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Quadratic Functions

Quadratic Equations

Problems/Quadratic Functions

Problems/Quadratic Equations

Radicals

Complex Numbers 1

Complex Numbers 2

Reciprocal Functions

Review&Test

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UNIT 3  : QUADRATIC FUNCTIONS & EQUATIONS

 LESSON 9:  REVIEW OF UNIT 3

 

Finding Maximum or Minimum Values of Quadratic Functions:

 

Example:

Write y = -3x2 – 15x – 13  in vertex form  y = a(x – h)2 + k

Solution:

           

 

Because a < 0, the graph is a parabola which opens down and has vertex at .

  Since the parabola opens down, it will have a maximum value at the vertex. 

The maximum value is .  See graph at left.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Quadratic Equations:

 

        

Text Box: Methods of Solving Quadratic Equations:
·	Factoring – see review notes below
·	Completing the square – don’t do this unless you have to!!
·	Quadratic Formula ΰ
 

 

 

 

 

 

 

 


Review of Basic Factoring methods:

 

1. Common Factoring:

Factor 6x3 – 15x

Solution:  6x3 – 15x = 3x(2x2 – 5)                   ** Find the HIGHEST COMMON FACTOR for each term ---  “3x”

                                                                        ** Divide “3x” into each term to get the second factor  --- “2x2 – 5”

                                                                        ** Check by expanding

 

2. Difference of Squares: Formula --  a2 – b2 = (a – b)(a + b)

Factor 49x2 – 64y2

Solution:  49x2 – 64y2 = (7x – 8y)(7x + 8y)

 

Factor x2 – 9y2

Solution: x2 – 9y2 = (x – 3y)(x + 3y)

 

3. Simple Trinomials: Form  x2 + bx + c  [Coefficient of x2 is 1]

a) Factor x2 + 5x + 6

 

Solution: Recall  x2 + 5x + 6 = (x + __ )(x + __ )                     ** We need two numbers that multiply to +6 and add to +5

                                                                                                ** Check all pairs of factors of 6:          {1, 6}  adds to 7

                                                                                                                                                            {2, 3}  adds to 5

               Hence  x2 + 5x + 6 = (x + 3 )(x + 2)                          ** Check by expanding

 

4. Hard Trinomials: Form  ax2 + bx + c  [Coefficient of x2  does not equal 1]

a) Factor 6m2 – 5m – 4

Solution:

We use the method of decomposition (although there are other methods)

We decompose the middle term “-5m” into two parts using the two clues:

            Multiply to (6)(-4) = -24   and

            Add to “-5”

Strategy: List all pairs of factors of “-24” and find the pair that adds to “-5”

            ** {1, 24}  cannot obtain “-5” with these two factors; 1+24=25; 1-24= -23

                                                                                                                        **  {2, 12} cannot obtain “-5” with these two factors

                                                                                                                        **  {3, 8}  -8 + 3 = -24  Choose these two factors

Hence  6m2 – 5m – 4 = 6m2 – 8m + 3m –4                                                       ** “-5m” broken into two parts “-8m + 3m”

                                    = 2m(3m – 4) + 1(3m – 4)                                           ** Group by twos and common factor

                                    = (3m – 4)(2m + 1)                                                      ** check by expanding                                                                                                

 

Zeros [x-intercepts]of Quadratic Functions:

Example 1:

Find the zeros (x-intercepts) of the quadratic function f(x) = 2x2 + x – 6.

 

Solution:

We are trying to find the points where the graph of the function crosses the x-axis – ie –the x-intercepts.

Let y = f(x) = 0  yielding the quadratic equation    2x2 + x – 6 = 0.  We solve by factoring.

 

                      2x2 + x – 6 = 0

            2x2 + 4x – 3x – 6 = 0                           ** See #4 above – Hard Trinomials – method of decomposition

        2x(x + 2) – 3(x + 2) = 0                           ** Common factor by grouping first two terms and last two terms

                (x + 2)(2x – 3) = 0                           ** Common factor (x + 2)

Hence either x + 2 = 0   or   2x – 3 = 0  and

           

Therefore the zeros are –2 and 3/2.

 

Note:  Finding zeros of a quadratic function always yields a quadratic equation to solve.

 

 

Example:

Solve 3x2 – 7x + 1 = 0 by completing the square.

