notes on quadratics functions with solutions.pdf

1 QUADRATICS
A quadratic function has the form y = ax2
+ bx + c where a 6= 0.
The simplest quadratic function is y = x2
. Its graph can
be drawn from a table of values.
x ¡3 ¡2 ¡1 0 1 2 3
y 9 4 1 0 1 4 9
The graph of a quadratic function is called a parabola.
The parabola is one of the conic sections, the others being circles, hyperbolae,
and ellipses. They are called conic sections because they can be obtained by
cutting a cone with a plane. A parabola is produced by cutting the cone with
a plane parallel to its slant side.
There are many examples of parabolas in everyday life, including water fountains, suspension bridges,
and radio telescopes.
TERMINOLOGY
The graph of a quadratic function y = ax2
+ bx + c,
a 6= 0 is called a parabola.
The point where the graph ‘turns’ is called the vertex.
The vertical line that passes through the vertex is called
the axis of symmetry. Every parabola is symmetrical
about its axis of symmetry.
The point where the graph crosses the y-axis is the
y-intercept.
The points (if they exist) where the graph crosses the x-axis are called the x-intercepts. They correspond
to the roots of the equation y = 0.
QUADRATIC FUNCTIONS
y
2
-2
x
2
4
6
8
y = x2
x
y
vertex
axis
of
symmetry
zero
y-intercept
parabola
minimum
zero
If the graph opens upwards, the y-coordinate of the
vertex is the minimum or minimum turning point and
the graph is concave upwards.
If the graph opens downwards, the y-coordinate of the
vertex is the maximum or maximum turning point and
the graph is concave downwards.
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Y:HAESEIB_SL-3edIB_SL-3ed_01028IB_SL-3ed_01.cdr Monday, 20 February 2012 4:45:35 PM BEN

GRAPHING y = a(x ¡ h)2
+ k
INVESTIGATION 2
GRAPHING y = a(x ¡ p)(x ¡ q)
INVESTIGATION 1
QUADRATICS 2
This investigation is best done using a graphing package or graphics calculator.
What to do:
1 a Use technology to help you to sketch:
y = (x ¡ 1)(x ¡ 3), y = 2(x ¡ 1)(x ¡ 3), y = ¡(x ¡ 1)(x ¡ 3),
y = ¡3(x ¡ 1)(x ¡ 3) and y = ¡1
2 (x ¡ 1)(x ¡ 3)
b Find the x-intercepts for each function in a.
c What is the geometrical significance of a in y = a(x ¡ 1)(x ¡ 3)?
y = 2(x ¡ 1)(x ¡ 4), y = 2(x ¡ 3)(x ¡ 5), y = 2(x + 1)(x ¡ 2),
y = 2x(x + 5) and y = 2(x + 2)(x + 4)
c What is the geometrical significance of p and q in y = 2(x ¡ p)(x ¡ q)?
y = 2(x ¡ 1)2
, y = 2(x ¡ 3)2
, y = 2(x + 2)2
, y = 2x2
c What is the geometrical significance of p in y = 2(x ¡ p)2
?
4 Copy and complete:
² If a quadratic has the form y = a(x ¡ p)(x ¡ q) then it ...... the x-axis at ......
² If a quadratic has the form y = a(x ¡ p)2
then it ...... the x-axis at ......
This investigation is also best done using technology.
What to do:
y = (x ¡ 3)2
+ 2, y = 2(x ¡ 3)2
+ 2, y = ¡2(x ¡ 3)2
+ 2,
y = ¡(x ¡ 3)2
+ 2 and y = ¡1
3 (x ¡ 3)2
+ 2
b Find the coordinates of the vertex for each function in a.
c What is the geometrical significance of a in y = a(x ¡ 3)2
+ 2?
y = 2(x ¡ 1)2
+ 3, y = 2(x ¡ 2)2
+ 4, y = 2(x ¡ 3)2
+ 1,
y = 2(x + 1)2
+ 4, y = 2(x + 2)2
¡ 5 and y = 2(x + 3)2
¡ 2
b Find the coordinates of the vertex for each function in a.
c What is the geometrical significance of h and k in y = 2(x ¡ h)2
+ k?
3 Copy and complete:
If a quadratic has the form y = a(x ¡ h)2
+ k then its vertex has coordinates ......
The graph of y = a(x ¡ h)2
+ k is a ...... of the graph of y = ax2
with vector ......
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Y:HAESEIB_SL-3edIB_SL-3ed_01029IB_SL-3ed_01.cdr Friday, 10 February 2012 2:20:19 PM BEN

