Quadratics
Definition or Depiction
• Linear Function:
– f(x) = mx+b
– Makes a line
– Constant change in y for each change in x
• Quadratic Function:
– Standard Form: f(x) = x2+bx+c or
– Vertex Form: f(x) = a(x-h)2+k
– Makes a parabola
Parent Function
• f(x) = x2
– Centered at (0, 0)
– Symmetric – (graph)
• Can move (Transform)
– Left or Right
– Up or Down
– Stretching or Compressing
Transformations (Alg 2.1)
• In Vertex Form
• F(x) = a(x – h)2 + k
– a is vertical stretch/compression (skinnier/fatter)
• Reflection over x & Direction of opening (- opens down (max),
+ opens up(min))
– h is translation left/right (x coordinate) (it is –h)
– k is translation up/down (y coordinate)
– Horizontal stretch/compression is with the x
• Vertex is (h, k)
• Domain & Range Impacts
Example: Translating Quadratic Functions
Use the graph of f(x) = x2 as a guide, describe the
transformations and then graph each function.
g(x) = (x – 2)2 + 4
Identify h and k.
g(x) = (x – 2)2 + 4
h
k
Because h = 2, the graph is translated 2 units right.
Because k = 4, the graph is translated 4 units up.
Therefore, g is f translated 2 units right and 4 units up.
Example: Translating Quadratic Functions
Use the graph of f(x) = x2 as a guide, describe the
transformations and then graph each function.
g(x) = (x + 2)2 – 3
Identify h and k.
g(x) = (x – (–2))2 + (–3)
h
k
Because h = –2, the graph is translated 2 units left.
Because k = –3, the graph is translated 3 units down.
Therefore, g is f translated 2 units left and 4 units down.
Domain & Range
• Domain is possible x values
– In quadratic functions – domain is typically All
Real Numbers (ARN)
– Written as [x: (-∞, ∞)] or ARN
• Range is possible y values
– Range will be limited by min/max in vertex
– Range will be [y: (k, ∞)] or [y: (-∞, k)]
Domain & Range
Caution!
The minimum (or maximum) value is the y-value at
the vertex. It is not the ordered pair that represents
the vertex.
Transformations
Reflections and Stretch/Compression
• Across the
x or y axis
• a or b value
stretches or
compresses
the graph
Example: Reflecting, Stretching, and
Compressing Quadratic Functions
Using the graph of f(x) = x2 as a guide, describe the
transformations and then graph each function.
1 2
(
)
g x =- x
4
Because a is negative, g is a
reflection of f across the x-axis.
Because |a| = , g is a
vertical compression of f by
a factor of .
Example: Reflecting, Stretching, and
Compressing Quadratic Functions
Using the graph of f(x) = x2 as a guide, describe the
transformations and then graph each function.
g(x) =(3x)2
Because b = , g is a
horizontal compression of
f by a factor of .
Application Example! Example 5
The minimum braking distance d in feet for a vehicle
on dry concrete is approximated by the function (v) =
0.045v2, where v is the vehicle’s speed in miles per
hour.
The minimum braking distance dn in feet for a
vehicle with new tires at optimal inflation is dn(v) =
0.039v2, where v is the vehicle’s speed in miles per
hour. What kind of transformation describes this
change from d(v) = 0.045v2, and what does
this transformation mean?
Example 5 Continued
Examine both functions in vertex form.
d(v)= 0.045(t – 0)2 + 0
dn(t)= 0.039(t – 0)2 + 0
The value of a has decreased from 0.045 to 0.039.
The decrease indicates a vertical compression.
Find the compression factor by comparing the new
a-value to the old a-value.
a from dn(t)
a from d(v)
=
0.039
0.045
=
13
15
Example 5 Continued
The function dn represents a vertical compression of d
by a factor of
. The braking distance will be less
with optimally inflated new tires than with tires
having more wear.
Check
Graph both functions on a graphing calculator.
The graph of dn appears to be vertically
compressed compared with the graph of d.
15
0
15
0
Transformations – All Together
• Example: f(x) = -4(x + 3)2 – 2
– Transformations:
• Graph:
Vertex Form from a Graph/Point
• Vertex Form Equation: f(x) = a (x – h)2 + k
• From Graph or given another point:
– Determine vertex
• Plug x coordinate for h and y coordinate for k
– Identify another point on graph
– Plug x and y coordinates into new vertex form
• Solve for a
• Plug a, h, & k into Vertex Form Equation
Example
• Vertex is (4, 4)
– Write f(x) = a(x-4)2+4
• Additional point
– (8, 2)
• Insert
– 2 = a(8-4)2 + 4
• Solve for a
• Rewrite Vertex Equation:
Standard Form (Alg 2.2)
• F(x) = ax2 + bx + c
– a determines which way parabola opens
• a is the same in vertex and standard form
• + opens up (happy) - opens down (sad)
– Axis of Symmetry x = - b/(2a)
– Y coordinate of vertex is when x=-b/(2a)
– Vertex is (-b/(2a), f(-b/(2a)))
Developing the standard form of a quadratic
Another useful form of writing quadratic functions is
the standard form. The standard form of a quadratic
function is f(x)= ax2 + bx + c, where a ≠ 0.
