Quadratic Functions Lesson #3
Quadratic Functions
• A quadratic function is a relationship between two variables
which can be written in the form y = ax 2 + bx + c , where x and y
are the variables and a, b, and c represent constants with
a ≠ 0.
• Using function notation, f(x) = ax 2 + bx + c can be written as
f : x 6 ax 2 + bx + c .
• As with linear functions, for any value of x a corresponding
value of y can be found by substituting into the function
equation.
• For example, if y = 2x 2 − 3x + 5 , and x = 3 , then
y = 2 × 32 − 3 × 3 + 5 = 14 . Therefore, the ordered pair (3, 14)
satisfies the function y = 2x 2 − 3x + 5 .
• Similarly, using function notation we could write,
o if f(x) = 2x 2 − 3x + 5 and x = 3 , then
f(3) = 2 × 3 2 − 3 × 3 + 5 = 14 .
• Example 1:
If y = −2x 2 + 3x − 4 , find the value of y when:
(a.) x = 0 (b.) x = 3
(a.)
When x = 0 ,
y = −2(0)2 + 3(0) − 4
y =0+0−4
y = −4
(b.)
When x = 3 ,
y = −2(3)2 + 3(3) − 4
y = −2(9) + 9 − 4
y = −18 + 9 − 4
y = −13
• Example 2:
If f(x) = −2x 2 + 3x − 4 , find: (a.) f(2) (b.) f( −4)
(a.)
f(2) = −2(2)2 + 3(2) − 4
f( −4) = −2( −4)2 + 3( −4) − 4
f(2) = −2(4) + 6 − 4
f(2) = −8 + 6 − 4
f(2) = −6
(b.)
f( −4) = −2(16)2 − 12 − 4
f( −4) = −32 − 12 − 4
f( −4) = −48
• Example 3:
State whether the following quadratic functions are satisfied by
the given ordered pairs:
(a.) y = 3x 2 + 2x (2, 16)
(b.) f(x) = − x 2 − 2x + 1 (-3, 1)
(a.)
y = 3(2)2 + 2(2)
y = 12 + 4
y = 16
i.e., when x = 2 , y = 16
Therefore, (2, 16) does
not satisfy y = 3x 2 + 2x .
(b.)
f( −3) = −( −3)2 − 2( −3) + 1
f( −3) = −9 + 6 + 1
f( −3) = −2
i.e., f( −3) ≠ 1
Therefore, (-3, 1) does not
satisfy f(x) = − x 2 − 2x + 1.
Finding x Given y
• It is also possible to substitute a value for y to find a
corresponding value for x. However, unlike linear functions, with
quadratic functions there may be 0, 1 or 2 possible values for x
for any one value of y.
• Example 1:
If y = x 2 − 6x + 8 , find the value(s) of x when:
(a.) y = 15 (b.) y = −1
(a.) If y = 15 ,
(b.) If y = −1,
x 2 − 6x + 8 = 15
x 2 − 6x − 7 = 0
(x + 1)(x − 7) = 0 {factorising}
Therefore, x = −1 or x = 7 , i.e., two solutions.
x 2 − 6x + 8 = −1
x 2 − 6x + 9 = 0
{factorising}
(x − 3)2 = 0
Therefore, x = 3 , i.e., only one solution.
• Example 2:
If f(x) = x 2 + 4x + 11, find x when:
(a.) f(x) = 23 (b.) f(x) = 7
If f( x) = 23 ,
x 2 + 4x + 11 = 23
x 2 + 4x − 12 = 0
(x + 6)(x − 2) = 0
{factorising}
Therefore, x = −6 or x = 2 , i.e. two solutions.
(b.) If f(x) = 7 ,
x 2 + 4x + 11 = 7
x 2 + 4x + 4 = 0
{factorising}
(x + 2)2 = 0
Therefore, x = −2 , i.e. only one solution.
• Example 3:
A stone is thrown into the air and its height in metres above the
ground is given by the function h(t) = −5t 2 + 30t + 2 , where t is
the time (in seconds) from when the stone is thrown.
(a.) How high above the ground is the stone at time t = 3 seconds?
(b.) How high above the ground was the stone released?
(c.) At what time was the stone’s height above the ground 27 m?
(a.)
(a.)
(b.)
(c.)
h(3) = −5(3)2 + 30(3) + 2
h(3) = −45 + 90 + 2
h(3) = 47
Therefore, the stone is 47 m above ground.
The stone is released when t = 0 seconds.
h(0) = −5(0)2 + 30(0) + 2
h(0) = 2
Therefore, it is released 2 m above ground level.
When h(t ) = 27 ,
−5t 2 + 30t + 2 = 27
−5t 2 + 30t − 25 = 0
t 2 − 6t + 5 = 0
{dividing each term by -5}
(t − 1)(t − 5) = 0
{factorising}
Therefore, t = 1 or t = 5 , i.e. after 1 second and after 5
seconds.
Graphs Of Quadratic Functions
• The graphs of all quadratic functions are parabolas. The
parabola is one of the conic sections.
• Conic sections are curves which can be obtained by cutting a
cone with a plane. The Ancient Greek mathematicians were
fascinated by conic sections.
• You may like to find the conic sections for yourself by cutting an
icecream cone.
• Cutting parallel to the side produces a parabola (as
seen to the right):
• There are many examples of parabolas in
every day life. The name parabola comes
from the Greek word for thrown because
when an object is thrown its path makes a
parabolic shape.
• Parabolic mirrors are used in car headlights,
heaters, radar discs and radio telescopes
because of their special geometric properties.