3 QUADRATICS
From Investigations 1 and 2 you should have discovered that a, the coefficient of x2
, controls the width
of the graph and whether it opens upwards or downwards.
For a quadratic function y = ax2
+ bx + c, a 6= 0:
² a > 0 produces the shape called concave up.
a < 0 produces the shape called concave down.
² If ¡1 < a < 1, a 6= 0 the graph is wider than y = x2
.
If a < ¡1 or a > 1 the graph is narrower than y = x2
.
Summary:
Quadratic form, a 6= 0 Graph Facts
² y = a(x ¡ p)(x ¡ q)
p, q are real
x-intercepts are p and q
axis of symmetry is x = p+q
2
vertex is
¡p+q
2 , f(p+q
2 )
¢
² y = a(x ¡ h)2
h is real
touches x-axis at h
axis of symmetry is x = h
vertex is (h, 0)
² y = a(x ¡ h)2
+ k axis of symmetry is x = h
vertex is (h, k)
² y = ax2
+ bx + c y-intercept c
axis of symmetry is x =
¡b
2a
vertex is
µ
¡
b
2a
, c ¡
b2
4a
¶
x-intercepts for ¢ > 0 are
¡b §
p
¢
2a
where ¢ = b2
¡ 4ac
x = h
x
V ,
(h 0)
p q
x
x =
p + q
2
x = h
V ,
(h k)
-b -
p
¢
x
x =
-b
2a
p
¢
-b +
2a
2a
¡b ¡
p
¢
2a
and
¡b +
p
¡
b
2a
is the average of
¢
2a
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QUADRATICS 4
Using axes intercepts only, sketch the graphs of:
a y = 2(x + 3)(x ¡ 1) b y = ¡2(x ¡ 1)(x ¡ 2) c y = 1
2 (x + 2)2
a y = 2(x + 3)(x ¡ 1)
has x-intercepts ¡3, 1
When x = 0,
y = 2(3)(¡1)
= ¡6
) y-intercept is ¡6
b y = ¡2(x ¡ 1)(x ¡ 2)
has x-intercepts 1, 2
When x = 0,
y = ¡2(¡1)(¡2)
= ¡4
) y-intercept is ¡4
c y = 1
2 (x + 2)2
touches x-axis at ¡2
When x = 0,
y = 1
2 (2)2
= 2
) y-intercept is 2
EXERCISE 1.1
1 Using axes intercepts only, sketch the graphs of:
a y = (x ¡ 4)(x + 2) b y = ¡(x ¡ 4)(x + 2)
c y = 2(x + 3)(x + 5) d y = ¡3x(x + 4)
e y = 2(x + 3)2
f y = ¡1
4 (x + 2)2
2 State the equation of the axis of symmetry for each graph in question 1.
3 Match each quadratic function with its corresponding graph.
a y = 2(x ¡ 1)(x ¡ 4) b y = ¡(x + 1)(x ¡ 4)
c y = (x ¡ 1)(x ¡ 4) d y = (x + 1)(x ¡ 4)
e y = 2(x + 4)(x ¡ 1) f y = ¡3(x + 4)(x ¡ 1)
g y = ¡(x ¡ 1)(x ¡ 4) h y = ¡3(x ¡ 1)(x ¡ 4)
A B C D
E F G H
Example 1
y
x
1
-3
-6
y
x
1 2
-4
y
x
2
-2
The axis of symmetry
is midway between
the -intercepts.
x
y
x
-4
4
1
y
x
4
4
1
y
x
8
4
1
y
x
-12
4
1
y
x
12
1
-4
y
x
-8
1
-4
y
x
-4
4
-1
y
x
4
4
-1
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5 QUADRATICS
Use the vertex, axis of symmetry, and y-intercept to graph y = ¡2(x + 1)2
+ 4.
The vertex is (¡1, 4).
The axis of symmetry is x = ¡1.
When x = 0, y = ¡2(1)2
+ 4
= 2
a < 0 so the shape is
4 Use the vertex, axis of symmetry, and y-intercept to graph:
a y = (x ¡ 1)2
+ 3 b y = 2(x + 2)2
+ 1 c y = ¡2(x ¡ 1)2
¡ 3
d y = 1
2 (x ¡ 3)2
+ 2 e y = ¡1
3 (x ¡ 1)2
+ 4 f y = ¡ 1
10 (x + 2)2
¡ 3
5 Match each quadratic function with its corresponding graph:
a y = ¡(x + 1)2
+ 3 b y = ¡2(x ¡ 3)2
+ 2 c y = x2
+ 2
d y = ¡(x ¡ 1)2
+ 1 e y = (x ¡ 2)2
¡ 2 f y = 1
3 (x + 3)2
¡ 3
g y = ¡x2
h y = ¡1
2 (x ¡ 1)2
+ 1 i y = 2(x + 2)2
¡ 1
A B C
D E F
G H I
Example 2
y
x
V ,
(-1 4)
2
x= -1

3
3
x
y
-3
2 x
y
-6
-3
x
y
-4
2
x
y
4
2
x
y
-2
3
x
y
3
4
x
y
2
2
x
y
2
2
-2
1
x
y
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QUADRATICS 6
Determine the coordinates of the vertex of y = 2x2
¡ 8x + 1.
y = 2x2
¡ 8x + 1 has a = 2, b = ¡8, and c = 1
)
¡b
2a
=
¡(¡8)
2 £ 2
= 2
) the axis of symmetry is x = 2
When x = 2, y = 2(2)2
¡ 8(2) + 1
= ¡7
) the vertex has coordinates (2, ¡7).
6 Locate the turning point or vertex for each of the following quadratic functions:
a y = x2
¡ 4x + 2 b y = x2
+ 2x ¡ 3 c y = 2x2
+ 4
d y = ¡3x2
+ 1 e y = 2x2
+ 8x ¡ 7 f y = ¡x2
¡ 4x ¡ 9
g y = 2x2
+ 6x ¡ 1 h y = 2x2
¡ 10x + 3 i y = ¡1
2 x2
+ x ¡ 5
7 Find the x-intercepts for:
a y = x2
¡ 9 b y = 2x2
¡ 6 c y = x2
+ 7x + 10
d y = x2
+ x ¡ 12 e y = 4x ¡ x2
f y = ¡x2
¡ 6x ¡ 8
g y = ¡2x2
¡ 4x ¡ 2 h y = 4x2
¡ 24x + 36 i y = x2
¡ 4x + 1
j y = x2
+ 4x ¡ 3 k y = x2
¡ 6x ¡ 2 l y = x2
+ 8x + 11
Consider the quadratic y = 2x2
+ 6x ¡ 3.
a State the axis of symmetry. b Find the coordinates of the vertex.
c Find the axes intercepts. d Hence, sketch the quadratic.
y = 2x2
+ 6x ¡ 3 has a = 2, b = 6, and c = ¡3.
a > 0 so the shape is
a
¡b
2a
=
¡6
4
= ¡3
2
The axis of symmetry is x = ¡3
2 .
b When x = ¡3
2 ,
y = 2(¡3
2 )2
+ 6(¡3
2 ) ¡ 3 = ¡71
2
The vertex is (¡3
2 , ¡71
2 ).
c When x = 0, y = ¡3
) y-intercept is ¡3.
When y = 0, 2x2
+ 6x ¡ 3 = 0
Using technology,
x ¼ ¡3:44 or 0:436
d
Example 4
Example 3
The vertex is the
or the depending
on whether the graph is
concave down or concave up.
maximum
minimum
0.436
-3
-3.44
y
x
V( , )
- -
1 7
1
2
1
2
x =- 3
2
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7 QUADRATICS
8 For each of the following quadratics:
i state the axis of symmetry ii find the coordinates of the vertex
iii find the axes intercepts, if they exist iv sketch the quadratic.
a y = x2
¡ 2x + 5 b y = x2
+ 4x ¡ 1 c y = 2x2
¡ 5x + 2
d y = ¡x2
+ 3x ¡ 2 e y = ¡3x2
+ 4x ¡ 1 f y = ¡2x2
+ x + 1
g y = 6x ¡ x2
h y = ¡x2
¡ 6x ¡ 8 i y = ¡1
4 x2
+ 2x + 1
SKETCHING GRAPHS BY ‘COMPLETING THE SQUARE’
If we wish to find the vertex of a quadratic given in general form y = ax2
+bx+c then one approach
is to convert it to the form y = a(x ¡ h)2
+ k where we can read off the coordinates of the vertex
(h, k). One way to do this is to ‘complete the square’.
Consider the simple case y = x2
¡ 4x + 1, for
which a = 1.
To obtain the graph of y = x2
¡ 4x + 1 from the
graph of y = x2
, we shift it 2 units to the right
and 3 units down.
y = x2
¡ 4x + 1
) y = x2
¡ 4x + 22
| {z } + 1 ¡ 22
| {z }
) y = (x ¡ 2)2
¡ 3
Write y = x2
+ 4x + 3 in the form y = (x ¡ h)2
+ k by ‘completing the square’. Hence
sketch y = x2
+ 4x + 3, stating the coordinates of the vertex.
y = x2
+ 4x + 3
) y = x2
+ 4x + 22
+ 3 ¡ 22
) y = (x + 2)2
¡ 1
shift 2
units left
shift 1
unit down
The vertex is (¡2, ¡1)
and the y-intercept is 3.
Example 5
x
y
y = x2
y = x - 4x + 1
2
1
(2 -3)
,
+2
+2
-3
-3
-1
-1
-2
-2
y = x2
y = x + 4x + 3
2
vertex ,
(-2 -1)
y
x
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QUADRATICS 8
EXERCISE 1.2
1 Write the following quadratics in the form y = (x ¡ h)2
+ k by ‘completing the square’. Hence
sketch each function, stating the coordinates of the vertex.
a y = x2
¡ 2x + 3 b y = x2
+ 4x ¡ 2 c y = x2
¡ 4x
d y = x2
+ 3x e y = x2
+ 5x ¡ 2 f y = x2
¡ 3x + 2
g y = x2
¡ 6x + 5 h y = x2
+ 8x ¡ 2 i y = x2
¡ 5x + 1
a Convert y = 3x2
¡ 4x + 1 to the form y = a(x ¡ h)2
+ k without technology.
b Hence, write down the coordinates of its vertex and sketch the quadratic. Use technology to
check your answer.
a y = 3x2
¡ 4x + 1
= 3[x2
¡ 4
3 x + 1
3 ] ftaking out a factor of 3g
= 3[x2
¡ 2(2
3 )x + (2
3 )2
¡ (2
3 )2
+ 1
3 ] fcompleting the squareg
= 3[(x ¡ 2
3 )2
¡ 4
9 + 3
9 ] fwriting as a perfect squareg
= 3[(x ¡ 2
3 )2
¡ 1
9 ]
= 3(x ¡ 2
3 )2
¡ 1
3
b The vertex is (2
3 , ¡1
3 )
and the y-intercept is 1.
2 For each of the following quadratics:
i Write the quadratic in the form y = a(x ¡ h)2
+ k without using technology.
ii State the coordinates of the vertex.
iii Find the y-intercept.
iv Sketch the graph of the quadratic.
v Use technology to check your answers.
Example 6
x
x = We
y
1.5
1
-0.5
1
-1
V ,
(We Qe
- )
y= x - x+
¡ ¡ ¡ ¡ ¡ ¡
3 4 1
2
TI-84 Plus
Casio fx-CG20
TI- spire
n
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Y:HAESEIB_SL-3edIB_SL-3ed_01035IB_SL-3ed_01.cdr Monday, 20 February 2012 4:46:10 PM BEN

