
UNIT 3 : QUADRATIC FUNCTIONS & EQUATIONS
LESSON
9: REVIEW OF UNIT 3
Finding
Maximum or Minimum Values of Quadratic Functions:
Example:
Write y
= 3x^{2} 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:
_{}
Review of Basic Factoring methods:
1. Common Factoring:
Factor 6x^{3} 15x
Solution: 6x^{3} 15x = 3x(2x^{2} 5) **
Find the HIGHEST COMMON FACTOR for each term  3x
**
Divide 3x into each term to get the second factor  2x^{2} 5
**
Check by expanding
2. Difference of Squares: Formula  a^{2} b^{2 }= (a b)(a + b)
Factor 49x^{2} 64y^{2}
Solution: 49x^{2} 64y^{2} = (7x 8y)(7x + 8y)
Factor x^{2} 9y^{2}
Solution: x^{2} 9y^{2} = (x 3y)(x + 3y)
3. Simple Trinomials: Form x^{2} + bx + c [Coefficient of x^{2 }is 1]
a) Factor x^{2} + 5x + 6
Solution: Recall x^{2} + 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
x^{2} + 5x + 6 = (x + 3 )(x + 2) ** Check by expanding
4. Hard Trinomials: Form ax^{2} + bx + c [Coefficient of x^{2 } does not equal 1]
a) Factor 6m^{2} 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; 124= 23
** {2, 12} cannot
obtain 5 with these two factors
** {3, 8}
8 + 3 = 24 Choose these two
factors
Hence 6m^{2}
5m 4 = 6m^{2} 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
[xintercepts]of Quadratic Functions:
Example 1:
Find the
zeros (xintercepts) of the quadratic function f(x)
= 2x^{2} + x 6.
Solution:
We are
trying to find the points where the graph of the function crosses the xaxis
ie the xintercepts.
Let y =
f(x) = 0 yielding the quadratic equation
2x^{2} + x 6 = 0. We
solve by factoring.
2x^{2} + x 6 = 0
2x^{2} + 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 3x^{2}
7x + 1 = 0 by completing the square.
Solution:
_{}
Example:
Solve 3x^{2}
+ 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: b^{2}
4ac > 0 
One zero: b^{2} 4ac = 0 
No zeros: b^{2} 4ac < 0 



Here b^{2}
4ac > 0 and the graph of the quadratic
function cuts the xaxis in 2 distinct points. The
quadratic equation has 2 real, distinct
roots. 
Here b^{2}
4ac = 0 and the graph of the
quadratic function is tangent to the xaxis yielding 1 xintercept. The
quadratic equation has 2 real, equal
roots. Some texts say 1 real
root for this case. 
Here b^{2}
4ac < 0 and the graph of the
quadratic function does not cut the xaxis.
The
quadratic equation has no real roots or
imaginary roots. 
Example:
In each case,
calculate the value of the discriminant b^{2} 4ac and determine the
number of zeros of the quadratic function.
a) f(x) = 2x^{2} + 3x 1 b) g(x) = 16 x^{2} c) y = 3x^{2} 2x + 5 d) y = x^{2} 6x + 9
Solutions:
a = 2 b = 3 c= 1
a) f(x) = 2x^{2} + 3x 1
b^{2} 4ac = (3)^{2}
4(2)(1)
= 9 + 8
= 17
Since b^{2}
4ac > 0, there are 2 zeros.
The corresponding quadratic equation
2x^{2} + 3x 1 = 0 has 2
real distinct roots.
a = 3 b = 2 c = 5
c) y = 3x^{2} 2x + 5
b^{2}
4ac = (2)^{2} 4(3)(5)
= 4 60
= 56
Since b^{2}
4ac < 0, there are no zeros.
The corresponding quadratic equation
3x^{2} 2x + 5 = 0 has no real roots.
a = 1 b = 6 c = 9
d) y = x^{2} 6x + 9
b^{2} 4ac = (6)^{2}
4(1)(9)
= 36 36
= 0
Since b^{2} 4ac = 0, there is 1 zero. The corresponding quadratic equation x^{2} 6x + 9 = 0 has 1 real root or real and equal roots.
Example:
In each case, determine the number xintercepts of the quadratic function.
Solutions:
a) f(x) = 5x^{2} 3x + 2
a = 5 b = 3 c = 2
b^{2} 4ac = (3)^{2}
4(5)(2)
= 9 40
= 31
Since b^{2} 4ac < 0, there are no xintercepts or zeros. The corresponding quadratic equation 5x^{2} 3x + 2 = 0 has no real roots
Example:
For what
value of k does the equation kx^{2}
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 b^{2} 4ac = 0
(4)^{2} 4(k)(2) = 0
16 8k = 0
 8k =
16
k =
2
b) For 2
real distinct roots set b^{2} 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:
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 m^{2}. 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 m^{2}. 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.
Properties
of Radicals:
Example:
Write as mixed radicals.
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)
= x^{2} + 5x + 3x + 15
= x^{2}
+ 8x + 15
Example 8:
Simplify by rationalizing the
denominator.
Example 10: Simplify
Solving
Quadratic Equations with NonReal Roots:
Example 3: Solve 3x^{2}
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 
 Ό 
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) = x^{2} 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.