The coefficients a, b, and c can show properties of
the graph of the function. You can determine these
properties by expanding the vertex form.
f(x)= a(x – h)2 + k
f(x)= a(x2 – 2xh +h2) + k
Multiply to expand (x – h)2.
f(x)= a(x2) – a(2hx) + a(h2) + k
Distribute a.
f(x)= ax2 + (–2ah)x + (ah2 + k)
Simplify and group terms.
a=
b=
c=
Graphing Standard Form
No Calculator
•
•
•
•
•
•
•
Determine whether parabola opens up/down
Find AOS (x value of vertex)
Find y at the AOS (y value of vertex)
Plot Vertex
Plot y axis intercept = the value of c
Plot a point across AOS from y intercept
Fill in the curve
Example: Graphing Quadratic Functions in Standard
Form
Consider the function f(x) = 2x2 – 4x + 5.
a. Determine whether the graph opens upward
or downward.
Because a is positive, the parabola opens upward.
b. Find the axis of symmetry.
The axis of symmetry is given by
Substitute –4 for b and 2 for a.
The axis of symmetry is the line x = 1.
.
Example: Graphing Quadratic Functions in
Standard Form
Consider the function f(x) = 2x2 – 4x + 5.
c. Find the vertex.
The vertex lies on the axis of symmetry, so the
x-coordinate is 1. The y-coordinate is the value of
the function at this x-value, or f(1).
f(1) = 2(1)2 – 4(1) + 5 = 3
The vertex is (1, 3).
d. Find the y-intercept.
Because c = 5, the intercept is 5.
Example 2A: Graphing Quadratic Functions in
Standard Form
Consider the function f(x) = 2x2 – 4x + 5.
e. Graph the function.
Graph by sketching the axis of
symmetry and then plotting the
vertex and the intercept point
(0, 5). Use the axis of symmetry
to find another point on the
parabola. Notice that (0, 5) is 1
unit left of the axis of symmetry.
The point on the parabola
symmetrical to (0, 5) is 1 unit to
the right of the axis at (2, 5).
Summary of Standard Form
Properties of Quadratic Example
• Given: f(x) = -3x2 + 6x – 4
• Find:
– Opens Up or Down
– Maximum or minimum
– Axis of Symmetry
– Vertex
• Coordinate and Minimum or Maximum
– Y intercept
Real life Application: Example
The highway mileage m in miles per
gallon for a compact car is
approximately by
m(s) = –0.025s2 + 2.45s – 30, where
s is the speed in miles per hour. What
is the maximum mileage for this
compact car to the nearest tenth of a
mile per gallon? What speed results
in this mileage?
Example Continued
The maximum value will be at the vertex (s, m(s)).
Step 1 Find the s-value of the vertex using
a = –0.025 and b = 2.45.
2.45)
(
b
s === 49
2a
2 (-0.025)
Example Continued
Step 2 Substitute this s-value into m to find the
corresponding maximum, m(s).
m(s) = –0.025s2 + 2.45s – 30
Substitute 49 for r.
m(49) = –0.025(49)2 + 2.45(49) – 30
m(49) ≈ 30
Use a calculator.
The maximum mileage is 30 mi/gal at 49 mi/h.
Roots (Solutions or Zeros)
• Roots are what make the equation = zero
– f(x) = 0
or
y=0
• One method to determine
– Plug equation into calculator ( Y= ……..)