9 QUADRATICS
a y = 2x2
+ 4x + 5 b y = 2x2
¡ 8x + 3
c y = 2x2
¡ 6x + 1 d y = 3x2
¡ 6x + 5
e y = ¡x2
+ 4x + 2 f y = ¡2x2
¡ 5x + 3
3 Use the graphing package or your graphics calculator to determine the vertex of each
of the following functions. Hence write each function in the form y = a(x¡h)2
+k.
a y = x2
¡ 4x + 7 b y = x2
+ 6x + 3 c y = ¡x2
+ 4x + 5
d y = 2x2
+ 6x ¡ 4 e y = ¡2x2
¡ 10x + 1 f y = 3x2
¡ 9x ¡ 5
THE DISCRIMINANT AND THE QUADRATIC GRAPH
The discriminant of the quadratic equation ax2
+ bx + c = 0 is ¢ = b2
¡ 4ac.
We used ¢ to determine the number of real roots of the equation. If they exist, these roots correspond
to zeros of the quadratic y = ax2
+ bx + c. ¢ therefore tells us about the relationship between a
quadratic function and the x-axis.
The graphs of y = x2
¡ 2x + 3, y = x2
¡ 2x + 1, and y = x2
¡ 2x ¡ 3 all have the same axis
of symmetry, x = 1.
Consider the following table:
For a quadratic function y = ax2
+ bx + c, we consider the discriminant ¢ = b2
¡ 4ac.
If ¢ < 0, the graph does not cut the x-axis.
If ¢ = 0, the graph touches the x-axis.
If ¢ > 0, the graph cuts the x-axis twice.
y = x2
¡ 2x + 3 y = x2
¡ 2x + 1 y = x2
¡ 2x ¡ 3
¢ = b2
¡ 4ac
= (¡2)2
¡ 4(1)(3)
= ¡8
¢ = b2
¡ 4ac
= (¡2)2
¡ 4(1)(1)
= 0
¢ = b2
¡ 4ac
= (¡2)2
¡ 4(1)(¡3)
= 16
¢ < 0 ¢ = 0 ¢ > 0
does not cut the x-axis touches the x-axis cuts the x-axis twice
a is always
the factor to
be ‘taken out’.
y
x
1
1
y
x
3
1
x
1 3
-3
-1
y
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QUADRATICS 10
POSITIVE DEFINITE AND NEGATIVE DEFINITE QUADRATICS
Positive definite quadratics are quadratics which are positive for all
values of x. So, ax2
+ bx + c > 0 for all x 2 R .
Test: A quadratic is positive definite if and only if a > 0 and ¢ < 0.
Negative definite quadratics are quadratics which are negative for all
values of x. So, ax2
+ bx + c < 0 for all x 2 R .
Test: A quadratic is negative definite if and only if a < 0 and ¢ < 0.
Use the discriminant to determine the relationship between the graph of each function and the
x-axis:
a y = x2
+ 3x + 4 b y = ¡2x2
+ 5x + 1
a a = 1, b = 3, c = 4
) ¢ = b2
¡ 4ac
= 9 ¡ 4(1)(4)
= ¡7
Since ¢ < 0, the graph does not cut
the x-axis.
Since a > 0, the graph is concave up.
The graph is positive definite, and lies
entirely above the x-axis.
b a = ¡2, b = 5, c = 1
) ¢ = b2
¡ 4ac
= 25 ¡ 4(¡2)(1)
= 33
Since ¢ > 0, the graph cuts the x-axis
twice.
Since a < 0, the graph is concave
down.
EXERCISE 1.3
1 Use the discriminant to determine the relationship between the graph and x-axis for:
a y = x2
+ x ¡ 2 b y = x2
¡ 4x + 1 c y = ¡x2
¡ 3
d y = x2
+ 7x ¡ 2 e y = x2
+ 8x + 16 f y = ¡2x2
+ 3x + 1
g y = 6x2
+ 5x ¡ 4 h y = ¡x2
+ x + 6 i y = 9x2
+ 6x + 1
2 Show that:
a x2
¡ 3x + 6 > 0 for all x b 4x ¡ x2
¡ 6 < 0 for all x
c 2x2
¡ 4x + 7 is positive definite d ¡2x2
+ 3x ¡ 4 is negative definite.
3 Explain why 3x2
+ kx ¡ 1 is never positive definite for any value of k.
4 Under what conditions is 2x2
+ kx + 2 positive definite?
Example 7
x
x
x
x
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11 QUADRATICS
If we are given sufficient information on or about a graph we can determine the quadratic function in
whatever form is required.
Find the equation of the quadratic function with graph:
a b
a Since the x-intercepts are ¡1 and 3,
y = a(x + 1)(x ¡ 3).
The graph is concave down, so a < 0.
When x = 0, y = 3
) 3 = a(1)(¡3)
) a = ¡1
The quadratic function is
y = ¡(x + 1)(x ¡ 3).
b The graph touches the x-axis at x = 2,
so y = a(x ¡ 2)2
.
The graph is concave up, so a > 0.
When x = 0, y = 8
) 8 = a(¡2)2
) a = 2
The quadratic function is
y = 2(x ¡ 2)2
.
Find the equation of the quadratic function with graph:
The axis of symmetry x = 1 lies midway between the x-intercepts.
) the other x-intercept is 4.
) the quadratic has the form
y = a(x + 2)(x ¡ 4) where a < 0
But when x = 0, y = 16
) 16 = a(2)(¡4)
) a = ¡2
The quadratic is y = ¡2(x + 2)(x ¡ 4).
FINDING A QUADRATIC FROM ITS GRAPH
Example 9
Example 8
y
x
3
-1 3
y
x
2
8
-2
y
x
x = 1
16
-2
y
x
x = 1
16
3 units
3 units
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QUADRATICS 12
EXERCISE 2
1 Find the equation of the quadratic with graph:
a b c
d e f
2 Find the quadratic with graph:
a b c
Find the equation of the quadratic whose graph cuts the x-axis at 4 and ¡3, and which passes
through the point (2, ¡20). Give your answer in the form y = ax2
+ bx + c.
Since the x-intercepts are 4 and ¡3, the quadratic has the form y = a(x ¡ 4)(x + 3) where
a 6= 0.
When x = 2, y = ¡20
) ¡20 = a(2 ¡ 4)(2 + 3)
) ¡20 = a(¡2)(5)
) a = 2
The quadratic is y = 2(x ¡ 4)(x + 3)
= 2(x2
¡ x ¡ 12)
= 2x2
¡ 2x ¡ 24
Example 10
y
x
8
2
y
x
4
1 2
y
x
3
3
1
y
x
1
-3
y
x
3
-1 3
y
x
12
-2 3
y
x
x = 3
2
12
y
x
x =-3
-12
y
x
x =-1
-4
4
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13 QUADRATICS
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 ¡1
2 , 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.
Find the equation of each quadratic function given its graph:
a b
a Since the vertex is (3, ¡2), the
quadratic has the form
y = a(x ¡ 3)2
¡ 2 where a > 0.
When x = 0, y = 16
) 16 = a(¡3)2
¡ 2
) 16 = 9a ¡ 2
) 18 = 9a
) a = 2
The quadratic is y = 2(x ¡ 3)2
¡ 2.
b Since the vertex is (¡4, 2), the
quadratic has the form
y = a(x + 4)2
+ 2 where a < 0.
When x = ¡2, y = 0
) 0 = a(2)2
+ 2
) 4a = ¡2
) a = ¡1
2
The quadratic is
y = ¡1
2 (x + 4)2
+ 2.
4 If V is the vertex, find the equation of the quadratic function with graph:
a b c
d e f
Example 11
y
x
V(- )
4 2
,
-2
y
x
16
V ,
(3 -2)
y
x
V( )
2 4
,
y
x
V( - )
2 1
,
7
y
x
1
V( )
3 8
,
y
x
7
V( - )
4 6
,
y
x
V( )
2 3
,
(3 1)
,
y
x
V *
&Qw Ew
, -
& *
Ew Qw
,
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Y:HAESEIB_SL-3edIB_SL-3ed_01040IB_SL-3ed_01.cdr Wednesday, 15 February 2012 3:35:53 PM ED

