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Quadratic Function
Quadratic Graph
Quadratic Functions and Their Graphs
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Let
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ave
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Quadratic Function
Could you give me an example of a function?
Example
y = 4x + 3 y = x2 + 3
y = , where x 0 f(x) = , where x 0
f(x) = x2 + 2x + 1
y is a function of x if there exists a relationship between x and y
such that every value of the independent variable x determines
exactly one corresponding value of the dependent value of y.
xx
1
2
Quadratic Function
Consider f(x) = x2 + 2x + 1.
What is the index of x for the term x2?
What is the index of x for the term 2x?
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x2 + 2x + 1
Index of x = 2 Index of x = 1 ‧Index of x = 0 ‧Constant term
Quadratic function a b c
The following are examples of quadratic functions.
Quadratic Function
A quadratic function in x can be written in the form
f(x) = ax2 + bx + c, or
y = ax2 + bx + c
where a, b and c are constants with a 0.
y = –x2
y = x2 + 2x + 1
f(x) = 3x2 – 5x – 2
–1 0 0
1 1
3 –5 –2
2
4
How could we check whether a given
function is a quadratic function?
Quadratic Function
A quadratic function in x can be written in the form
y = ax2 + bx + c
where a, b and c are constants with a 0 .
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Function Quadratic
function? Reason
Quadratic Function
Check whether each of the following functions is a quadratic function.
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The index of x is not 2. No
Yes
No
y = 2x + 1
y = x2 + 1
No a = 0. It’s a linear function.
x2y = x2 +
y = x
1
For , the index of x is
neither 2 nor 1.
x2
a = 1, b = 0, c = 1
Quadratic Graph
It’s a parabola.
It opens upwards.
x-intercept is 0.
y-intercept is 0.
It is symmetrical about x = 0.
There is a vertex with coordinates (0, 0).
The graph of a quadratic function is called a quadratic graph.
Could you describe the quadratic graph of y = x2?
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Quadratic Graph
Could you describe the quadratic graph of y = –x2 + 4x – 5?
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It’s a parabola.
It opens downwards.
There is no x-intercept.
y-intercept is –5.
It is symmetrical about x = 2.
There is a vertex with coordinates
(2, –1).
Quadratic Graph
Let’s investigate a property of a parabola,
i.e. the graph of y = ax2 + bx + c.
When a > 0, the parabola opens upwards.
When a < 0, the parabola opens downwards.
Open upwards Open downwards Open upwards
y = x2 a > 0 y = –x2 + 2x +3 a < 0 y = x2 – 3x + 2 a > 0
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Quadratic Graph
Example 1
It opens downwards.
Find the direction of opening for each of the following quadratic graphs. (a) (b)
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It opens upwards.
Quadratic Graph
The parabola is symmetrical about a vertical line,
which is called the axis of symmetry of the parabola.
x = 0 x = 1 x = 1.5
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Let’s investigate another property of a parabola.
Quadratic Graph
Example 2
The axis of symmetry is x = 2.
Find the axis of symmetry for each of the following quadratic graphs. (a) (b)
x = 2 x = –1
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The axis of symmetry is x = –1.
Quadratic Graph
Consider the graph of the function y = x2.
As a > 0, the parabola opens upwards.
The axis of symmetry is x = 0.
The vertex of a parabola is the point of
intersection of the axis of symmetry and
the parabola.
In this case, the coordinates of the vertex
of the graph of y = x2 are (0, 0).
y = x2
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x = 0
vertex
Quadratic Graph
Consider the graph of the function y = x2.
As a > 0, the parabola opens upwards.
The graph has the lowest point (0, 0).
The y-coordinate of the lowest point is the
minimum value of the function.
In this case, the minimum value of the
function y = x2 is 0.
y = x2
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Quadratic Graph
y = –x2 + 2x + 3 15
Consider the graph of the function y = –x2 + 2x + 3.
As a < 0, the parabola opens downwards.
The graph has the highest point (1, 4),
which is the vertex of the graph.
The y-coordinate of the highest point is
the maximum value of the function.
In this case, the maximum value of the
function y = –x2 + 2x + 3 is 4.
Quadratic Graph
Example 3
The figure shows the graph of a
quadratic function. Find the
maximum or minimum value of
the function.
As the graph of the function
opens downwards, the
function has a maximum
value of –1.
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Quadratic Graph
Example 4
The figure shows the graph of a
quadratic function. Find the
maximum or minimum value of
the function.
As the graph of the function
opens upwards, the function
has a minimum value of –4.
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Quadratic Graph
Do you remember how to solve the quadratic equation
x2 – 3x + 2 = 0 by graphical method?
Step 1: Draw the graph of y = x2 – 3x + 2.
The graph cuts the x-axis at two points.
Step 2: When y = 0, x = 1 or x = 2.
These two values of x are the two x-intercepts
of the graph y = x2 – 3x + 2.
Step 3: These two values of x are the roots of the
quadratic equation x2 – 3x + 2 = 0.
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Quadratic Graph
Consider a quadratic equation ax2 + bx + c = 0.
