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Quadratic Functions

Quadratic Functions, SPM Practice (Short Questions)

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Question 10 (3 marks):
Diagram shows the graph y = a (x – p)2 + q, where a, p and q are constants. The straight line y = –8 is the tangent to the curve at point H.

Diagram

(a) State the coordinates of H.
(b) Find the value of a.

Solution:
(a)
Coordinate x of H = −1+7 2 = 6 2 =3 Thus, coordinates of H=( 3,−8 ).

(b)
y=a ( x−p ) 2 +q y=a ( x−3 ) 2 +( −8 ) y=a ( x−3 ) 2 −8 ......... ( 1 ) Substitute ( 7,0 ) into ( 1 ): 0=a ( 7−3 ) 2 −8 0=16a−8 16a=8 a= 1 2



Question 11 (3 marks):
Faizal has a rectangular plywood with a dimension 3x metre in length and 2x metre in width. He cuts part of the plywood into a square shape with sides of x metre to make a table surface.
Find the range of values of x if the remaining area of the plywood is at least (x2 + 4) metre2.

Solution:



Area of plywood – area of square ≥ (x2 + 4)
3x(2x) – x2 ≥ x2 + 4
6x2 – x2 – x2 ≥ 4
4x2 ≥ 4
x2 – 1 ≥ 0
(x + 1)(x – 1) ≥ 0
x ≤ –1 or x ≥ 1
Thus, x ≥ 1 (length is > 0)




Quadratic Functions, SPM Practice (Short Questions)

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Question 7:
The diagram below shows the graph of the quadratic function f(x) = (x + 3)2 + 2h – 6, where h is a constant.


(a) State the equation of the axis of symmetry of the curve.
(b) Given the minimum value of the function is 4, find the value of h.

Solution:
(a)
When x + 3 = 0
 x = –3
Therefore, equation of the axis of symmetry of the curve is x = –3.

(b)
When x + 3 = 0, f(x) = 2h – 6
Minimum value of f(x) is 2h – 6.
2h – 6 = 4
2h = 10
h = 5



Question 8 (4 marks):
The quadratic function f is defined by f(x) = x2 + 4x + h, where h is a constant.
(a) Express f(x) in the form (x + m)2 + n, where m and n are constants.

(b) Given the minimum value of f(x) is 8, find the value of h.

Solution:
(a)
f(x) = x2 + 4x + h
   = x2 + 4x + (2)2 – (2)2 + h
   = (x + 2)2 – 4 + h

(b)
Given the minimum value of f(x) = 8
– 4 + h = 8
h = 12



Question 9 (3 marks):
Find the range of values of x such that the quadratic function f(x) = 6 + 5x – x2 is negative.

Solution:
(a)
f(x) < 0
6 + 5x – x2 < 0
(6 – x)(x + 1) < 0
x < –1, x > 6



3.9.4 Quadratic Functions, SPM Practice (Long Question)

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Question 6:
Quadratic function f(x) = x2 – 4px + 5p2 + 1 has a minimum value of m2 + 2p, where m and p are constants.
(a) By using the method of completing the square, shows that m = p – 1.
(b) Hence, find the values of p and of m if the graph of the quadratic function is symmetry at x = m2 – 1.

Solution:
(a)
f( x )= x 2 −4px+5 p 2 +1 = x 2 −4px+ ( −4p 2 ) 2 − ( −4p 2 ) 2 +5 p 2 +1 = ( x−2p ) 2 + p 2 +1 Minimum value, m 2 +2p= p 2 +1 m 2 = p 2 −2p+1 m 2 = ( p−1 ) 2 m=p−1

(b)
x= m 2 −1 2p= m 2 −1 p= m 2 −1 2 Given m=p−1⇒p=m+1 m+1= m 2 −1 2 2m+2= m 2 −1 m 2 −2m−3=0 ( m−3 )( m+1 )=0 m=3 or −1 When m=3, p= 3 2 −1 2 =4 When m=−1, p= ( −1 ) 2 −1 2 =0

Quadratic Functions, SPM Practice (Long Questions)

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Question 4:
(a) Find the range of values of k if the equation x2 – kx + 3k – 5 = 0 does not have real roots.
(b) Show that the quadratic equation hx2 – (h + 3)x + 1 = 0 has real and distinc roots for all values of h.

