
= 45
So, the maximum height reached is 45 metres.
H (m)
45
c The ball hits the ground when H(t) = 0
) 30t ¡ 5t2 = 0
) 5t2 ¡ 30t = 0
) 5t(t ¡ 6) = 0 ffactorisingg
) t = 0 or 6
So, the ball hits the ground after 6 seconds.
3
t (s)
EXERCISE 12H
1 Andrew dives off a jetty. His height above the water t seconds after diving is given by
H(t) = ¡4t2 + 4t + 3 metres, t > 0.
a
b
c
d
How
How
How
How
high above the water is the jetty?
long does it take for Andrew to reach the maximum height of his dive?
far is Andrew above the water at his highest point?
long does it take for Andrew to hit the water?
2 Jasmine makes necklaces for her market stall. Her daily profit from making x necklaces
is given by P (x) = ¡x2 + 20x dollars.
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a How many necklaces should Jasmine make per day to maximise her profit?
b Find the maximum daily profit that Jasmine can make.
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\384IB_STSL-2_12.CDR Monday, 8 February 2010 12:38:05 PM PETER
IB_STSL-2ed
QUADRATIC FUNCTIONS
(Chapter 12)
385
3 The average cost of making x televisions per week
is C(x) = x2 ¡ 40x + 500 dollars per television.
a How many televisions should be made per week
to minimise the average cost?
b Find the minimum average cost.
c How many televisions are made per week if the
average cost is $200 per television?
4 Answer the questions in the Opening Problem on page 366.
Example 20
Self Tutor
A farmer has 20 metres of fencing to create a
rectangular enclosure next to his barn.
xm
The two equal sides of the enclosure are x metres
long.
a Show that the area of the enclosure is given
by A = x(20 ¡ 2x) m2 .
barn
b Find the dimensions of the enclosure of
maximum area.
a The side XY has length y = 20 ¡ 2x m
The area of the enclosure = length £ width
) A = x(20 ¡ 2x) m2
2
xm
X
2
b A = 20x ¡ 2x = ¡2x + 20x
which has a = ¡2, b = 20
barn
Since a < 0, the shape is
.
Z
So, the maximum area occurs when
¡20
¡b
=
=5m
x=
2a
¡4
The area is maximised when YZ = 5 m and XY = 20 ¡ 2(5) = 10 m.
Y
5 A rectangular plot is enclosed by 200 m of fencing and
has an area of A square metres. Show that:
a A = 100x ¡ x2 where x m is the length of one of
its sides
b the area is maximised when the rectangle is a square.
xm
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6 A rectangular paddock to enclose horses is to be made, with one side being a straight
water drain. If 1000 m of fencing is available for the other 3 sides, what dimensions
should be used for the paddock so that it encloses the maximum possible area?
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\385IB_STSL-2_12.CDR Tuesday, 9 February 2010 9:56:17 AM PETER
IB_STSL-2ed
386
QUADRATIC FUNCTIONS
(Chapter 12)
ym
7 1800 m of fencing is available to fence six
identical pens as shown in the diagram.
a Explain why 9x + 8y = 1800.
b Show that the total area of each pen is
given by A = ¡ 98 x2 + 225x m2 .
xm
c If the area enclosed is to be maximised,
what is the shape of each pen?
8 500 m of fencing is available
to make 4 rectangular pens of
identical shape.
Find the dimensions that
maximise the area of each
pen if the plan is:
a
b
Example 21
Self Tutor
A manufacturer of pot-belly stoves has the following situation to consider.
¶
µ
400
dollars and the total income
If x are made per week, each one will cost 50 +
x
per week for selling them will be (550x ¡ 2x2 ) dollars.
How many pot-belly stoves should be made per week to maximise the profit?
The total profit
P = income ¡ costs
µ
¶
400
x
) P = (550x ¡ 2x2 ) ¡ 50 +
x
cost for one
x of them
) P = 550x ¡ 2x2 ¡ 50x ¡ 400
) P = ¡2x2 + 500x ¡ 400 dollars
This is a quadratic in x, with a = ¡2, b = 500, c = ¡400.
Since a < 0, the shape is
P is maximised when x =
¡b
¡500
=
= 125
2a
¡4
) 125 stoves should be made per week to maximise the profit.
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1 2
9 The total cost of producing x toasters per day is given by C = ( 10
x + 20x + 25) euros,
1
and the selling price of each toaster is (44 ¡ 5 x) euros. How many toasters should be
produced each day in order to maximise the total profit?
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\386IB_STSL-2_12.CDR Monday, 8 February 2010 12:38:10 PM PETER
IB_STSL-2ed
QUADRATIC FUNCTIONS
387
(Chapter 12)
10 A manufacturer of barbecues knows that if x of
them
are made
µ
¶ each week then each one will cost
800
60 +
pounds and the total income per
x
week will be (1000x ¡ 3x2 ) pounds.
How many barbecues should be made per week
for maximum profits?
INVESTIGATION 4
TUNNELS AND TRUCKS
A tunnel is parabolic in shape with the
dimensions shown.
A truck carrying a wide load is 4:8 m high
and 3:9 m wide and needs to pass through
the tunnel. Your task is to determine if the
truck will fit through the tunnel.
8m
6m
What to do:
1 If a set of axes is fitted to the parabolic tunnel as
shown, state the coordinates of points A, B and C.
A
y
2 On your graphics calculator:
a enter the x-coordinates of A, B and C into List 1
b enter the y-coordinates of A, B and C into List 2.
3 Draw a scatterplot of points A, B and C.
4 Set your calculator to display 4 decimal places. By
fitting a quadratic model to the data, determine the
equation of the parabolic boundary of the tunnel in the
form y = ax2 + bx + c.
x
C
B
5 What is the equation of the truck’s roofline?
6 Graph the equations from 4 and 5 on the same set of axes. Calculate the points of
intersection of the graphs of these functions.
