Page 54 - Mathematics_Form_3

Functions

Alternatively, the range can be found y
by completing the square as follows: 2
2
y = x − 4x + 5
-2 0 2 4 6 8 x
2
y = x − 4x + 4 + 5 − 4 -2
FOR ONLINE READING ONLY
y = (x − 2) + 5 − 4 -4 f(x)=-x +4x-5
2
2
y = (x − 2) + 1 -6
2
2
Since the term ( x − ) 2 is a squared -8
expression which is always non- -10 f(x)=-10
negative ( 0≥ ) . Adding 1 to it makes -12
the smallest possible value of y = 1
(when x = ) 2 . As x increases from 2, To solve − x + 4x − 5 = − 10, draw
2
y increases without bound. the line represented by y = − 10 on

Therefore, range = { :yy ≥ } 1 . the same graph. The equation of line
is obtained as follows:
Subtracting y = − x + 2 4x − 5 and
Example 2.20
− 10 = − x + 2 4x − 5 gives y = − 10 .

Without using a table of values, draw
the graph of f (x ) = − x + 4x − 5 and Read the x-coordinates at the
2

use it to solve the equation intersection of the graph and the
− x + 4x − 5 = − 10. straight line. Therefore, the solutions
2
to the equation − x + 4x − 5 = − 10 are
2
Solution
y = − ( x − 4x ) − 5 x = − 1 and x = 5.
2

2
= − ( x − 4x + 4 ) − 5 + 4
2
Therefore, y = − (x − 2) − 1 . Example 2.21
Thus, the maximum point is (2,−1) and Find the maximum or minimum value
the axis of symmetry of the quadratic of the function f (x ) = 4 − 3x − 2 x .
2
function is x = 2. In order to find the
point(s) at which the graph crosses the Solution Mathematics for Secondary Schools
2
y-axis, set x = 0 in the function to get From f (x ) = 4 − 3x − 2 x it follows
= −5. Since the value of a is negative, that a = − 2, b = − 3, and c = 4.
the graph opens downwards. Also, since Since the coefficient of x is negative,
2
the turning point is below the -axis, the function has a maximum value at
the graph does not cross the -axis. the turning point. Thus, the maximum
value is given by
The graph of ()fx = − x + 2 4x − 5 is 2
_
shown in the following figure. y = 4ac − b

4a

Student\s Book Form Three 47

18/09/2025 09:58:58
MATHEMATIC F3 SB.indd 47
MATHEMATIC F3 SB.indd 47 18/09/2025 09:58:58