– Look at Table (2nd Graph) for when y equals 0
– Take the x value
– Works when roots are whole numbers
• Example: Find roots of f(x) = x2 + 5x - 6
Factoring (Alg 2.3)
• Set equation equal to 0 (f(x) = 0)
• Factor - Separate Sheet
– Common Factors
– Product-Sum
– Perfect Square
– Difference of Squares
• Example: f(x) = x2 - 4x - 12
Factoring (Alg 2.3)
•
•
•
•
Lead coefficient not 1 (a not 1)
Product – Sum – Group – Reduce(combine)
Box Method
Example: f(x) = 6x2 - 13x + 6
Roots from Factors
•
•
•
•
Find the factors (from previous)
Set each factor = 0
Solve for x
Example: f(x) = x2 + 3x – 18
• Example: (2x-3)(x+5)
Roots From Square Roots
• Find roots by taking square root of both sides
– If there is no linear term (no bx)
– Using square roots to find solutions
• Example: x2 – 25 = 0
• Example:
3x2 – 4 = 68
• Example:
25 = (x + 3)2
– If in form of an equation set f(x) = 0
• Example: f(x) = 2x2 - 32
Solutions by Square Roots
• If written in vertex form (f(x) = a(x-h)2 + k)
• Move k over
• Divide by a
• Take square root of both sides
– Example: f(x) = 2(x-2)2 - 8
• Can work the same technique by setting up
vertex form
– Done by working to vertex form & setting up
perfect square
Complex Numbers
• The square root of -1 is an imaginary number
– No number multiplied by itself can be negative
• Negative * negative = positive
• Positive * positive = positive
– i represents square root of -1
• To take square root of a negative number
– Factor into -1 * positive number
– Factor positive portion with greatest square (reduce)
– Solve the positive portion and replace -1 with i
Complex Numbers (cont)
• 2 Parts of number (5 + 2i)
– Real part (constant without the i : 5)
– Imaginary part (part with the i : 2i)
• Square root of negative can lead to both parts
– Example: x2 = -25
– Example: 2x2 + 144 = 0
– Example: (x – 5)2 = -16
Complex Numbers (cont)
• Complex Conjugates
–
–
–
–
Write in form a + bi
Leave the sign with the constant
Flip the sign with the imaginary (i) part
Allows multiplication (FOIL) to achieve quadratic
• Examples:
9 – i
i + 6
 -4i
Setting up a Perfect Square
Complete the Square (Alg 2.4)
• Add term to make a perfect trinomial
– x2 + bx + _____
– Need to fill in blank (c) with (b/2)2
– Factors to (x + (b/2))2
– Example: x2 – 2x + ____
– Example: x2 + 5x + _____
Setting up the Square (cont.)
• If there is already a value for c
– Move c to opposite side of equation
– Add (b/2)2 to both sides of the equation
– Example: x2 + 4x – 6 = 0
– Example: y = 2x2 + 6x - 8
Using CTS to Solve Quadratics
Check It Out! Example
Solve the equation by completing the square.
3x2 – 24x = 27
x2 – 8x = 9
x2 –8x +
=9+
Divide both sides by 3.
Set up to complete the
square.
Add
to both sides.
Simplify.
Check It Out! Example Continued
Solve the equation by completing the square.
Factor.
Take the square root
of both sides.
Simplify.
x – 4 =–5 or x – 4 = 5
x =–1 or x = 9
Solve for x.
Using CTS to set up Vertex Form
• If equation is in f(x) = x2 + bx + c
– Parenthesize variables: f(x) = (x2+bx ) + c
– Add (b/2)2 inside parentheses (x2 + bx + (b/2)2)
– Subtract (b/2)2 from outside () + c – (b/2)2)
– Rewrite parentheses into square (x+(b/2))2
– Combine constants outside
• Example: (a=1) f(x) = x2 + 10x -13
Using CTS - Vertex Form: a not 1
• If equation is in f(x) = ax2 + bx + c
– Parenthesize variables: f(x) = (ax2+bx ) + c
– Factor out the a: f(x) = a(x2+(b/a)x ) + c
– Add (b/2a)2 inside: a(x2 + (b/a)x + (b/2a)2)
– Subtract a(b/2a)2 from outside () + c – a(b/2a)2)
– Rewrite parentheses into square (x+(b/2a))2
– Combine constants outside
• Example: f(x) = 4x2 – 8x + 5
Quadratic Formula
• Keep track of numbers – always works
Example
Find the zeros of f(x)= x2 – 8x + 10 using the Quadratic Formula.
x2 – 8x + 10 = 0
Set f(x) = 0.
Write the Quadratic Formula.
Substitute 1 for a, –8 for b, and 10
for c.
Simplify.
Write in simplest form.
Example Continued
Check Solve by completing the square.
x2 – 8x + 10 = 0
x2 – 8x = –10
x2 – 8x + 16 = –10 + 16
(x + 4)2 = 6

Quadratic Formula
• The Discriminant:
– Value under the square root (b2 – 4ac)
• Tells how many real roots the equation has
– Discriminant is > 0 : there are 2 real roots
– Discriminant = 0 : there is one real root (double)
– Discriminant is < 0 : 2 COMPLEX roots (No Real)
The graph shows
related functions.