14 Solutions- QUADRATICS
EXERCISE I.I
' • y = (x -4)(x+2)
has x-intercepts
-2 and 4
andy-intercept
-8
y b y = -(x -4)(x+2)
has x -
intercepts
-2 and 4
and y -
intercept 8
y=(x-4)(:i:+2) y= -(x - 4){x + 2)
d y = -3x(x+4)
c y = 2(x+3)(x+5)

has x-intercepts
-5 and -3
andy-intercept
30
e y=2(x+3)2
has x-intercept
-3
andy-intercept
18
y=2(x+3)(x+5)
---5 -3
y=2(x+3)2
y
X
18
has x -
intercepts
0 and -4
and y-intercept 0
'y=-¼(:i::+2)2
has x-intercept
-2
and y-intercept
-1
t. a The average ofthe x -
intercepts is 1, so the axis of synnn�try is x = 1.
b The average of the x-intercepts is 1, so the axis ofsymmetry is x = 1.
c The average ofthe x -
intercepts is -4, so the axis of symmetry is x = -4.
d The average of the x-intercepts is -2, so the axis of symmetry is x = -2.
e The only x-intercept is -3, so the axis ofsymmetry is x =-3.
I The only x-intercept is -2, so the axis of symmetry is .x = -2.
3 1 C b E c I
y
d F • G I H I A
4 a The vertex is
(1, 3).
The axis of
symmetry is
X = 1.
They-intercept
is 4.
V(l,3)
!
x=l
X
b The vertex is
(-2,1).
The axis of
synnnetry is
X = -2.
The
y-intercept
is 9.
x=-2
X
4
X
y
X
h D
y
9

