      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.  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

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

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

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 : 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) 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:

à 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

à 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

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

à 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.

à  We are told the area is 300 m2.  Write this statement as an equation. à  Write this as an equation in one variable.  Since the perimeter is 80, we have a secondary relation between the two variables.     Solutions:   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:  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 - ¼ -1 - 3 -1/3 0 -2 -½ 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. 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. <% End If %>