Download Quadratic Functions and Transformations

Quadratic Functions and Transformations
• Quadratic Function (or Parabola)
 Standard Form: 𝑓 𝑥 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 ; 𝑎 ≠ 0
 Vertex Form: 𝑓 𝑥 = 𝑎(𝑥 − ℎ)2 +𝑘 ; 𝑎 ≠ 0
• The vertex is at (h,k) and the axis of symmetry is
at x=h.
• The parent function is f(x) = x2 and we can make
any other parabola by applying the
transformations we have studied.
• For a>0, y is a minimum at k, for a<0, y is a
maximum at k.
Interpreting the Vertex Form
For the following parabolas, find:
 Vertex and Axis of Symmetry
 Minimum or Maximum value
 Domain and Range
1) y = 3(x-4)2 – 2
2) y = -2(x+1)2 + 4
Using Vertex Form
• For the following parabolas:
 Graph by hand:




Identify a, h, k
Plot the vertex
Plot two symmetric points
Sketch by hand
 Describe a series of transformations that result in
the parabola, starting from y=x2.
1) f(x) = -2(x-1)2 + 3
2) f(x) = -2(x+1)2 + 3
3) f(x) = 2(x+2)2 - 5
Using Vertex Form (cont’d)
• Write the equation of the following parabolas in
vertex form:
Hint: Identify (h,k), substitute in the given point and
solve for “a” in the vertex form of the equation:
y = a(x-h)2+k
1) Vertex (1,2) through point (2, -5)
2) Vertex (0,5) through point (1, -2)
• Do Problem 39 page 199, using a graphing
calculator.
Hwk 16
Hwk 16: Pages 199-201: 28, 30, 32, 36, 38,
40, 42
Standard Form of a Quadratic Function
Standard Form: f(x) = ax2+bx+c ; a ≠0
Properties of the parabola:
• If a>0, opens upward; if a<0, opens downward.
• Axis of symmetry: x = - (b/2a)
• Vertex: (h, k) is: h = -(b/2a) and k = f(h)
• y-intercept is at (0,c)
 Exercise: Verify the above by converting:
f(x) = a(x - h)2 + k (vertex form) into standard form.
(In other words: find a, b, and c in terms of a, h, k.)
Using Standard Form
•