Solutions- QUADRATICS 14
C Thevertex is
y x=l
(1, -3).
The axisof
symmetry is V(l,-
3)
X =1. _,,
I
"
The y-intercept
i
is-5 . y=-2(x-1)2
-3
• Thevertex is Y,
(1,4). 3! i V(l,4)
'
The axisof
symmetry is
X =1. x=l
The y-intercept
y=-½{x-1)2+4
. 3'
IS 3·
5 a G b A
6 II y=x
2
- 4z+2
c I d B
has a=l, b=-4, c=2
-� = _
(-4)
= 2
· ·
2a 2(1)
:. the axisofsymmetry is x=2.
When x=2,
y=22
- 4 X 2+2 =-2
:. thevertex isat (2,-2).
C y=2x
2
+4
has a=2, b=O, c=4
b 0
.. -- = -= = 0
2a 2(2)
:. the axisofsymmetry is x=0.
When x=O, y=4
:. thevertex isat (0, 4).
e y=2x
2
+8x-7
has a=2, b=B, c=-7
b 8
:. -- =-=,,- =-2
2a 2(2)
. . the axisofsymmetry is x=-2.
When x=-2,
• I
y =2(-2)
'+8(-2)- 7=-15
:. thevertex isat (-2,-15).
9 y=2x
2
+6x- 1
has a=2, b= 6, c=-1
b 6 3
-- --
• •
2a 2(2) 2
:. the axisofsymmetry is x=-J.
When x=-J, y=2(-J)
2
+6(-J)- 1
=£-9- 1
-ll
· th rte
.
t ( 3 '')
. . eve x 1sa -2,-2 .
d Thevertex is y=½(x-3)2
+2
(3, 2).
The axisof
symmetry is
X =3.
The y-intercept !V(3,2)
isJ/-. X=3 X
I Thevertex is l y
(-2, -3). V(-2,-3) l
The axisof i
symmetry is
X =-2. x=-2
The y-intercept
. 3'
IS - 5·
y=-lo{x+2)2
_3
I C 9 D h F I H
b y=x
2
+2x- 3
has a=l, b=2, c=-3
b 2
. . -- = -��=-1
2a 2(1)
.·. the axisofsymmetryis x=-1.
When X = -1,
y =(-1)
2
+2(-1)- 3 =-4
:. thevettexisat (-1,-4).
d y=-3x
2
+1
has a=-3, b=O, c=l
. -� = -
0
= 0
· ·
2a 2(-3)
.·. the axisofsymmetry is x=0.
When x=O, y=1
:. the Vettexisat (0, 1).
f y=-x2
-4z-9
h
has a=-1, b=-4, c=-
9
. -� = - (-4) = -2
· ·
2a 2(-1)
. . the axisofsymmetry is x=-2.
When x=-2, y=-(-2)
2
- 4(-2)-9
=-4+8-9
=-5
• • thevertex isat (-2,-5).
y=2x2
- 10x
+3
has a=2, b= -10, c=3
b (-10) 5
-- -
• •
2a 2(2) 2
. . the axisofsymmetry is x=£.
When x= �, y = 2(�)
2
- 10(�)+ 3
-M_§!!.+3
- 2 2
''
-,
:. thevertex isat (£,-¥).

15 S
I y=-�x2
+x-5
has a=-½, b=l, c=-5
b I
:. --= - 1 =1
2a 2(-,)
.·. the axisof symmetry isx=1.
7 I Wheny=O, x 2
-9=0
:. (x+3)(x-3)=0
· x=±3
:. the
x-intercepts are±
3
c When y=O, x2
+7x+10=0
:. (x +5)(x +2) =0
:. x=-5 or-2
· the
x -
intercepts are -5 and -2
e When y=O, 4x-x 2
=0
:. x(4-x)=0
:. x=O or4
:. the
x -
intercepts areO and4
9 When y=O, -2x 2
-4x- 2=0
:. x2
+ 2
x+l=O
:. (x+l)'=O
· x =-I
:. the
x-intercept is-
1 (touching)
I When y=O, x2- 4
x+l=O
a=l, b=-4, and c=l
k
•
• •
•
• •
x=
-(-4)± ✓C-4)'- 4(1)(1)
2(1)
4±/U
2
4±2v'3
2
= 2± v'3
the
a: -
intercepts are 2± /'3
When y=0, x2
-6
x- 2=0
a=1, b=-6, and c=-2
•
• •
•
• •
x=
-C-6)± ✓c-6)'- 4(1)c-2)
2(1)
6± v'44
2
6± 2ffi
2
=3±ffi
the
x -
intercepts are3± vTI
When
x=l, y=-�(1)2
+1-5=-�
:. thevertex isat (1, -J).
b Wheny=O, 2a:2
-6=0
:. a:2
-3=0
:. (x+ v'3)(x- v'3)=0
:. X
=±v'3
:. the
a:-intercepts are
±/'3
d When y=O, a:2
+x-12=0
:. (x+ 4)(x-3)=0
:. x =-4 or 3
· the
x-intercepts are-4 and3
f Wheny=O, -a: 2
-6a:- 8=0
:. x2
+6x+8=0
:. (x +4)(x +2) =0
:. X=-4or-2
:. the
x-intercepts are
-4 and-2
h When y=O, 4x 2
-24x+3
6=0
:. a: 2
-6a:+9=0
:. (x-3)'=0
:. X =3
:. the
x-intercept is3 (touching)
I Wheny=O, x2+ 4
x-3=0
a=l, b=4, and c=-3
I
•
• •
•
. .
-4± ✓4'- 4(1)(-3)
x=
2(1)
-4±¥'28
2
-4±2y'7
2
=-2±v'7
the
x-interceptsare
-2± ./7
Wheny=0, x2 +Bx+
1
1=0
a=1, b= 8, and C
=1
1
•
• •
•
. .
-8± ✓82- 4(1)(1
1)
x=
2(1)
-8±v20
2
-8±2v's
2
=-4±v's
the
x-intercepts are
-4 ± v5

8 •
b
C
QUADRATICS 16
I y=:i:2
-2:i:+5
has a=l, b=-2, c=5
__i,__ = _(-2) = 1
. .
2a 2(1)
. . the ax is
ofsymmetry is x = 1
Ill When :i:=0, y=5,
so the y-intercept is5
When y=O, :i:2
-2:i:+5=0
x=
•
••
-(-2) ± J(-2)'- 4(1)(5)
2(1)
2±J4-20
2
Thishas no real soluti ons,
so there are no :i: -
in
tercepts.
I y=:i:2
+4:i:- 1
has a=l, b=4, c=-1
b 4
.. --=
- =
-2
2a 2(1)
. . the axis
ofsymmetryis x = -2
Ill When x=0, y=-1,
so the y-intercept is-
1.
I
When y=O, :i:2
+4:i:-1=0
•• x=
-4 ± J42 - 4(1)(-1)
2(1)
-4±v'25
2
-4± 2v'5
2
=
-2±v'5
. . the x -
in tercepts are -2 ± v'5
y=2:i:2
- 5:i:+2
has a=2, b=-5, c=2
b (-5) 5
-- -
2a 2(2) 4
. . the axisofsymmetry is
Ill When x=O, y=2,
so the y-intercept is2.
When y=O, 2:i:2
- 5:i:+2=0
:. (2x- l)(x- 2)=0
•• x=}or2
. . the x-intercepts are } and 2
II When x = 1,
y=12
- 2(1)+5
=
1- 2+5
=4
•
Iv
the VeiteA isat (1, 4)
lV(l,4)
'
'
a:=1
II When x = -2,
X
y=(-2)2
+4(
-2)-1
=4-8-1
=
-5
:. thevertexisat (-2,-5)
Iv
Z= -2
y
-2-/s -1
�
-2+./5
V(-2,-5)!
II When
y=2(!)2
- 5( !)+2
=M- M+2
' '
=-¾
.·. thevertexisat
Iv
2
y --2
•-,
'v(' ')
' i•S
X

