Page 24 - Focus SPM KSSM Tg 4.5 - Add Maths

Additional Mathematics SPM Chapter 2 Quadratic Functions

E Sketching the graph of quadratic Therefore, the graph of f(x) intersects the x-axis at two
functions different points.
f(x) = 2x + 3x – 2
2
Sketching the graph of a quadratic function  3 
2
f(x) = ax + bx + c = 2 x + x – 1
2
2
3
3
3
2

Determine the shape of the graph of the = 2 x + x +     2 – 1 
–
2
quadratic function by identifying the value of a. 2 2 4 4
9

= 2 x + 3  – 16 – 1 
4

Determine the position of the graph of the = 2 x + 3  2 – 25
quadratic function by calculating the value of 4 8
3

2
the discriminant b – 4ac. Minimum point is – , – 25  . Form 4
4
8
When f(x) = 0, 2x + 3x – 2 = 0
2
(2x – 1)(x + 2) = 0
Determine the maximum or minimum point
by expressing the quadratic function in the 2x – 1 = 0 or x + 2 = 0
1
2
vertex form f(x) = a(x – h) + k where the x = x = –2
2
vertex is (h, k) and the equation of the axis of
symmetry is x = h. When x = 0, f(0) = –2.
3 f(x)
x = – —
4
x
1
Solve the quadratic equation f(x) = 0 to –2 0 —
determine the points of intersection on –2 2
3
the x-axis, which are the real roots of the ( – , – 25 )
—
—
4
8
quadratic equation, if the roots exist.
3
The equation of the axis of symmetry is x = – .
4
Determine the point of intersection of the Try Question 29 in ‘Try This! 2.3’
graph with the y-axis by finding the value of
f(x) when x = 0.
F Solving problems involving quadratic
functions
Draw a smooth parabola which is symmetrical 27
at the line x = h and passes through all the The curve of the quadratic function f(x) = –2(x – p) + 2q
2
points that have been determined. cuts the x-axis at points (1, 0) and (5, 0). The line y = 8
touches the curve at its maximum point.
(a) Find the values of p and q.
26 (b) Sketch the graph of f(x) for 0 < x < 6.
Sketch the graph for the quadratic function (c) If the graph is reflected on the x-axis, write the
f(x) = 2x + 3x – 2. Hence, state the axis of symmetry equation of the curve.
2
for the graph of the quadratic function. Solution
Solution (a) f(x) = –2(x – p) + 2q
2
f(x) = 2x + 3x – 2
2
a = 2 . 0 The coordinates of vertex of f(x) are (p, 2q).
The graph is in the shape of with a minimum point. p = 1 + 5 2q = 8
2
b – 4ac = 3 – 4(2)(–2) = 3 q = 4
2
2
= 25 . 0
47