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Quadratic Functions, SPM Practice (Short Questions)

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Question 5:
Find the range of values of k if the quadratic equation 3(x2 – kx – 1) = k – k2 has two real and distinc roots.

Solution:
3( x 2 −kx−1 )=k− k 2 3 x 2 −3kx−3−k+ k 2 =0 3 x 2 −3kx+ k 2 −k−3=0 a=3,b=−3k,c= k 2 −k−3 In cases of two real and distinc roots, b 2 −4ac>0 is applied. ( −3k ) 2 −4( 3 )( k 2 −k−3 )>0 9 k 2 −12 k 2 +12k+36>0 −3 k 2 +12k+36>0 − k 2 +4k+12>0 k 2 −4k−12<0 ( k+2 )( k−6 )<0 k=−2,6



The range of values of k is −2<k<6.



Question 6:
Given that the quadratic equation hx2 – (h + 2)x – (h – 4) = 0 has real and distinc roots.  Find the range of values of h.

Solution:
The quadratic equation h x 2 −( h+2 )x−( h−4 )=0 has real and distinc roots. b 2 −4ac>0 is applied. ( −h−2 ) 2 −4( h )( −h+4 )>0 h 2 +4h+4+4 h 2 −16h>0 5 h 2 −12h+4>0 ( 5h−2 )( h−2 )>0 The coefficient of  h 2  is positive,  the region above the x-axis should be shaded. ( 5h−2 )( h−2 )=0 h= 2 5 ,2



The range of values of h for ( 5h−2 )( h−2 )>0 is h< 2 5  or h>2.

3.5b Nature of the Roots (Combination of Straight Line and a Curve)

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Example 2 (Combination of Straight Line and a Curve)
The straight line y = 2 k + 1   intersects the curve y = x + k 2 x   at two distinct points.  Find the range of values of k.



Example 3 (Straight Line does not intersect the curve)
Find the range of values of m for which the straight line y = m x + 6   does not meet the curve 2 x 2 − x y = 3  .

3.4 Quadratic Inequalities (Part 2)

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(C) Linear Inequality

Example 1:

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

Solution:

(D) Quadratic Inequality

Example 2:

Find the range of values of x which satisfy the following inequalities:

  1. (2x + 1) (3x – 1) < 14
  2. (x – 2) (5x – 4) + 1 > 0

Solution:

 

 

 

 

3.4 Quadratic Inequalities (Part 1)

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Solving Quadratic Inequality

  1. The solution of the quadratic inequalities are the ranges of values which satisfied the inequalities itself. 
  2. There are 2 methods to solve the inequalities. 
    1. Graph method
    2. Number line method


Step to solve a Quadratic Inequalities
Step 1 : Rewrite the inequality with zero on one side and make sure a > 0
Step 2 : Find where it crosses the x-axis  (Put y = 0)
Step 3 : Sketch the graph and shade the region to find the range of values of  x


Example 
Find the range of values of x for each of the following :
(a) x 2 − 4 x + 3 < 0
(b) 12 + 10 x − 2 x 2 < 0





Revision - Linear Inequality



Solving Quadratic Inequality - Graph Method



Solving Quadratic Inequality - Number Line Method

Suggested Video

Solving Quadratic Inequalities - The Basics

3.3b Sketching the Graph of Quadratic Functions

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Graphing Quadratic Function

If you are asked to sketch the graph of a quadratic function, you need to show
a. the shape of the graph
b. the maximum/minimum point of the graph
c. the x-intercept of the graph
d. the y-intercept of the graph


Example
Sketch the curve of the quadratic function f ( x ) = x 2 − x − 12

Answer:
The shape of the graph
Since the coefficient of x2 is positive, hence the graph is a U shape parabola with a minimum point.

The minimum point of the graph
By completing the square
f ( x ) = x 2 − x − 12 f ( x ) = x 2 − x + ( 1 2 ) 2 − ( 1 2 ) 2 − 12 f ( x ) = ( x − 1 2 ) 2 − 1 4 − 12 f ( x ) = ( x − 1 2 ) 2 − 12 1 4 Minimum point =  ( 1 2 , − 12 1 4 )

For y-intercept, x = 0
f ( 0 ) = ( 0 ) 2 − ( 0 ) − 12 = − 12

For x-intercept, f(x) = 0
f ( x ) = x 2 − x − 12 0 = x 2 − x − 12 ( x + 5 ) ( x − 6 ) = 0 x = − 5  or  x = 6



Suggested Video

Graphs of Quadratic Function - khanacademy

Algebra - Quadratic Functions (Parabolas) - yaymath

3.3a Finding the Maximum and Minimum Points of Quadratic Function using Completing the Square (Examples)

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Example 1 
Find the maximum or minimum value of each of the following quadratic function by completing the squares.  In each case, state the value of x at which the function is maximum or minimum.  And also, state the maximum or minimum point and axis of symmetry for each case.

(a) f ( x ) = x 2 + 6 x + 7
(b) f ( x ) = 2 x 2 − 6 x + 7
(c) f ( x ) = 5 − 2 x − x 2
(d) f ( x ) = 4 + 12 x − 3 x 2













3.3 Finding the Maximum and Minimum Points of Quadratic Function using Completing the Square

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Steps to convert general form of Quadratic Function into completing the square form

General form of quadratic function : f ( x ) = a x 2 + b x + c

Completing the square form : f ( x ) = a ( x + p ) 2 + q

Step 1 : Make sure the coefficient x 2 is 1, if not factorize.
Step 2 : Insert + ( c o e f f i c i e n t ​ ​ − o f ​ x 2 ) 2 − ( c o e f f i c i e n t ​ ​ − o f ​ x 2 ) 2
Step 3 : Completing the square [convert f ( x ) = a x 2 + b x + c into f ( x ) = a ( x + p ) 2 + q .]