17 QUADRATICS
d
•
I
•
I y=-a:
2 +3:z:- 2
has a=-1, b=3, c=-2
b 3 3
:. -- =-c-,--,-,=-
2a 2(-1) 2
. . the axis ofsymmetryis a: = ;
Ill When a:=0, y =-2,
sothey -
intercept is -2.
When y=O, - a:
2
+3:z:- 2 =0
:. a:
2
- 3:z:+2 =0
· (x-l)(x- 2)=0
•
.. a:=lor2
the a:-intercepts are 1 and 2
I y=-3:z:
2 +4x- 1
has a=-3, b=4, c=-1
•
• •
b 4 2
--=- =-
2a 2(-3) 3
. . the axis ofsymmetry is a: = ;
Ill When a:=0, y=-1,
sothey -
intercept is -1.
When y=O, -3x
2+4x-1=0
:. 3x"-4x+l=O
.. (3x-l)(x- I} = 0
.. a:=iorl
:. the a:-intercepts are ½ and1
I y=-2x
2 +x+l
has a=-2, b=l, c=l
b 1 1
• •
2a 2(-2) 4
. . the axis ofsymmetry is a: = ¼
Ill When :t=O, y=l,
sothey-intercept is 1.
When y=0, -2x
2
+x+1 =0
.. 2x2-x- 1=0
:. (2x+l)(x-1)=0
· x=-½or1
. . the x-intercepts are -� and 1
I y=6x-x
2
has a=-1, b=6, c=O
• •
_.!?._ =-
6 =3
2a 2(-1)
. . the axis ofsymmetry is x= 3
Ill When x=O, y=O,
so
they-intercept is 0.
When y=O, 6x-x
2
=0
:. x(6-x)= 0
• • x=Oor6
:. the x-intercepts are O and 6
II When x=;, y=-(;)2
+3(; )- 2
Iv
=-9+9- 2
' '
=¼
:. the vertexis at (f, ¼)
Y Y<!,¼>
X
-2
�-�
•-,
II When x=;, y =-3(; )
2
+4(;) - 1
= _.! +.!! -1
' '
Iv
-1
.·. the verte,,. is at ( ' ')
3' 3
II When a:=¼, y=-2(¼)
2
+¼+1
= _!+!+1
8 '
Iv
=I
. . the vertex is at (¼, �)
-½
x=¼
X
II When x=3, y=6x3 - 3 2
=9
Iv
:. the vertexis at (3, 9)
y
-V(3,9)
x=3

h
I
QUADRATICS 18
I
Ill
I
y=-x2-6x-8
ha, a= -1, b=-6, c=-8
_.!:_=_(-6)
=-3
2a 2(-1)
• • theaxis of symmetryis x=-3
When X= 0, y=-
8,
sothe y-intercept is -
8.
When y=o, -x2
-Ba::-8 =0
• • x2+6x+8 =0
(x+ 4)(x+2) =0
•
x=-4or-2
• •
• • thex -
intercepts are-4 and -
2
y=-!x2
+2x+l
has a=-¼, b=2, c=l
. -� = - 2 - 4
· · 2a 2(-¼)
-
.. theaxis of symmetryis x= 4
Ill When x=O, y=1,
sothe y-intercept is 1.
When y=0, -¼x2
+2x+1 =0
:. x2
-8x- 4 =0
• • x=
-(-8) ± ✓<-8)'- 4(1)(- 4)
8 ± v'so
2
8 ±4v'5
2
=4±2v'5
2(1)
.. thex-intercepts are 4 ±2v'5.
II When x =-3,
Iv
y =-
(-3)'-6(-3) -8
=-9+18 -8
=1
:. thevertexis at (-3,1)
y
X
I
x=-3
-2
-8
II When x=4, y=-!(4)2
+2( 4)+1
=-4+8 +1
Iv
=5
:. thevertex is at ( 4, 5)
Y i V(4,5)
x=4
X
4+ 2./5
EXERCISE 1.I
I a y=x2
-2x+ 3
.. y=x2
-2x+1 2
+ 3 -1 2
:. y=(x-1)'+2
. . vertexis (1,2), y-intercept is 3
y
3 '--,./
V(l,2)
C y=x2
- 4x
:. y=x2 _4x+2 2
-22
.. y=(x-2) 2
- 4
X
:. vertex is (2,-4), y -
intercept is 0
b y=x2
+ 4x-2
y
.. y=x2
+ 4x+22
-2-22
.. y=(x+2)'-6
:. vertex is (-
2,-
6), y-intercept is -2
y
-2
V(-2,-6)
X
V(2,-4)

19 QUADRATICS
d y=x2
+3x
:. y=x2
+3x+(J)2
-(J)2
:. y=(x+J)2
-¾
.. vertex is (-J, -¾), y -
intercept is 0
y
X
V(-J,-!)
I y=x2
-3x+2
:. y=x2
-3x+(f)2
+2-(j)2
:. y=(x-�)2
-!
:. vertexis (�,-!), y-interceptis 2
y
2
h y=x2
+8x-2
•
:. y=x2
+8x+ 42
-2- 42
:. y=(x + 4)'-18
:. vertexis (
-4,-18), y-interceptis-2
Y=(x+4)2-I8
y
-2
V(-4,-18)
I y=2x2
+ 4x+5
=2[x2
+2x+!J
=2[x2
+2x+12
- 12
+j]
= 2[(x + 1)2
+ ¾I
= 2(x+ 1)2
+3
II The vertex is (-1,3).
• y=x2
+5x-2
y=x2
+5x+(J)2
-2-(J)2
.. y=(x+J)2
-¥
.. vertexis (-
;,-s;), y-interceptis-
2
y
9 y=x2
-6x+5
.. y=x2
-6x+32
+5-32
.. y = (x-3)'- 4
.. vertexis (3,-4), y -
interceptis 5
y
5
V(3, -4)
I y=x2
-5x+l
Ill
Iv
.. y=x2
-5x+(J)2
+1-(J)2
.. y=(x-�)2 _ 2
4
1
.. vettexis (;,-s!), y-interceptisl
y
I
V(j,-5)
When x=O, y=S
•
the y -
interceptis 5
' .
y
y=2x2
+4x+5
V(-1,3)

b
C
d
•
I
20
I
II
I
II
I
II
y=2x2
-Bx+3
=2[x2
-4x+�]
=2[x2
-4x+22
- 22
+
�]
-2[(x- 2)'- ii
-2(x-2)'- 5
The vertex is (2,-5).
y=2x2
-6x+l
=2[x2
- 3x+ ½J
=2[x2
- 3x+(j)2
- (j)2
+½]
=2[(x-�)
2
- !l
=2(x-�)2
-�
The vertex is (j, -½)-
y=3x2
-6x+5
= 3[x2
- 2x+£]
=3[x2
- 2x+12
- 12
+£]
-3[(x-1)' + :1
-3(x-1)'+2
The vertex is (1, 2).
I y=-x2
+
4x+2
- -Ix' -<x- 2]
=-[x2
-4x+22
- 22
- 2]
--[(x- 2)'- 6]
--(x- 2)' +6
II The vertex is (2,6).
I
II
y=-2x
2
- 5x+3
=-2[x
2
+jx- j]
=-2[x
2
+
�x+(!)
2
- (!)2
-�]
=_2[(x+
5
)2 _ 25 _ 24]
4 16 16
=-2[(x+!)2
- *]
=-2(x+!)2
+4:
The vertex is (-!, �9).
Ill When x=O, y=3
• • the y-interceptis 3
Iv
X
y=2x
2 -8x+3
V(2,-5)
Ill When x=O, y=l
• • the y-interceptis 1
Iv
y
y=2x
2 -&:+l
V(j, -l)
Ill When x=O, y=5
the y -
interceptis 5
Iv
V(l,2)
Ill When x=O, y=2
. . the y -
interceptis 2
Iv
y V(2,6)
2
X
y=-x2 +4x+2
Ill When x=0, y=3
the y -
interceptis 3
Iv V(-j,f) y
,
y=-2x2 - 5x+3 3
X

