~ chapter 5 ~ solving & applying proportions algebra i lesson 5-1 relating graphs to events...
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~ Chapter 5 ~Solving & Applying
Proportions
Algebra I
Lesson 5-1 Relating Graphs to Events
Lesson 5-2 Relations & Functions
Lesson 5-3 Function Rules, Tables & Graphs
Lesson 5-4 Writing a Function Rule
Lesson 5-5 Direct Variation
Lesson 5-6 Describing Number Patterns
Chapter Review
Algebra I
~ Chapter 5 ~Cumulative Review Ch 1-
4
Algebra I Algebra I
Relating Graphs to EventsNotesLesson 5-1
Interpreting Graphs
You can use a graph to show the relationship between two variables…
What does each section of the graph represent?
Sketching a Graph
A plane is flying from NY to London. Sketch a graph of the planes altitude during the flight. Label each section…
Relating Graphs to EventsNotesLesson 5-1
Sketch a graph of the distance from a child’s feet to the ground as the child jumps rope. Label each section.
Analyzing Graphs
A car travels at a steady speed.
Which graph could you use?
CA B
Relating Graphs to Events
HomeworkLesson 5-1
Homework ~ Practice 5-1
Relations & Functions Practice 5-1
Lesson 5-2
Relations & Functions Practice 5-1
Lesson 5-2
Relations & Functions NotesLesson 5-2
Relation – a set of ordered pairs
The domain of a relation is the first set of coordinates (x values)
The range of a relation is the second set of coordinates (y values)
Find the domain & range of the relation represented by the data in the table.
{-2,-1,4}{-2,1,3} list in order from least to greatest.
A function [f(x)] is a relation that assigns exactly one value in the range to each value in the domain.
One way to tell whether a relation is a function is to use the vertical-line test. If a vertical line passes through more than one point… the relation is NOT a function.
Using a mapping diagram
If the domain maps to more than one range… then the relation is not a function. If the domain only maps to one range then the relation is a function.
Relations & Functions NotesLesson 5-2
A function rule is an equation that describes a function.(Input – x, output – y)
Evaluating a function rule…
For x = 2.1
y = 2x + 1 f(x) = x2 – 4 g(x) = -x + 2
y = 5.2 f(2.1) = 0.41 g(2.1) = -0.1
Finding the Range
You can use a function rule and a given domain to find the range of the function…
Find the range of each function for the domain {-2,0,5}
f(x) = x – 6
Range = {-8,-6,-1}
y = -4x
Range = {-20,0,8}
g(t) = t2 + 1 y = ¼ x
Relations & Functions
HomeworkLesson 5-2
Homework ~ Practice 5-2
Function Rules, Tables & Graphs NotesLesson 5-3
Independent variable – x – the inputs are values for this variable.
Dependent variable – y – the outputs are values for this variable.
The independent variable graphs on the x-axis, the dependent variable graphs on the y-axis.
Model the function rule y = ½ x + 3 using a table of values and a graph…
To make a table, choose input values for x and evaluate to find y.
To graph, plot points for the ordered pairs from your table… (x,y)
Join the points to form a line or a curve.
Your turn… Model the function rule f(x) = 3x + 4
A recording company charges $300 for making a master CD and designing the art. It charges $2.50 for burning each CD. Use the function rule P(c) = 300 + 2.5 c. Make a table of values and a graph.
Graphing functions
Graph the function y = |x| + 1
(hint: make a table of values and then graph)
Function Rules, Tables & Graphs NotesLesson 5-3
Graph the function f(x) = x2 - 1
Homework ~ Practice 5-3
Writing a Function Rule Practice 5-2
Lesson 5-4
Writing a Function Rule
Practice 5-3Lesson 5-4
Writing a Function Rule
Practice 5-3Lesson 5-4
Practice 5-3
Practice 5-3
Writing a Function Rule NotesLesson 5-4
Writing a rule from a table…
Ask what can be done to 1 to get to -1… Then see if that rule
applies to get from 2 to 0. If not then try again…
So f(x) = x – 2
Rule?
y = 2x
Write a function rule to calculate the cost of buying apples at$1.25 a pound.
Write a function rule to calculate the total distance d(n) traveled after n hours at a constant speed of 45 miles per hour.