•


Find the axis of symmetry, vertex, and graph:
f(x) = x2 + 2x + 3
f(x) = -2x2 + 2x -5
Confirm with a graphing calculator
Convert the following parabolas to vertex form:
f(x) = 2x2 + 10x +7
f(x) = -x2 + 4x -5
Modeling the New River Gorge Bridge
• An arch bridge spans a river using an arch that
can be modeled as y = -.000498x2+.847x where y
= height above some level on both river banks
from which the bridge starts to rise, and x is the
horizontal distance from the left bank from where
the bridge starts to rise. (Both x and y are in feet.)
Find the maximum height of the bridge.
Find the horizontal extent of the bridge above the
arch.
Modeling with Quadratic Functions
How many examples of quadratic functions have
you seen in Physics?
Give one from kinematics.
Give one from the work-energy unit.
Writing an Equation of a Parabola
Standard form of a parabola: y = ax2 + bx + c
How many points on a parabola would we need in
order to be able to write its equation?
Write the equation of the following parabolas :
1) (0,0), (-1,-2), (1,6)
2) (0,0), (1,-2), (-1,-4)
Quadratic Regression
• Quadratic Regression finds the best fit parabola
for a set of data points.
• Here our temperatures at various times during a
day: Time (8:00) (10:00) (12:00) (2:00P) (4:00P) (6:00P)
(24
hour
clock)
(x)
8
10
12
14
16
18
Temp
(◦F) (y)
52
64
72
78
81
76
Using the TI-nspire for Quadratic Regression
1)
2)
3)
4)
Start a new Document:Prob3Pg211
Choose Lists and Spreadsheets (4)
Enter data
Menu: Statistics(4)-> Stat Calculations(1)
>Quadratic Regression(6)->Put in x,y, tab to OK.
5) Now Let’s plot the points and the best fit parabola: Add a
Page: ctrl +
1) Menu: Add data&Statistics(5)
2) Menu: Plot Properties(2) Add x, Plot Properties(2) Add y
3) Superimpose best fit: Menu: Analyze(4)->Plot
Function(4)-> Enter function given by best fit.
Stopping Distance
The following data relates speed to stopping
distance:
speed
(mph)
10
stopping 15.1
distance
(ft)
20
30
40
50
39.9
75.2
120.5
175.9
Let’s use the TI-nspire to construct a Quadratic
Regression model for this data.
Can we estimate the coefficient of kinetic friction
(µk ) from this data?
Identifying Linear and Quadratic Data
• Linear Data: “first differences” in the y-values are
the same as we increment x by the same amount.
Here the first differences are 3.
x
-2
-1
0
1
2
y=3x+1
-5
-2
1
4
7
• Quadratic Data: “second differences” (differences
in the first differences” in the y-values are the
same as we increment x by the same amount.
Here, the second differences are 6.
x
-2
-1
0
1
2
y=3x2+1
13
4
1
4
13
Which of the following are sets of data are
Quadratic?
x
1
2
3
4
5
6
7
y
6
12
22
36
54
76
102
x
-2
-1
0
1
2
3
4
y
-8
-1
0
1
8
27
64
Factoring Quadratic Expressions
• We can factor trinomial quadratic expressions into the
product of two binomial expressions.
• When a = ±1: We need to find the numbers whose
product is c and sum is b in x2+bx+c.
 x2+ 9x +20, x2+14x-72, -x2+13x+12
• Special case 1: “Perfect square trinomial” A quadratic
expression that can be factored into a form:
f(x) = x2+2kx+k2 = (x+k)2 , f(x) = x2-2kx+ k2 = (x-k)2
(Note: b = 2k, c = k2 =
𝑏 2
2
• Special case2: “Difference of two squares”
f(x) = x2-k2 = (x-k) (x+k) = x2-kx +xk – k2 = x2-k2
Completing the Square
• This technique allows us to:
rewrite any quadratic expression as a perfect square plus
a constant term.
solve any quadratic equation by using one formula
• Ex1: Rewrite f(x) = 3x2 -12x as a perfect square plus
a constant term.
Factor out 3: f(x) = 3(x2-4x
)
Add a term to the part in parentheses to turn it into a
perfect square and then subtract the same amount to
keep f(x) unchanged. We need to add: c =
𝑏 2
2
or
−4 2
:
2
f(x) = 3(x2-4x +4 -4 ) = 3( x2-4x +4 -4 )= 3( (x-2)2 -4) =
3(x-2)2 -12
Completing the Square
• Ex 2: f(x) = 3x2 -12x =0 two ways
• 1) By factoring: f(x) = 3x( x-4) = 0 So x =0, +4 are
roots.
• 2) By completing the square: From last slide we
saw: f(x) = 3(x-2)2 -12 =0
Divide by 3: (x-2)2 - 4=0
Then: (x-2)2 = 4
Then x-2 = ±2 Which gives two linear equations:
x -2 = 2 giving x=4
x-2 = -2 giving x=0
 But completing the square will work when you can’t factor.
The Quadratic Formula
• By completing the square, we can come up with a
formula that will solve any quadratic equation of
the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
The Quadratic Formula
• By completing the square, we can come up with a
formula that will solve any quadratic equation of
the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
The Quadratic Formula
• The result is
−𝑏 ± 𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
Solve the following equations using the
quadratic formula.
x2+6x+9=0
2x2 –x = 4
The Discriminant
We call the value: 𝑏2 − 4𝑎𝑐 the discriminant of a
quadratic equation. Since it appears within a radical and
appears after the ± sign, its value tells us the following:
If 𝑏2 − 4𝑎𝑐 > 0, we have two real solutions.
If 𝑏2 − 4𝑎𝑐 = 0, we have one real solution.
If 𝑏2 − 4𝑎𝑐 < 0, we have no real solutions.
We will soon define a number: 𝑖 = −1 which will
expand our work into “complex” numbers and we’ll see
that when 𝑏2 − 4𝑎𝑐 < 0, we have complex solutions.
The Different Numbers
•
•
•
•
Natural or Counting Numbers: 1, 2, 3, ….
Whole Numbers: 0, 1, 2, 3, ….
Integers: …., -3, -2, -1, 0, 1, 2, 3, ….
Rational Numbers: numbers that can be expressed
as ratios of two integers (fractional form); includes
decimal numbers that have an infinitely repeating
pattern.
• Irrational Numbers: decimals that can’t be
expressed in fractional form. (e.g.: π, 2 )
• Real Numbers: Includes rational and irrational
numbers.
Complex Numbers
• Invented by Gerolamo Cardano (1500’s) and given
practical usage by Leonhard Euler (1700’s). Initially
invented to solve polynomial equations that otherwise
couldn’t be solved. (ex: 𝑥 2 + 1 = 0).
• We define: 𝑖 2 = -1 or 𝑖 = −1
 Then z = a + bi is a complex number where:
a = “the real part” and b = “the imaginary part”
 If a=0 , and b≠0 then z is a pure imaginary number.
 Complex numbers can be represented graphically with
a plotted on the horizontal (real) axis and b plotted on
the vertical (imaginary) axis.
 The absolute value of a complex number is:
𝑧 = 𝑎2 + 𝑏2
Graphically Represent and find the
Absolute Value:
• -5 + 3i
• 6i
• 4-3i
• 3+4i
Algebra with Complex Numbers
• Addition/Subtraction: Add or Subtract real and
imaginary components separately. (What type of
addition/subtraction is this similar to in Physics?)
 (4-3i) + (-4+3i) =
 (2+4i) - (6+2i) =
• Multiplication: Multiply the same way as you
would binomials, but: when you get a term
involving i2 it becomes real and negated since:
𝑖 2 = -1.
 (3 + 4i) (4 - 2i) =
 (2 + 3i) (2 - 3i) =
 (a + bi) (a - bi) =
Algebra with Complex Numbers (cont’d)
Note that the product: (2 + 3i) (2 - 3i) or, in general: (a +
bi)(a - bi) gave us a real result. We call such a pair of
numbers complex conjugates.
• Division: Multiply the numerator and the denominator
by the complex conjugate of the denominator. The new
denominator is real and you can finish the problem with
distribution of the denominator.