21
3 a Using technology, the graph is
•
y=a:2 -4a:+7
(2, 3)
Since the vertex is at (2, 3),
the function must be of the form
y = a(x-2)2
+3 for some a.
.-. when x =0,
y = a(-2)2
+3= 4a+3
but the y-intercept is 7
4a+3=7
• • a=l
:. the equation is y = (x- 2)2
+3
c Using technology, the graph is
• (2, 9)
Since the vertex is at (2, 9),
y = a(x- 2)2
+9 for some a.
.·. when x =0,
y = a(-2)2+9=4a+9
:. 4a+9=5
. . a =-1
the equation is y =-(x- 2)2
+ g
e Using technology, the graph is
(-' '') .
,, '
y=-2a:2
-10a:+1
b Using technology, the graph is
•
(-3, -6)
3
Since the vertex is at (-3,-6),
y = a(x+3)2
-6 for some a.
.-. when :r = 0,
y= a X 32
-6=9a-6
:. 9a-6=3
:. a=1
the equation is y = (x +3)2
-6
d Using technology, the graph is
•
X
-4
(-!, _1;)
Since the vertex is at (-< -ll)
2' 2 '
y = a(x + !)2 - 1J for some a.
.-. when :r = 0,
Y - a(")2
- !1 = .!!a - 11
2 2 4 2
but the y -
intercept is -4
'a-17
=-4
· · 4 2
• •
• •
!la=2
. '
a=2
. . the equation is y = 2(:r + !)2 - 1J
Since the vertex is at (-½, f),
y = a(x + ;)2 + 2
; for some a.
:. when :r = 0,
Y-a(')2+ 21 = 25a+ 21
2 2 4 2
but the y -
intercept is 1
''a+ 27
=1
• • 4 2
•
• •
• • a=-2
.. the equation is y =-2(x + J)2
+ ¥

I Using technology, the graph is
y
X
-5
22
S. th �- . t (' '')
mce e veu..,x ts a 2,
-
4 ,
y = a(x- ¾)2 - !] for some a.
y=&l:2 -9:r;-5
:. when x=O, y=a(-!)2 -4J=:a-4J
but the y -
intercept is -5
9a- 47 = _5
• •
(' -'')
,, .
• •
•
• •
• •
la=li
• 4
a=3
. . the equation is y=3(x- ¾)2 - !]
EXERCISE 1.3
I I y=x
2
+x- 2
has a=1, b=1, c=- 2
:. 6.=b
2
-4ac
=1' -4(1)(-2)
=9>0
.
·. the graph cuts the x-axis twice, and since
a>0, the graph is concave up.
C y=-x
2
- 3
has a=-1, b=O, c=-3
•
• • .6.=b2-4ac
=0'-4(-1)(-3)
= -12 < 0
:. the graph does not cut the x-axis, and
since a< 0, the graph is concave down.
.
·. it is entirely below the x-axis.
:. the graph is negative definite.
e y=x
2
+8x+16
has a=l, b=B, c=16
:. .6.=b
2
-4ac
=s' -4(1)(16)
=0
:. the graph touches the x -
axis, and since
9 y=6x
2 +5x- 4
has a=6, b=5, c=-
4
:. 6.=b2-4ac
=s' -4(6)(-4)
=121>0
.. the graph cuts the x-axis twice, and since
I y=9x
2 +6x+l
has a=9, b=6, c=l
.. 6.=b
2
-4ac
=6' -4(9)(1)
=0
:. the graph touches the x -
axis, and since
a> 0, the graph is concave up.
b y=x
2
-4x+l
has a=l, b=-4, c=l
:. .6.=b2
-4ac
= (-4)' -4(1)(1)
= 12>0
. . the graph cuts the x-axis twice, and since
d y=x
2 +7x- 2
has a=l, b=7, C=-2
.. 6.=b2
-4ac
=7' -4(1)(-2)
=57 > 0
.·. the graph cuts the x-axis twice, and since
I y=-2x
2 +3x+l
has a=-2, b=3, c=1
:. .6.=b2
-4ac
= 3' -4(-2)(1)
=17>0
.. the graph cuts the x -
axis twice, and since
a< 0, the graph is concave down.
h y=- x2 +x+6
has a=-1, b=l, c=6
6.=b2
-4ac
= 1' -4(-1)(6)
=25>0
. . the graph cuts the x-axis twice, and since
a< 0, the graph is concave down.

23 QUADRATICS
I •
C
x2
- 3x+6
bas a= 1, b= -3, c=6
•• .6..=b2
-4ac
=(-3)2- 4(1)(6)
=-15
Since a>O and 6.
< 0,
x2
- 3x+6 is positive definite.
.
•• x2
- 3x+6>0 for all x.
2x2
-4x+ 7
bas
•
. .
a=2, b=-4, c=7
.6.. =b2
-4ac
=(-4)2
-4(2)(7)
=-40
Since a>O and .6..<0,
2z2
-4x+ 7 is positive definite.
b
d
4x- x2
-6
has
• •
a= -1, b= 4, c= -
6
6.=1l-4ac
=
42-4(-1)(-
6)
=-8
Since a<O and 6.
< 0,
4x- x2
-6 is negative definite.
.
•• 4x- x2
-6<0 for all x.
-2x2
+ 3x-4
has a=-2, b=3, c=-
4
6.=b2
-4ac
=3 2
-
4(-2)(-4)
=-23
Since a< 0 and .6..< 0,
-2x2
+ 3x-4 is negative definite
.
3 3x2
+kx- 1 4 2x2
+kx+ 2
has a=3, b=k, c=-1
.. 6.=b2
-4ac
• •
= k2
-
4(3)(-1)
= k2
+12
6.>0 for allk
{as k2 �0 for allk}
. . 3x2
+kx- 1 has two real distinct
roots for allk .
:. it can neverbe positive definite.
has
. .
a=2, b=k, c=2
6.=b2
-4ac
= k2
-4(2)(2)
= k2
- 16
Now 2z2
+kx+2 has a>O.
•
•• it is positive definite provided k2
-16< 0
:. k2
<1
6
. . -
4<k<4
EXERCISE 2
1 I Thex-intercepts are1 and2 .
:. y = a(x- l)(x- 2)
for some a ':ft0 .
But the y-intercept is4.
:. a(-1)(-2) = 4
••
• •
2a=4
a=2
.. y =2(x- l)(x- 2)
d Thex-intercepts are-1
and3.
:. y = a(x+l)(x- 3)
for some a ':ft0 .
But the y -
intercept is3.
:. a(l)(-3) =3
••
•
• •
-3a=3
a= -1
.. y =-(x+l)(x- 3)
b Thegraph touches the
x-axis when x =2.
:. y
=a(x- 2)2
for some a ':ft0.
:. a(-2)2
=8
. . 4a =8
•
••
• Thegraph touches the
x -
axis when x =1.
:. y=a(x-1)2
for some a ':ft0.
But the y -
intercept is-3 .
a(-1)2
=-3
••
•
• •
a=-3
y=-3(x- 1)2
c Thex-interceptsare1 and3.
:. y = a(x- l)(x- 3)
for some a ':ft0.
.. a(-1)(-3)=3
.. 3a =3
•
• • a=l
:. y =(x- l)(x- 3)
f Thex-interceptsare -2
and3.
:. y = a(x +2)(x- 3)
for some a ':ft0.
But the y-intercept is12 .
a(2)(-3) =12
• •
••
-6a=12
a=-2
:. y =-2(x+2)(x- 3)