• Below is a single span parabolic bridge. Other suspension
bridges, such as the Golden Gate bridge in San Francisco, also
form parabolic curves.
The Simplest Quadratic Function
• The simplest quadratic function is y = x 2 and its
graph can be drawn from a table of values.
x
y
-3
9
-2
4
-1
1
0
0
1
1
2
4
3
9
• Note:
o The curve is a parabola and it opens
upwards.
o There are no negative y values, i.e., the curve does not
go below the x-axis.
o The curve is symmetrical about the y-axis because, for
example, when x = −3 , y = ( −3)2 and when x = 3 , y = 32
have the same value.
o The curve has a turning point or vertex at (0, 0).
• Example 1:
Draw a graph of y = x 2 + 2x − 3 from a table of values.
Consider f(x) = x 2 + 2x − 3 .
f( −3) = ( −3)2 + 2( −3) − 3 = 9 − 6 − 3 = 0
f( −2) = ( −2)2 + 2( −2) − 3 = 4 − 4 − 3 = −3
f( −1) = ( −1)2 + 2( −1) − 3 = 1 − 2 − 3 = −4
f(0) = (0)2 + 2(0) − 3 = 0 + 0 − 3 = −3
f(1) = (1)2 + 2(1) − 3 = 1 + 2 − 3 = 0
f(2) = (2)2 + 2(2) − 3 = 4 + 4 − 3 = 5
f(3) = (3)2 + 2(3) − 3 = 9 + 6 − 3 = 12
Table of Values:
x
y
-3
0
-2
-3
-1
-4
0
-3
1
0
2
5
3
12
• Graphs of the form y = x 2 + k have exactly the same shape as
the graph of y = x 2 . In fact k is the vertical translation factor.
Every point on the graph of y = x 2 is translated k units vertically
to give the graph of y = x 2 + k .
• Example 2:
Sketch y = x 2 on a set of axes and then use it to sketch
y = x 2 + 3 . Mark the vertex of y = x 2 + 3 .
Draw y = x 2 and translate it 3 units
upwards.
• Graphs of the form y = (x − h)2 have exactly the same shape as
the graph of y = x 2 .
• In fact, h is the horizontal translation factor. Every point on the
graph of y = x 2 is translated h units horizontally to give the
graph of y = (x − h)2 .
• Example 3:
Sketch y = x 2 on a set of axes and then use it to sketch
y = (x + 3)2 . Mark the vertex of y = (x + 3)2 .
Draw y = x 2 and translate it 3 units
to the left.
• Graphs of the form y = (x − h)2 + k have the same shape as the
graph of y = x 2 and can be obtained from y = x 2 by using a
horizontal shift of h units and a vertical shift of k units.
• The vertex is at (h, k).
• Example 4:
Sketch y = x 2 on a set of axes, then sketch the following,
stating the coordinates of the vertex:
(a.) y = (x − 2)2 + 3 (b.) y = (x + 2)2 − 5
(a.)
(b.)
Draw y = x 2 and translate is 2 units to the right and 3
units up.
The vertex is at (2, 3).
Draw y = x 2 and translate is 2 units to the left and 5 units
down.
The vertex is at (-2, -5).
• Based on the previous information and examples, we find that
we could write quadratics of the form y = x 2 + bx + c in the form
y = (x − h)2 + k . We could then easily sketch such graphs by
hand without needing a table of values.
Completing The Square
• Recall that in our unit on quadratic algebra that we used the
process of completing the square to assist us in solving
quadratic equations which did not factorise.
• This same process can be used here to convert quadratics into
the form y = (x − h)2 + k .
• Consider y = x 2 − 4x + 1
y = x 2 − 4x + 22 + 1 − 22 {keeping the equation balanced}
y = x 2 − 4x + 22 − 3
y = (x − 2)2 − 3
• So, y = x 2 − 4x + 1 is really y = (x − 2)2 − 3 and
therefore the graph of y = x 2 − 4x + 1 can be
considered as the graph of y = x 2 after it has
been translated 2 units to the right and 3 units
down.
• Example 1:
Write y = x 2 + 4x + 3 in the form y = (x − h)2 + k using
completing the square and then sketch y = x 2 + 4x + 3 , stating
the coordinates of the vertex.
y = x 2 + 4x + 3
y = x 2 + 4x + (2)2 + 3 − (2)2
y = x 2 + 4x + 4 − 1
y = (x + 2)2 − 1
So, we move the graph of y = x 2 2 units
to the left and 1 unit down.
• If a > 0 , then y = ax 2 opens upwards i.e.,
•
•
•
•
If a < 0 , then y = ax 2 opens downwards i.e.,
Graphs of the form y = ax 2 are ‘wider’ or ‘thinner’ than y = x 2 .
If a < −1 or a > 1, then y = ax 2 is ‘thinner’ than y = x 2 .
If −1 < a < 1 ( a ≠ 0 ), then y = ax 2 is ‘wider’ than y = x 2 .
• Example 2:
Sketch the graph of y = x 2 on a set of axes and then sketch:
(a.) y = 3x 2 (b.) y = −3x 2
(a.) y = 3x 2 is ‘thinner’ than y = x 2 .
(b.) y = −3x 2 is the same shape as
y = 3x 2 , but opens downwards.
• Summary:
• Example 3:
Sketch the graph of y = −(x − 2)2 − 3 from the graph of y = x 2
and then state the coordinates of the vertex.
y = −(x − 2)2 − 3 has a reflection in the x-axis, a horizontal shift
of +2 units and a vertical shift of 3 units down.