9+12𝑖
3𝑖

5−2𝑖
3+4𝑖
=

8−7𝑖
8+7𝑖
=
=
Quadratic Equations and Complex Solutions
Every quadratic equation has two complex solutions
which are sometimes real.
Look at the quadratic formula:
−𝑏 ± 𝑏 2 − 4𝑎𝑐
𝑥=
2𝑎
 When does the solution have an imaginary part?
(Hint: look at the discriminant.)
 When the solution has an imaginary part, how do
the two solutions relate to each other?
 So if the solutions are not real, they occur in
complex conjugate pairs.
Some Exercises
 Solve using factoring: 5𝑥 2 +20 = 0
 Solve using the quadratic formula:
 2𝑥 2 − 3𝑥 + 5 = 0
Powers of 𝒊
Notice:
𝑖0 = 1
𝑖1 = 𝑖
𝑖 2 = −1
𝑖 3 = −𝑖
𝑖4 = 1
𝑖 5 = 𝑖 etc.
SO: When you have a power: 𝑖 𝑛 you can divide n
by 4 and get the answer from the remainder.
Hwk 20
Pages 253-255: 22, 24, 28, 30, 40, 45 (just do
a,b,c,d for point A), 50, 56, 60, 66, 68.
Solving Quadratic Inequalities
 Factoring and reasoning applied to possible signs
of the factors:
 2𝑥 2 − 14𝑥 < 0
 2𝑥 2 − 14𝑥 > 0
 Use the quadratic formula to find roots, then
look at the boundaries on either side of the
roots.
 𝑥 2 − 6𝑥 + 5 < 0
 Use a graphing calculator to plot the quadratic
function and identify the zeros and regions of
desired inequality.
Solving Quadratic Systems
• Linear-Quadratic System of Equations
• Quadratic System of Equations
• Solve by:
• Graphing and identifying points of intersection
• Algebra using substitution.
• Solve:
 𝑦 = −𝑥 2 + 5𝑥 + 6 and 𝑦 = 𝑥 + 6
 𝑦 = −𝑥 2 − 𝑥 + 6 and 𝑦 = 𝑥 + 3
 𝑦 = −𝑥 2 − 𝑥 + 12 and 𝑦 = 𝑥 2 + 7𝑥 + 12
 𝑦 = 𝑥 2 − 4𝑥 + 5 and 𝑦 = −𝑥 2 + 5
Quadratic System of Inequalities
• Graph and identify regions of overlap:
• Solve: 𝑦 < −𝑥 2 − 9𝑥 − 2 and 𝑦 > 𝑥 2 − 2
Hwk 21
Pages 262-264: 18, 29, 30, 47, 61