If the parabola y = ax2 + bx + c cuts the x-axis at
(, 0) and (, 0), and are called the x-intercepts.
and are also the roots of the quadratic equation
ax2 + bx + c = 0.
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Quadratic Graph
Example 5
Find the x-intercepts of the
quadratic graph shown.
The x-intercepts are 1
and –3.
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Quadratic Graph
Example 6
Find the x-intercept of the
quadratic graph shown.
The x-intercept is 2.
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Quadratic Graph
Example 7
Find the x-intercepts of the
quadratic graph shown.
The graph has no x-intercepts.
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Quadratic Graph
A parabola must cut the y-axis at only one point.
Let the point be (0, c). c is called the y-intercept.
y = –x2 + 2x + 3
The y-intercept of the graph of
y = –x2 + 2x + 3 is 3.
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Quadratic Graph
Example 8
Find the y-intercept of the
quadratic graph shown.
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The y-intercept is –3.
Quadratic Graph
Let’s describe the geometric
features of the quadratic
graph of y = –x2 + 2x + 3.
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Quadratic Graph
Graph
It’s a parabola.
It opens downwards.
The x-intercepts are –1 and 3.
The y-intercept is 3.
It is symmetrical about x = 1.
The coordinates of the vertex
are (1, 4).
Function
The maximum value of the
function y = –x2 + 2x + 3 is 4.
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Quadratic Graph
a > 0 a < 0
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Quadratic Graph
Example 9
Find the direction of opening, the y-intercept and the x-intercept(s) of the
graph of each of the following functions.
(a) y = 3x2 – 5x – 2 (b) y = –9 – x2 + 6x (c) y = (x + 1)(4 – x) – 7
(a) Consider y = 3x2 – 5x – 2.
The coefficient of x2 is 3 > 0. The graph opens upwards.
When x = 0, y = 3(0)2 – 5(0) – 2 = –2.
∴ The y-intercept is –2.
When y = 0, 3x2 – 5x – 2 = 0
(3x + 1)(x – 2) = 0
3x + 1 = 0 or x – 2 = 0
x = or x = 2
∴ The x-intercepts are and 2.
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3
1
3
1
Quadratic Graph
Example 9
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Find the direction of opening, the y-intercept and the x-intercept(s) of the
graph of each of the following functions.
(a) y = 3x2 – 5x – 2 (b) y = –9 – x2 + 6x (c) y = (x + 1)(4 – x) – 7
(b) Consider y = –9 – x2 + 6x = –x2 + 6x – 9.
The coefficient of x2 is –1 < 0. The graph opens downwards.
When x = 0, y = –(0)2 + 6(0) – 9 = –9.
∴ The y-intercept is –9.
When y = 0, –x2 + 6x – 9 = 0
x2 – 6x + 9 = 0
(x – 3)2 = 0
x = 3 (repeated)
∴ The x-intercept is 3.
Quadratic Graph
Example 9
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Find the direction of opening, the y-intercept and the x-intercept(s) of the
graph of each of the following functions.
(a) y = 3x2 – 5x – 2 (b) y = –9 – x2 + 6x (c) y = (x + 1)(4 – x) – 7
(c) Consider y = (x + 1)(4 – x) – 7 = –x2 + 3x – 3.
The coefficient of x2 is –1 < 0. The graph opens downwards.
When x = 0, y = –(0)2 + 3(0) – 3 = –3.
∴ The y-intercept is –3.
When y = 0, –x2 + 3x – 3 = 0.
= (3)2 – 4(–1)(–3)
= –3
< 0
∴ The graph has no x-intercepts.
Quadratic Graph
Example 10
Consider the function y = –x2 + 2x + 5.
(a) Complete the following table.
(b) Plot the graph of y = –x2 + 2x + 5 from x = –2 to x = 4.
(c) Find the maximum or minimum value of the function.
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x
y
–1 0 1 2 3 4 –2
Quadratic Graph
Example 10
Consider the function y = –x2 + 2x + 5.
(a) Complete the following table.
x
y
–1 0 1 2 3
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(a) x
y
–2 0 1 2 4
–3 5 6 5 –3
–1
2
3
2
4 –2
Quadratic Graph
Example 10
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(b) Plot the graph of y = –x2 + 2x + 5 from x = –2 to x = 4.
(b) y = –x2 + 2x + 5
Quadratic Graph
Example 10
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(c) Find the maximum or minimum value of the function.
(c) As the graph opens
downwards, the
function has a
maximum value of 6.
y = –x2 + 2x + 5
Quadratic Graph
Example 11
The figure shows the graph of y = –3x2 + 12x – (k + 2) with the axis of symmetry x = 2. Find (a) the value of k, (b) the coordinates of the vertex.
(a) Since the y-intercept is –5,
–(k + 2) = –5
k + 2 = 5
k = 3
(b) Put x = 2 and k = 3 into the function.
y = –3(2)2 + 12(2) – (3 + 2)
= 7
∴ The coordinates of the vertex are (2, 7).
O x
y
–5
y = –3x2 + 12x – (k + 2)
x = 2
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