3
24
a As the axis of symmetry is x= 3,
the other x-intercept is 4.
:. y=a(x - 2)(x-4) for some a=/:-0.
But they -
intercept =12
. . a(-2)(-4)=12
•
Sa =12
· a=ll=A
· · 8 2
:. y = !(x- 2)(x- 4)
c The graph touches the x-axis at x= -3,
:. y=a(x+3)2 for some a=j:.O.
b As the axis ofsymmetry is x= -1,
the other x-intercept is 2.
:. y = a(x+4)(x- 2) for some a=/:-O.
But the y -
intercept = 4
.. a(4)(-2)=4
-8a=4
. . a=-�
:. y=-½(x+4)(x- 2)
But they-intercept is -12, so a(3)2
= -12
• • 9a = -12
•
• • a =_.!2, = _J.
9 3
•
• • y=-½(x+3)
2
a Since the x -
intercepts are 5 and 1, the
equation is y =
a(x - 5)(x- 1)
for some a =/:- 0.
• •
Butwhen x=2, y=-9
:. -9=a(2- 5)(2-1)
:. -9=a(-3)(1)
• •
•
• •
-3a =-9
a =3
the equation is y=3(x- 5)(x- 1)
y=3(x2
- 6x+5)
y =3x
2
- 1&:+15
• •
•
• •
c Since the graph touches the x-axis at 3,
its equation is y = a(x- 3)
2
,
for some a #- O.
Butwhen x= -2, y =-25
:. -25 = a(-2 - 3)'
•
• •
• •
-25 =25a
a=-1
•
the equation is y =-(x- 3)2
y=-(x
2
-6x+9)
y=-x
2
+6x- 9
• •
•
..
c Since the graph cuts the x-axis at 3
and has axis ofsymmetry x=2,
it must also cut the x-axis at 1.
:. the x-intercepts are 3 and 1, and
the equation is y =a(x - 3)(x- 1)
Butwhen
for some a #- 0.
x=5, y=12
.. 12 =a(5 - 3)(5 - 1)
:. 12=a(2)(4)
•
• •
•
• •
Sa=12
a-ll-1
- 8 - 2
b Since the x -
intercepts are 2 and -½,the
equation is y =a(x - 2)(x+!)
for some a=/:-0.
Butwhen x=3, y=-14
:. -14=a(3 - 2)(3+½)
:. -14 = a(l)(½)
.. ½a =-14
:. a=-4
:. the equation is y =-4(x- 2)(x+!)
:. y=-4(x
2
-lx- 1)
.. y=-4x
2
+6x+4
d Since the graph touches the x-axis at -2,
its equation is y =a(x + 2) 2,
• •
for some a #- O.
Butwhen x=-1, y=4
.. 4=a(-1 +2)'
4=a
• • the equation is y=4(x+2)2
y=4(x2
+4x+4)
y =4x
2
+16x+16
the equation is
• •
•
• •
•
. .
• •
y=!(x- 3)(x-1)
y=!(x2
- 4x+3)
Y = 3
x2 - 6x+.l!.
' '

25
f Since thegraph cuts the
x-axis at
5
an d has axis ofsymmetry x= 1,
it mustalso cut the
x -
axis at-3.
:
. the
x-intercepts are
5 and
-3, and
the equationis y=a(x-5)(x+3)
for somea =/:-0.
Butwhen
x=2, y=5
•
.-. 5 = a(2 - 5)(2 +3)
. . 5 = a(-3)(5)
-15a=5
•
• •
I Theveitex. is(2,4),
sothe quadratic has equation
y
= a(x-2) 2 +4 for somea =f:.0.
Butthegraph passes through the origin
0 = a(O- 2)' +4
.. 4a+4= 0
•
• • a=-1
:. theequationis y=-(x-2) 2
+4
c Thevertex is(3,8),
y
= a(x-3) 2
+8 for somea =f:.0.
But thegraphpasses through(1,0)
.. O=a(l-3)2+8
• •
•
• •
0= 4a+8
a=-
2
:. theeq_uationis y=-
2(x-3) 2
+8
e Thevertex is(2,3),
y
= a(x-2) 2
+3 for somea =f:.0.
Butthegraphpasses through(3,1)
.. l=a(3-2)'+3
•
• •
•
• •
l=a+3
a=-
2
:. theequationis y=-
2(x-2) 2
+3
•
the equationis
•
•
. .
y =-½(x- 5)(x+ 3)
y=
-½(x 2
-
2x-15)
Y-
-1x 2
+ lx+5
- 3 3
b Thevertex is(2,
-1),
so the quadratic hasequation
y= a(x-2) 2 -1 for somea =f:.0.
But thegraph passes through(0, 7)
:. 7 =a(O - 2)' - 1
•
• • 7=4a-1
a=2
.. theequationis y= 2(x-2)2
- 1
d Thevertex is(4,
-6),
y= a(x-4) 2 -6 for somea =f:.0.
But thegraph passes through(7,0)
• •
:. 0 = a(7 -4)2 - 6
9a- 6= 0
.. a= f
:. theequationis y= �
(x-4) 2 -6
f Thevertex is(½, -½),
y=a(x -½)2 -J for some a#- O.
But thegraph passes through(;, �)
½ =a(½-½)
2
-¾
1-a-A
, - 2
:. a= 2
.. theequationis y= 2(x- ½)2 - ½

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