 

Solution:

 

Example:

Solve 3x2 + 6x + 1 = 0 using the quadratic formula.  Express roots to the nearest hundredth

 

Solution:

 

Finding the number of roots of a Quadratic Equation :

 

Text Box: Method:   use the Discriminant   b2 – 4ac
·	If b2 – 4ac > 0 then the equation has 2 real, distinct roots and the corresponding quadratic function will have two zeros or x-intercepts.  The parabola will cut the x-axis in two distinct points 
·	If b2 – 4ac = 0 the roots are real and equal [1 real root].  The corresponding quadratic function will have one zero or x-intercept.  The parabola will be tangent to the x-axis.
·	If b2 – 4ac < 0 there are no real roots [or 2 imaginary roots].  The corresponding quadratic function will have no zeros or x-intercepts. The parabola is above or below the x-axis.
·	See graphs below to illustrate these cases.
         

 

 

 

 

 

 

 

 

 

 

 

 

 

Two zeros:  b2 – 4ac > 0

One zero: b2 – 4ac = 0

No zeros: b2 – 4ac < 0

 

 

 

 

 

Here b2 – 4ac > 0 and  the graph of the quadratic function cuts the x-axis in 2 distinct points.

The quadratic equation has 2 real, distinct roots.

Here b2 – 4ac = 0 and  the graph of the quadratic function is tangent to the x-axis yielding 1 x-intercept.

The quadratic equation has 2 real, equal roots.

Some texts say 1 real root for this case.

Here b2 – 4ac < 0 and  the graph of the quadratic function does not cut the x-axis. 

The quadratic equation has no real roots or imaginary roots.

 

Example:

In each case, calculate the value of the discriminant b2 – 4ac and determine the number of zeros of the quadratic function.

a)  f(x) = 2x2 + 3x – 1               b)  g(x) = 16 – x2                      c)  y = 3x2 – 2x + 5                  d)  y = x2 – 6x + 9

 

Solutions:

 

a = 2

b = 3

c= -1

 
a)  f(x) = 2x2 + 3x – 1                                                              

 

   b2 – 4ac = (3)2 – 4(2)(-1)

                 = 9 + 8

                 = 17

Since b2 – 4ac > 0, there are 2 zeros.  The corresponding quadratic equation  2x2 + 3x – 1 = 0 has 2 real distinct roots.

 

a = 3

b = -2

c = 5

 
c)  y = 3x2 – 2x + 5

 

b2 – 4ac = (-2)2 – 4(3)(5)

              = 4 – 60

              = -56            

Since b2 – 4ac < 0, there are no zeros.  The corresponding quadratic equation  3x2 – 2x + 5 = 0 has  no real roots.

 

a = 1

b = -6

c = 9

 
d)  y = x2 – 6x + 9

 

   b2 – 4ac = (-6)2 – 4(1)(9)

                 = 36 – 36

                 = 0             

Since b2 – 4ac = 0, there is 1 zero.  The corresponding quadratic equation  x2 – 6x + 9 = 0 has  1 real root or real and equal roots.

 

Example:

In each case, determine the number x-intercepts of the quadratic function.

Solutions:

 

a)  f(x) = 5x2 – 3x + 2              

a = 5

b = -3

c = 2

 
                       

 

  b2 – 4ac = (-3)2 – 4(5)(2)

                = 9 – 40

                = -31

Since b2 – 4ac < 0, there are no x-intercepts or zeros.  The corresponding quadratic equation  5x2 – 3x + 2 = 0 has  no real roots

           

Example:

For what value of k does the equation   kx2 – 4x + 2 = 0 have:

a) exactly one root [real equal roots] ?

b) 2 real distinct roots ?

 

Solutions:

a = k

b = -4

c = 2

 
a) For 1 real root set b2 – 4ac = 0                                

                       (-4)2 – 4(k)(2) = 0

                                 16 – 8k = 0

                                       - 8k = -16

                                           k = 2

 

b)  For 2  real distinct roots set b2 – 4ac > 0

                    (-4)2 – 4(k)(2) > 0

                                6 – 8k > 0

                                   - 8k > -16

                                        k < 2                                    ** The inequality reverses when dividing by a negative

             

 

2.  Quadratic Function Problems (Max/Min Problems)

Text Box: Strategies for Solving Max/Min Problems.
·	Read carefully – draw a diagram where possible
·	Determine what is unknown or what you are asked to find.  Assign variable(s) to these unknowns.
·	Determine the quantity to be maximized or minimized.   This should be clear from a careful reading of the problem.  
·	Write a formula for this quantity.  This is your main function in the problem.
·	If your main function involves two variables, write it as a function of one variable.  There should be a secondary relation in the problem involving the two variables.  Isolate one of the variables and substitute in the main function.
·	Simplify fully and complete the square to find the maximum or minimum value.
 