Solution:
(a)
x 2 −kx+( 3k−5 )=0 If the above equation has no real root, ∴  b 2 −4ac<0. k 2 −4( 3k−5 )<0 k 2 −12k+20<0 ( k−2 )( k−10 )<0

Graph function y = (k – 2)(k – 10) cuts the horizontal line at k = 2 and k = 10 when b2 – 4ac < 0.



The range of values of k that satisfy the inequality above is 2 < k < 10.

(b)
h x 2 −( h+3 )x+1=0 b 2 −4ac= ( h+3 ) 2 −4( h )( 1 ) = h 2 +6h+9−4h = h 2 +2h+9 = ( h+ 2 2 ) 2 − ( 2 2 ) 2 +9 = ( h+1 ) 2 −1+9 = ( h+1 ) 2 +8

The minimum value of (h + 1) + 8 is 8, a positive value. Therefore, b2 – 4ac > 0 for all values of h.
Hence, quadratic equation hx2 – (h + 3)x + 1 = 0 has real and distinc roots for all values of h.

Quadratic Functions, SPM Practice (Long Questions)

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Question 3:
Given that the quadratic function f(x) = 2x2 – px + p has a minimum value of –18 at x = 1.
(a) Find the values of p and q.
(b) With the value of p and q found in (a), find the values of x, where graph f(x) cuts the x-axis.
(c) Hence, sketch the graph of f(x).

Solution:
(a)
f( x )=2 x 2 −px+q =2[ x 2 − p 2 x+ q 2 ] =2[ ( x+ −p 4 ) 2 − ( −p 4 ) 2 + q 2 ] =2[ ( x− p 4 ) 2 − p 2 16 + q 2 ] =2 ( x− p 4 ) 2 − p 2 8 +q


∴ p 4 =1−−−−−−( 1 ) and − p 2 8 +q=−18−−−( 2 ) From( 1 ),p=4. Substitute p=4 into ( 2 ): − ( 4 ) 2 8 +q=−18   − 16 8 +q=−18  q=−18+2    =−16


(b)
f( x )=2 x 2 −4x−16 f( x )=0 when it cuts x-axis 2 x 2 −4x−16=0 x 2 −2x−8=0 ( x−4 )( x+2 )=0 x=4,−2 Graph f( x ) cuts x-axis at x=−2 and x=4.

(c)

Quadratic Functions, SPM Practice (Short Questions)

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Question 3:
The straight line y = 5x – 1 does not intersect with the curve y = 2x2 + x + h.
Find the range of values of h.

Solution:
y=5x−1         ...... (1) y=2 x 2 +x+h ...... (2) Substitute (1) into (2), 5x−1=2 x 2 +x+h 2 x 2 +x+h−5x+1=0 2 x 2 −4x+h+1=0                   b 2 −4ac<0 ( −4 ) 2 −4( 2 )( h+1 )<0              16−8h−8<0                             8<8h                             h>1

Question 4:
Find the maximum value of the function 5 – x – 2x2 , and the corresponding value of x.

Solution:
5−x−2 x 2 =−2 x 2 −x+5 =−2[ x 2 + 1 2 x− 5 2 ] =−2[ x 2 + 1 2 x+ ( 1 4 ) 2 − ( 1 4 ) 2 − 5 2 ] =−2[ ( x+ 1 4 ) 2 − 1 16 − 5 2 ] =−2[ ( x+ 1 4 ) 2 − 41 16 ] =−2 ( x+ 1 4 ) 2 +5 1 8


5−x−2 x 2  has a maximum value when 2 ( x+ 1 4 ) 2 =0               x=− 1 4 The maximum value of 5−x−2 x 2  is 5 1 8 .

Quadratic Functions, SPM Practice (Short Questions)

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Question 1:

Find the minimum value of the function f (x) = 2x2 + 6x + 5. State the value of xthat makes f (x) a minimum value.

Solution:

By completing the square for f (x) in the form of f (x) = a(x + p)2 + q to find the minimum value of f (x).

f ( x ) = 2 x 2 + 6 x + 5 = 2 [ x 2 + 3 x + 5 2 ] = 2 [ x 2 + 3 x + ( 3 × 1 2 ) 2 − ( 3 × 1 2 ) 2 + 5 2 ]
= 2 [ ( x + 3 2 ) 2 − 9 4 + 5 2 ] = 2 [ ( x + 3 2 ) 2 + 1 4 ] = 2 ( x + 3 2 ) 2 + 1 2

Since a = 2 > 0,
Therefore f (x) has a minimum value when x = − 3 2 . . The minimum value of f (x) = ½. 

