Transcript y = x 2

4.1 Linear Applications
Phone Charges: The monthly cost C, in dollars for international calls
on a certain cellular phone plan is given by the function C(x) = .38x + 5
where x is the number of minutes used.
(a) What is the cost if you talk on the phone for x = 50 minutes.
(a) Suppose that your monthly bill is $29.32.
How many minutes did you use the phone?
(c)
Suppose what you budget yourself $60 per monthe for the phone.
What is the maximum number of minutes that you can talk?
Application: Supply & Demand
Suppose that the quantitu supplied S and quantity demanded D of hot dogs
at the baseball game are given by the following functions:
S(p) = -2000 + 3000p
p is the price of a hot dog.
D(p) = 10,000 – 1000p
(a) Find the equilibrium price for hot dogs at the baseball game. What is the
equilibrium quantity?
(a) Determine the prices for which quantity demanded is less than quantitu
supplied.
(a) What do you think will eventually happen to the price of hot dogs if
quantity demanded is less than quantity supplied?
Application: Taxes
The function T(x) = .15(x – 7300) + 730 represents the tax bill T of a single
person whose adjusted gross income is x dollars for income between $7,300
and $29,700, inclusive in 2005. [ Source: Internal Revenue Service]
(a) What is the domain of this linear equation
(a) What is a single filer’s tax bill if adjusted gross income is $18,000?
(a) Which variable is independent and which is dependent?
(a) Graph the linear function over the domain specified in part (a)
(a) What is a single filer’s adjusted gross income if the tax bill is $2860?
4.3 Quadratic Functions & Graphs
y = x2
x
0
1
-1
2
-2
y
0
1
1
4
4
A Parabola
Do you know what an axis of symmetry is?
Vertex
Lowest (Minimum) point if
The parabola opens upward,
And highest (Maximum) point
if parabola opens downward.
Quadratic Functions & Graphs
y = x2 - 2
x
0
1
-1
2
-2
y
-2
-1
-1
2
2
Vertex
Notice this graph is shifted down 2 from the origin.
Y = x2 – k (shifts the graph down k units)
Y = x2 + k (shifts the graph up k units)
To shift the graph to the right or to the left
y = (x – h)2 (shifts the graph to the right)
y = (x + h)2
(shifts the graph to the left)
General Form of a Quadratic
y = ax2 + bx + c
a, b, c are real numbers & a 0
A quadratic Equation: y = x2 + 4x + 3a = _____ 1b = _____ 4c = ______3
x
-5
-4
-3
-2
-1
0
1
Where is the vertex?
Where is the axis of symmetry?
y
8
3
0
-1
0
3
8
Formula for
Vertex:
X = -b
2a
Plug x in to
Find y
The Role of “a”
a0
a0
f ( x)  x 2
f ( x)   x 2
f ( x )  2 x
2
f ( x)  2 x
2
f ( x )  3 x
2
f ( x)  3x
2
Quadratic Equation Forms
• Standard Form:
2
y  ax  bx  c
• Vertex Form:
y  a ( x  h)  k
2
Vertex = (h, k)
Examples
Find the vertex, axis of symmetry, and graph
each
a.
y  3( x  5)  2
b.
y  2( x  2)  3
c.
2
2
1
2
y  ( x  4)  6
2
Vertex (5, 2)
Vertex (-2, -3)
Vertex (4, -6)
Convert from Vertex Form to
Standard Form
Vertex Form:
y = 2(x + 2)2 + 1
To change to standard form, perform multiplication,
add, and combine like terms.
y = 2 (x + 2) (x + 2) + 1
y = 2 (x2 + 2x + 2x + 4) + 1
y = 2 (x2 + 4x + 4) + 1
y = 2x2 + 8x + 8 + 1
y = 2x2 + 8x + 9 (Standard Form)
Convert from Standard Form to Vertex
Form
(Completing the Square – Example 1)
Step 1: Check the coefficient of the x2 term. If 1 goto step 2
If not 1, factor out the coefficient from x2 and x terms.
Step 2: Calculate the value of : (b/2)2
Step 3: Group the x2 and x term together, then add (b/2)2 and subtract (b/2)2
Step 4: Factor & Simplify
Example 1: y = x2 –6x – 1 (Standard Form)
y = (x2 –6x + 9) – 1 -9
y = (x – 3) (x – 3) – 1 – 9
y = (x – 3)2 – 10 (Vertex Form)
(b/2)2 = (-6/2)2 = (-3)2 = 9
Convert from Standard Form to Vertex
Form
(Completing the Square – Example 2)
Step 1: Check the coefficient of the x2 term. If 1 goto step 2
If not 1, factor out the coefficient from the x2 and x term.
Step 2: Calculate the value of : (b/2)2
Step 3: Group the x2 and x term together, then add (b/2)2 and subtract (b/2)2
Step 4: Factor & Simplify
Example 2: y = 2x2 +4x – 1 (Standard Form)
y = 2( x2 + 2x) –1
(2/2)2 = (1)2 = 1
y = 2(x2 +2x + 1) – 1 -2 (WHY did we subtract 2 instead of 1?)
y = 2(x + 1) (x + 1) – 1 – 2
y = 2(x + 1)2 – 3 (Vertex Form)
Solving Quadratic Equations
General Form of a Quadratic Equation
y = ax2 + bx + c
0 = ax2 + bx + c (If y = 0, we can solve for the x-intercepts)
A quadratic Equation: y = x2 + 4x + 3
1
4
3
a = _____
b = _____
c = ______
Graphical Solution
Numerical
Solution
Vertex (-2, -1)
Formula for
Vertex:
X = -b
2a
x
-5
-4
-3
-2
-1
0
1
y
8
3
0
-1
0
3
8
Symbolic/Algebraic Solution
x2 + 4x + 3 = 0
(x + 3) (x + 1) = 0
x+3 =0
x+1=0
x = -3
x = -1
Number of Solutions
y = x2 + 4x + 3
2 Real Solutions
x = -3
x = -1
y = x2 - 4x + 4
1 Real Solution
x=2
y = 2x2 + 1
NO Real Solutions
(No x-intercepts)
Quadratic Graphing Things to Know
Standard Form of a quadratic: y = ax2 + bx + c
Vertex Form of a quadratic: y = a(x – h)2 + k
Using either form above:
• Find and graph the vertex
•
Use an x/y chart to plot points & graph the parabola
•
Use and describe transformations to graph the parabola
•
Use intercepts and symmetry while graphing the parabola.
•
Show and give the equation for the line of symmetry.
•
Convert from standard to vertex form and vice vera
4.4 Quadratic Modeling Application
Insurance Claims: The years 1999 to 2005 were particularly costly for
insurance companies, with 7 of the 10 most costly catastrophes in U.S.
history (as of 2005). For the United States Automobile Association
(USAA) and its affiliates, the total cost of claims for catastrophic losses,
in millions, can be approximated by C(x) = 34.87x2 – 98.1x + 258.3,
where x = 0 for 1999, x = 1 for 2000, x = 2 for 2001, and so on.
(a) Estimate the total cost of claims for the year 2003
(b) According to the model during which year were catastrophic loss
claims at a minimum?
(c) Would C(x) be useful for predicting total catastrophic loss claim for
the year 2015? Why or why not?