 

 

 

 

 

 

 

 

 

 

 

 


Area Problems:

Example:

Farmer Al has 240 m of fencing available.  He wishes to enclose a rectangular garden with this fence.  One side borders a stream bank and requires no fence.  Find the dimensions he should use to enclose a field of maximum area.

Solution:

ΰ Determine what is unknown or what you are asked to find?  Assign variables to the unknowns.  Here the dimensions of the field are unknown.

     Let the length be x and the width be y.

 

                          Stream bank     

 

 
 

 


   

           y                                            y

 

 

                                 x                                                                                                                                                          

 

ΰ Determine the quantity to be maximized or minimized.

     We are asked to maximize the area.  Hence our formula is:

                       

                        ** This is your main function for the problem

 

ΰ  Write this formula as a  function of  one variable.  Since the available fencing is240 m, we have a secondary relation between the two variables.

 

           

3.  Quadratic Equation Problems

Example:

A farmer has 80 m of fencing available.  He wishes to enclose a rectangular field of area 300 m2.  Find the dimensions of the field.

 

                                   x

 

 
 

 


           y

 

 

 

 

Solution:

ΰ Determine what is unknown ?  Assign variables to the unknowns.  Here we are asked to find the width and length of the field.

Let the length of the field be x and the width be y.

ΰ  We are told the area is 300 m2.  Write this statement as an equation.

                                   

ΰ  Write this as an equation in one variable.  Since the perimeter is 80, we have a secondary relation between the two variables.

 

           

 

 

 

Text Box: Strategies for Solving Quadratic Equations Problems.
·	Determine what is unknown or what you are asked to find. Draw a diagram if possible.
·	Assign variables to these unknowns.  You can use a single variable as in examples 3 & 4 or use two variables as in examples 1 & 2.  These are your let statements.
·	Translate the information in the problem into an algebraic equation.
·	Write this as an equation in one variable if needed  [ see examples 1 & 2 ].
·	Put all the terms on one side of the equation and solve by factoring or the quadratic formula.
 

 

 

 

 

 

 

 

 

 

 


Properties of Radicals:

 

Text Box: 	        
 
                       

 

 

 

 

 

 

 

 

 

Example:  Write as mixed radicals.

                                                                                                                                  

 

Solutions:

                                                           

                                               

 

 

Text Box: Tips for Changing to Mixed Radicals:
·	Break the radical into two factors
·	Use the following set of perfect squares to get the first factor 
·	{4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, …}
·	Take the square root of the first number to get the first factor of the mixed radical.                                             
                       

 

 

 

 

 

 

 

Example:

 

Example:  Simplify

                                            

 

Solutions:

Again we follow the rules of ordinary algebra for expanding

 

Recall from ordinary algebra the product of two binomials:

           

            (x + 3)(x + 5) = x(x + 5) + 3(x + 5)

                                  = x2 + 5x + 3x + 15

                                  = x2 + 8x + 15

 

 

Example 8:  Simplify by rationalizing the denominator.

Example 10: Simplify

           

 

 

Solving Quadratic Equations with Non-Real Roots:

 

Example 3:    Solve 3x2 – 4x + 10 = 0  where x is a complex number.  Round roots to nearest hundredth.

 

Solution:

           

 

Text Box: Recall roots of quadratics in factored form:

Theorem: If m and n are the roots of a quadratic equation, then the equation in factored form is  

						(x – m)(x – n) = 0
 

 

 

 

 

 

 

 

 


Example 3:  Find the quadratic equation in factored form whose roots are –4, 3

Solution:

                        (x - (-4))(x – 3) = 0

                           (x + 4)(x – 3) = 0

 

Example 4:  Find the quadratic equation in factored form if one root is 2 + 3i

Solution:

Since complex numbers occur in conjugate pairs, the other root is 2 – 3i.  Hence the equation is

                        [x - (2 + 3i)][x – (2 – 3i)] = 0    or

                          (x – 2 – 3i)(x – 2 + 3i) = 0

 

Operations with Complex Numbers:

 

Example 2:

                                         

                                               

Solutions:

 

                                   

 

                                   

Complex Conjugates:

 

Complex conjugates are often denoted using the notation  .

Example 3:

                                         

 

Note that in all cases the product of a complex number and its conjugate is a real number.