Notice that the
number of real
solutions for the
equation can be
changed by changing
the value of the
constant c.
Quadratic Formula – Finding Roots
• When the Discriminant is > 0
– Example: x2 + 5x – 6
• When the Discriminant = 0
– Example: x2 – 12x + 36
• When the Discriminant is < 0
– Example: x2 + 4x + 8
Complex Number Operations
• Adding Complex Numbers
– Add the real numbers
– Add the imaginaries
– Keep them separate
• Expansion of i
– i2 = -1
– i3 = -i
– i4 = 1
– And so on ………….
Examples
Complex Number Operations
• Multiplying Complex Numbers
– FOIL the terms – remember i2 = -1
– Combine like terms (reals, i’s)
• Dividing Complex Numbers
– Multiply top & bottom by complex conjugate
(from Intro to Complex Numbers)
– Eliminates complex on bottom
– Results in multiplying rules on top
– Simplify
Writing Equations from Roots
•
•
•
•
•
•
Write roots as solutions
Set roots = 0
Combine factors & set = 0
Multiply binomials (FOIL)
Check via calculator
Example: Roots 2 & -4
• Example: Roots -3/2, 3
Quadratic Inequalities
• Change to Equality
• Make = 0
– Add/Subtract term to make standard equation
• Factor (if able)
• Find the roots from the factors (or CTS, Quad Form)
– These set the boundaries of the inequality
– If Greater Than or Less Than
• Open Circles
– If Greater Than/Equal To or Less Than/Equal To
• Closed Circles
Quadratic Inequalities
• Plot on Number Line
– Open vs Closed Circle
• Take value above, between & below
– Check for True or False
– And is between – written low root<x<high root
– Or is outside – written x<low root or x>high root
Quadratic Inequalities
• Using a calculator
– Graphing
• Plot Y1 = Quadratic Equation
• Plot Y2 = Value
• Look at graph where equation meets criteria
– Table
• Plot Y1 = Quadratic Equation
• Look at where table meets criteria
Real World Example
A business offers educational tours to Patagonia, a
region of South America that includes parts of Chile and
Argentina . The profit P for x number of persons is P(x)
= –25x2 + 1250x – 5000. The trip will be rescheduled if
the profit is less $7500. How many people must have
signed up if the trip is rescheduled?
Setup
1
Understand the Problem
The answer will be the number of people signed up for
the trip if the profit is less than $7500.
List the important information:
• The profit will be less than $7500.
• The function for the profit is
P(x) = –25x2 + 1250x – 5000.
Setup to Solve
2
Make a Plan
Write an inequality showing profit less than
$7500. Then solve the inequality by using
algebra.
Apply Appropriate MATH Concept
3
Solve
Write the inequality.
–25x2 + 1250x – 5000 < 7500
Find the critical values by solving the related equation.
–25x2 + 1250x – 5000 = 7500
–25x2 + 1250x – 12,500 = 0
–25(x2 – 50x + 500) = 0
Write as an equation.
Write in standard form.
Factor out –25 to simplify.
Do the Math
3
Solve
Use the Quadratic
Formula.
Simplify.
x ≈ 13.82 or x ≈ 36.18
Continue the Math
3
Solve
Test an x-value in each of the three regions formed by
the critical x-values.
Critical values
5
10
15
20
25
30
Test points
35
Finish the Math
3
Solve
–25(13)2 + 1250(13) – 5000 < 7500
7025 < 7500
–25(30)2 + 1250(30) – 5000 < 7500
10,000 < 7500
–25(37)2 + 1250(37) – 5000 < 7500
7025 < 7500

Try x = 13.
Try x = 30.
x
Try x = 37.

Write the solution as an inequality. The solution is
approximately x > 36.18 or x < 13.82. Because you cannot
have a fraction of a person, round each critical value to the
appropriate whole number.
Answer the Question
4
State the Answer
The trip will be rescheduled if the number of
people signed up is fewer than 14 people or
more than 36 people.
Focus & Directix
• Focus is a point inside the parabola
• Directrix is a straight line outside the parabola
• The parabola is formed by points equidistant
from both
• Visual Example: online at
– http://www.mathwarehouse.com/quadratic/para
bola/focus-and-directrix-of-parabola.php
Focus & Directrix
• Focus is on Axis of Symmetry
– h or x coordinate of the vertex
• Vertex is ½ way between focus & directrix
• The vertical stretch defines the distance from
the Vertex to the Focus
– If focus above vertex: positive p (+p)
– If focus below vertex: negative p (-p)
– Distance (p) = 1/(4a)
– Flatter parabola = farther focus/directrix