Write a function rule to calculate the area A(r) of a circle with radius r.
x f(x)
1 -1
2 0
3 1
4 2
x y
1 2
2 4
3 6
4 8
x f(x)
1 1
3 9
6 36
9 81
x f(x)
-8 64
-4 16
0 0
4 16
8 64
Writing a Function Rule
HomeworkLesson 5-4
Homework – Practice 5-4
Direct Variation
Practice 5-4Lesson 5-5
Direct VariationNotesLesson 5-5
Direct variation – a function in the form ~ y = kx, where k ≠ 0. x & y vary directly… meaning that if x increases in value, y increases in value, and vice versa.
The constant of variation, k, is the coefficient of x.
Determine if an equation is a direct variation…
5x + 2y = 0 Solve the equation for y
2y = -5x
y = -5/2 x (this is in the form y = kx so 5x + 2y = 0 is a direct variation)
What is k in 5x + 2y = 0?
k = -5/2
? 7y = 2x 3y + 4x = 8 y – 7.5x = 0
Writing an equation given a point
Write an equation of the direct variation that includes the point (4, -3)
Remember… y = kx substitute ~ so -3 = k(4) and solve for k
k = -3/4 so the equation is y = -3/4 x
Direct VariationNotesLesson 5-5
Write an equation of the direct variation that includes the point (-3, -6)
y = kx… -6 = k(-3)
k = 6/3 = 2 so… y = 2x
A recipe for one dozen muffins calls for 1 cup of flour. The number of muffins varies directly with the amount of flour you use. Write a direct variation for the relationship between the number of cups of flour and the number of muffins.
x = 1 y = 12 so… 12 = k(1) so k = 12
y = 12x
Direct Variations & Tables
You can rewrite a direct variation y = kx as y/x = k.
For each table, use the ratio y/x, to determine whether y varies directly with x.
In a direct variation, the ratio is the same for all pairs of data where x ≠ 0.
So the proportion x1/y1 = x2/y2 for (x1 ,y1) & (x2, y2)
x y
-2 3.2
1 2.4
4 1.6
x y
4 6
8 12
10 15
Direct Variation
HomeworkLesson 5-5
Homework – Practice 5-5 odd
Describing Number Patterns
Practice 5-5Lesson 5-6
Describing Number Patterns
NotesLesson 5-6
Extending number patterns… What are the next two numbers in the pattern?
1, 4, 9, 16, … rule?
3, 9, 27, 81, … rule?
2, -4, 8, -16, … rule?
Each number in a sequence is a term.
The common difference (d) is a fixed number in a sequence that is added to each previous term resulting in the next term in the sequence.
Determine the common difference in each sequence…
11, 23, 35, 47, …
8, 3, -2, -7, …
In a sequence the term is considered to be the output (y). The input(x) is the number of the term in the sequence.
Arithmetic sequence = A(n) = a + (n-1)d, where n = term number, a = first term, and d = common difference.
Describing Number Patterns
NotesLesson 5-6
Write the arithmetic sequence for 18, 7, -4, -15,…
a = d =
so A(n) = + (n - 1)
Use the equation to find the 10th term in the sequence above.
A(10) = 18 + (10-1)(-11)
A(10) = 18 + (-99) A(10) = -81
Your turn.. 1/2 , 1/3, 1/6, 0, … find the 12th term in the sequence.
d = - 1/6 so A(n) = 1/2 + (n – 1) (-1/6) and A(12) = 1/2+(12 – 1) (-1/6)
A(12) = 1/2+(-11/6) = 3/6 + (-11/6) = -8/6 = -4/3 = -1 1/3
Given the Arithmetic Sequence, find the 3rd, 5th, & 8th term…
A(n) = -2.1 + (n – 1)(-5) (plug and chug…)
A(3) = -2.1 + (3 – 1)(-5)
A(5) = -2.1 + (5 – 1)(-5)
A(8) = -2.1 + (8 – 1)(-5)
18 -11
18 (-11)
A(3) = -2.1 + (-10) = -12.1
A(5) = -2.1 + (-20) = -22.1
A(8) = -2.1 + (-35) = -37.1
Describing Number Patterns
HomeworkLesson 5-6
Homework – Practice 5-6 even
Chapter 5 Review due tomorrow
Describing Number Patterns
Practice 5-6Lesson 5-6
~ Chapter 5 ~Chapter Review
Algebra I Algebra I
~ Chapter 5 ~Chapter Review
Algebra I Algebra I
~ Chapter 5 ~Chapter Review
Algebra I Algebra I