 

Graphing functions and their reciprocals:

 

Example 1:

 

 

x

-4

- 6

-1/6 

-2

- 4

- Ό

-1

- 3

-1/3

0

-2

1

-1

-1

1.5

-0.5

-2

1.75

-0.25

-4

2

0

1/0 = undefined

2.25

0.25

4

2.5

0.5

2

3

1

1

4

2

½

5

3

1/3

 

 

 

 

 

 

 

 

Please note the following from the graph and table:

·        The graph of y = f(x) =  x – 2 (blue) is a line and its reciprocal (red) has 2 branches separated by the line x = 2 (dashed). It is called a hyperbola.

·        Where f(x) has a zero (x – intercept), the reciprocal has an asymptote (x = 2 dashed)

·        The behaviour near the asymptote is interesting;  as x approaches 2 from the right (x = 3, 2.5, 2.25 …), the reciprocal (red) gets very large in the positive direction;  as x approaches 2 from the left (x = 1, 1.5, 1.75 …), the reciprocal (red) gets very large in the negative direction.

·        As x takes on larger positive values, the reciprocal takes on smaller values approaching zero from above.  As x takes on larger negative values                       (-1,-2, -4, -10, …), the reciprocal takes on smaller values approaching zero again from below.

·        Where f(x) is positive, the reciprocal is positive;  where f(x) is negative, the reciprocal is negative.

·        Where f(x) = 1, the reciprocal equals 1;  where f(x) = -1, the reciprocal equals –1.

 

Text Box: Strategies for plotting linear reciprocal graphs:
·	Find the zero (x – int) of the original graph.  This is the asymptote of the reciprocal.
·	Make a table of values like above including points near the asymptote. 
·	Use the 1/x key on your calculator to get the reciprocal values in the third column of the table.
·	Where f(x) is positive, the reciprocal is positive;  where f(x) is negative, the reciprocal is negative.
 

 

 

 

 

 

 

 


Example 3:

Solution:

First find the zeros or x – intercepts.  Let y = 0 and solve for x.

           

Next find the vertex.

Now make a table of values as above including points near the asymptotes x = -2 and x = 4

 

           

x

-4

12

1/12

-3

7

1/7

-2.5

3.25

0.31

-2.1

0.61

1.64

-2.01

0.0601

16.64

-2

0

Undef.

-1.99

-0.0599

-16.69

-1.9

-0.59

-1.69

1

-9

-1/9

3.9

-0.59

-1.69

4

0

Undef

4.1

0.61

1.64

5

7

1/7

6

12

1/12

 

            Note values of x taken near asymptotes:

            x = - 2.1, -2.01, - 1.9, -1.99, 3.9, 4.1

 

 

 

 

 

 

Please note the following from the graph and table:

·        The graph of y = f(x) =  x2 – 2x - 8 (blue) is a parabola with vertex at (1, -9) and zeros –2, 4. Its reciprocal (red) has 3 branches separated by the lines

      x = - 2 and x = 4(dashed). 

·        Where f(x) has a zero (x – intercept), the reciprocal has an asymptote (x = - 2 and x = 4 dashed)

·        The behaviour near the asymptotes is interesting;  as x approaches - 2 from the right (x = -1.9, -1.99 in table), the reciprocal (red) gets very large in the negative direction (goes down);  as x approaches - 2 from the left (x = - 2.1, -2.01 in table), the reciprocal (red) gets very large in the positive direction (goes up).  Similar behaviour occurs near the other asymptote x = 4.

·        As x takes on larger positive values (x = 5, 6 in table), the reciprocal takes on smaller values approaching zero from above.  As x takes on larger negative values (- 3, - 4 in table), the reciprocal takes on smaller values approaching zero again from above.

·        Where f(x) is positive, the reciprocal is positive;  where f(x) is negative, the reciprocal is negative.

·        Where f(x) = 1, the reciprocal equals 1;  where f(x) = -1, the reciprocal equals –1.

 

 

 

Text Box: Strategies for plotting reciprocal quadratic graphs:
·	Find the zeros (x – int’s) of the original graph.  These are the asymptotes of the reciprocal
·	Find the vertex and its reciprocal.  This gives a point between the asymptotes.
·	Use the 1/x key on your calculator to get the reciprocal values in the third column of the table.
·	Where f(x) is positive, the reciprocal is positive;  where f(x) is negative, the reciprocal is negative.
·	 Where f has large positive values, the reciprocal will have small positive values near zero
·	 Where f has large negative values, the reciprocal will have small negative values near zero
·	Where f(x) = 1, the reciprocal equals 1;  where f(x) = –1, the reciprocal equals –1.
·	Make a table of values like above including points near the asymptotes. 
·	Use the mapping  (x, y) -------ΰ (x, 1/y)  to determine the points on the reciprocal graph.
 

 

 

 

 

 

 

 

 

 

 

 

 

 


           

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