Question 2:

The quadratic function f (x) = –x2 + 4x + k2, where k is a constant, has a maximum value of 8.
Find the possible values of k.

Solution:
f (x) = –x2 + 4x + k2
f (x) = –(x2 – 4x) + k2 ← [completing the square for f (x) in
the form of f (x) = a(x + p)2+ q]
f (x) = –[x2 – 4x + (–2)2 – (–2)2] + k2
f (x) = –[(x – 2)2 – 4] + k2
f (x) = –(x – 2)2 + 4 + k2

Given the maximum value is 8.
Therefore, 4 + k2 = 8
k2 = 4
k = ±2

 

 

 

 

3.4 Quadratic Inequalities (Part 2)

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(C) Linear Inequality 

Example 1
(a)  Given x = 6 − y 3  , find the range of values of x for which y > 9 .
(b) Given 2 x + 3 y − 6 = 0  , find the range of values of x for which y < 4 .






(D) Quadratic Inequalities 

Example 2
Find the range of values of x which satisfy  the following inequalities:
(a) (2x + 1) (3x – 1) < 14
(b) (x– 2) (5x – 4) + 1 > 0

3.2 Maximum and Minimum Value of Quadratic Functions

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Maximum and Minimum Point

  1. A quadratic functions f ( x ) = a x 2 + b x + c can be expressed in the form f ( x ) = a ( x + p ) 2 + q by the method of completing the square.
  2. The minimum/maximum point can be determined from the equation in this form f ( x ) = a ( x + p ) 2 + q .
Minimum Point
  1. The quadratic function f(x) has a minimum value if a is positive. 
  2. The quadratic function f(x) has a minimum value when (x + p) = 0
  3. The minimum value is equal to q.
  4. Hence the minimum point is (-p, q)

Maximum Point

  1. The quadratic function f(x) has a maximum value if a is negative.
  2. The quadratic function f(x) has a maximum value when (x + p) = 0
  3. The maximum value is equal to q.
  4. Hence the maximum point is (-p, q)



Example
Find the maximum or minimum point of the following quadratic equations
a. f ( x ) = ( x − 3 ) 2 + 7
b. f ( x ) = − 5 − 3 ( x + 15 ) 2

Answer:
(a)
f ( x ) = ( x − 3 ) 2 + 7 a = 1 , p = − 3 , q = 7 a > 0 ,  the quadratic function has a minimum point Minimum point = ( − p , q ) = ( 3 , 7 )

(b)
f ( x ) = − 5 − 3 ( x + 15 ) 2 a = − 3 ,   p = 15 ,   q = − 5 a < 0 ,  the quadratic function has a maximum point Maximum point = ( − p , q ) = ( − 15 , − 5 )

Quadratic Functions, SPM Practice (Long Questions)

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Question 5:


Diagram above shows the graphs of the curves y = x2 + x – kx + 5 and y = 2(x – 3) – 4h that intersect the x-axis at two points. Find
(a) the value of k and of h,
(b) the minimum value of each curve.


Solution:
(a)
y= x 2 +x−kx+5 = x 2 +( 1−k )x+5 = [ x+ ( 1−k ) 2 ] 2 − ( 1−k 2 ) 2 +5 axis of symmetry of the graph is x=− ( 1−k ) 2

y=2 ( x−3 ) 2 −4h axis of symmetry of the graph is x=3. ∴ − 1−k 2 =3 −1+k=6 k=7

Substitute k=7 into equation y= x 2 +x−7x+5   = x 2 −6x+5 At x-axis,y=0; x 2 −6x+5=0 ( x−1 )( x−5 )=0 x=1,5

At point ( 1,0 ) Substitute x=1,y=0 into the graph: y=2 ( x−3 ) 2 −4h 0=2 ( 1−3 ) 2 −4h 4h=2( 4 ) 4h=8 h=2

(b)
For y= x 2 −6x+5 = ( x−3 ) 2 −9+5 = ( x−3 ) 2 −4 ∴ Minimum value is −4. For y=2 ( x−3 ) 2 −8, minimum value is−8.