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7 Using the points of intersection found in 6, will the truck pass through the tunnel?
What is the maximum width of a truck that is 4:8 m high if it is to pass through the
tunnel?
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\387IB_STSL-2_12.cdr Tuesday, 9 February 2010 9:56:55 AM PETER
IB_STSL-2ed
388
QUADRATIC FUNCTIONS
(Chapter 12)
REVIEW SET 12A
1 If f (x) = x2 ¡ 3x ¡ 15 find:
a f (0)
c x such that f (x) = 3.
b f(1)
2 On the same set of axes, sketch y = x2 and the function:
b y = ¡ 12 x2
a y = 3x2
3 Consider the function y = ¡2(x ¡ 1)(x + 3).
a Find the:
i direction the parabola opens
ii y-intercept
iii x-intercepts
iv equation of the axis of symmetry.
b Sketch a graph of the function showing all of the above features.
4 Consider the function y = x2 ¡ 2x ¡ 15.
a Find the:
i y-intercept
ii x-intercepts
iii equation of the axis of symmetry iv coordinates of the vertex.
b Sketch a graph of the function showing all of the above features.
5 Find the x-intercepts of the following functions:
a y = 3x2 ¡ 12x
b y = 3x2 ¡ x ¡ 10
c f (x) = x2 ¡ 11x ¡ 60
6 Match the function with the correct graph:
y
a y = x2
A
B
b y = ¡ 12 x2
c y = 3x2
d y = ¡2x2
x
D
C
7 Find, in the form y = ax2 + bx + c, the equation of the quadratic whose graph:
a touches the x-axis at 4 and passes through (2, 12)
b has x-intercepts 3 and ¡2 and y-intercept 3.
8 Find the maximum or minimum value of the relation y = ¡2x2 + 4x + 3 and the
value of x for which the maximum or minimum occurs.
9 A stone was thrown from the top of a cliff 60 metres above sea level. The height of
the stone above sea level t seconds after it was released is given by
H(t) = ¡5t2 + 20t + 60 metres.
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a Find the time taken for the stone to reach its maximum height.
b What was the maximum height above sea level reached by the stone?
c How long did it take before the stone struck the water?
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IB_STSL-2ed
QUADRATIC FUNCTIONS
389
(Chapter 12)
REVIEW SET 12B
a Is f (x) = ¡2x2 + 13x ¡ 4 satisfied by the ordered pair (1, 5)?
1
b Find the value(s) of x for which g(x) = x2 ¡ 5x ¡ 9 takes the value 5.
2 Consider the function y = 3(x ¡ 2)2 .
a Find the:
i direction the parabola opens
ii y-intercept
iii x-intercepts
iv equation of the axis of symmetry.
b Sketch a graph of the function showing all of the above features.
3 Consider the function y = ¡x2 + 7x ¡ 10.
a Find the:
i y-intercept
ii x-intercepts
iii equation of the axis of symmetry iv coordinates of the vertex.
b Sketch a graph of the function showing all of the above features.
4 Find the equation of the axis of symmetry and the vertex of y = ¡3x2 + 8x + 7.
5 Find the equation of the quadratic relation with graph:
a
b
c
y
y
y
7
5
x
x
-2
2
(2, -20)
x
-3
x=4
6 Find the coordinates of the point(s) of intersection of the graphs with equations
y = x2 + 13x + 15 and y = 2x ¡ 3.
y
7 Determine the values of a and b, given the
diagram:
10
(b, 10)
1
(a, 0)
x
(3, -8)
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8 A farmer has 2000 m of fencing to enclose two identical adjacent fields.
a Find an expression for the total
area of the fields in terms of x.
b What is the maximum possible
xm
total area of the two fields?
c What dimensions give this
maximum total area?
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\389IB_STSL-2_12.cdr Tuesday, 9 February 2010 9:57:36 AM PETER
IB_STSL-2ed
390
QUADRATIC FUNCTIONS
(Chapter 12)
REVIEW SET 12C
1 If g(x) = x2 + 5x ¡ 2, find:
a g(¡1)
c x if g(x) = ¡6
b g(4)
2 Consider the quadratic function y = 2x2 + 6x ¡ 3.
a Find the coordinates of the vertex.
c Sketch the graph of the function.
b Find the y-intercept.
d Use technology to check your answers.
3 Draw the graph of y = ¡x2 + 2x.
4 Find the equation of the quadratic relation with graph:
y
a
y
b
(2.5, 24.5)
x
3
-3
x
-1
-27
5 Use your calculator to find the coordinates of the point(s) of intersection of the graphs
with equations y = 2x ¡ 9 and y = 2x2 ¡ 3x ¡ 8.
6 Find the vertex of y = 2x2 ¡ 6x + 3.
7 A retailer sells sunglasses for $15, and has 50 customers per day. From market
research, the retailer discovers that for every dollar increase in the price of the
sunglasses, he will lose 2 customers per day.
a Given that revenue = (price) £ (number of customers), show that the revenue
collected by the retailer each day is R = (15 + x)(50 ¡ 2x), where x is the
price increase of the sunglasses in dollars.
b Hence find the price the retailer should set for his sunglasses in order to maximise
daily revenue.
c How much revenue is made per day at this price?
8 Find an expression for the quadratic which cuts the x-axis at 3 and ¡2 and has
y-intercept 24. Give your answer in the form y = ax2 + bx + c.
9
Find:
a the axis of symmetry
b the second x-intercept.
y
x
-9
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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_12\390IB_STSL-2_12.cdr Friday, 12 March 2010 4:52:02 PM PETER
IB_STSL-2ed