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Representation of Functions
Equations and functions will be used "interchangeably" ... 8th grade only uses equations while Integrated uses functions. Since we have to learn 8th grade and Integrated simultaneously, both terms will be seen. Know that, for our purposes, y = f(x).
This unit includes writing an equation...
• from a graph
• from 2 points
• from a table
• from a scenario
This unit includes writing an equation...
• from a graph
• from 2 points
• from a table
• from a scenario
...from a graph
• Know that m = rise/run (from one point to another)
• b = the y-coordinate of where the line crosses the y-axis
• b = the y-coordinate of where the line crosses the y-axis
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fromgraph1.pdf |
fromgraph2.pdf |
fromgraph3.pdf |
...from 2 points
• Know that m = y2-y1/x2-x1
• Take a point (x,y) and your newly-found m and plug back into y=mx+b t solve for b
• Rewrite y=mx+b, substituting in m and b
• Take a point (x,y) and your newly-found m and plug back into y=mx+b t solve for b
• Rewrite y=mx+b, substituting in m and b
from2points1.pdf |
from2points2.pdf |
from2points3.pdf |
from2points4.pdf |
from2points5.pdf |
...from a table
In the table below, we have x and f(x). If f(x)=y, then we end up with x and y. When we have x and y, we have an ordered pair (x,y).
• When x=2, f(x)=7, so we can represent f(2)=7 as the point (2,7)
• When x=4, f(x)=9, so we can represent f(4)=9 as the point (4,9)
• When x=10, f(x)=15, so we can represent f(10)=15 as the point (10,15)
You might have been able to figure out that the function being described here is f(x) = x+5
• When x=2, f(x)=7, so we can represent f(2)=7 as the point (2,7)
• When x=4, f(x)=9, so we can represent f(4)=9 as the point (4,9)
• When x=10, f(x)=15, so we can represent f(10)=15 as the point (10,15)
You might have been able to figure out that the function being described here is f(x) = x+5
fromtable1.pdf |
fromtable2.pdf |
...from a scenario
Let m = the repeated change of value (look for words like per, every, each)
Let b = the initial value (what you start with before the "problem" begins
Example 1
Mr. Lawrence gives Mr. Guesto $50 to teach Integrated Math. (Not true, sadly.) He decided to give him an additional $20 for each student that gets an A at the end of 1st Quarter.
Let x = the number of students that get an A
Let y = the amount of money that Mr. Guesto makes.
In this situation, m = 20 because $20 is earned for each student (and can be earned repeatedly). If 6 students get an A, it's 20+20+20+20+20+20 or (20*6).
In this situation, b = 50 because the $50 is earned at the beginning (before any students receive an A) and is only given one time.
Thus, the situation can be represented by the equation y = 20 x + 50
We can easily turn this into a function if we let:
a = the number of students receiving an A for 1st Quarter
m(a) = the amount of money given to Mr. Guesto
...then our function would be m(a) = 20a + 50
Example 2
Ms. L sells some artwork at a local market. It costs her $85 to set up a tent, but she the work for $12 per piece. She keeps all profits.
Let x = the number of pieces of art sold
Let y = the amount of money that Ms. L earns
• In this situation, m = 12 because $12 is the price per piece. (Selling 3 pieces would yield $36 because 3*12=36)
• In this situation, b = -85 because she must pay $85 and is NOT making $85.
• Thus, the situation can be represented by the equation y = 12x - 85
• As a function, if we let:
a = the number of pieces of art
P(a) = the profit made
...then our function would be P(a) = 12a - 85
• To extend this problem a bit, Ms. L would have to sell at least 8 pieces of art to make a profit.
P(7) = -1 means that selling 7 pieces of art would yield a "profit" of $-1...which means she actually lost money :(
P(8) = 11 means that selling 8 pieces of art would yield a profit of $11
• Mathematically, the domain would be (-∞,+∞) because you can put any real number in for x
• Contextually the domain would be [0,n] where n = the total number of pieces available to sell. The lower bound is 0 because she can't sell negative pieces and the upper bound is n because she can't sell more pieces than what she has.
• Contextually, the lower bound of the range would be [-85] because she can't LOSE any money after paying the $85 tent fee. The upper bound of the range would be dependent on the amount of pieces she has available to sell.
• We could set up an inequality to represent the question, "How many pieces of art would Ms. L have to sell in order to at least break even?
• If we define breaking even as having a "profit" of $0 or more, then we establish the inequality 12a - 85 ≥ 0
• Solving the inequality, we could get a ≥ 7.083, but since a can only be an integer (she can't sell partial pieces), we would have to round up to 8.
Let b = the initial value (what you start with before the "problem" begins
Example 1
Mr. Lawrence gives Mr. Guesto $50 to teach Integrated Math. (Not true, sadly.) He decided to give him an additional $20 for each student that gets an A at the end of 1st Quarter.
Let x = the number of students that get an A
Let y = the amount of money that Mr. Guesto makes.
In this situation, m = 20 because $20 is earned for each student (and can be earned repeatedly). If 6 students get an A, it's 20+20+20+20+20+20 or (20*6).
In this situation, b = 50 because the $50 is earned at the beginning (before any students receive an A) and is only given one time.
Thus, the situation can be represented by the equation y = 20 x + 50
We can easily turn this into a function if we let:
a = the number of students receiving an A for 1st Quarter
m(a) = the amount of money given to Mr. Guesto
...then our function would be m(a) = 20a + 50
Example 2
Ms. L sells some artwork at a local market. It costs her $85 to set up a tent, but she the work for $12 per piece. She keeps all profits.
Let x = the number of pieces of art sold
Let y = the amount of money that Ms. L earns
• In this situation, m = 12 because $12 is the price per piece. (Selling 3 pieces would yield $36 because 3*12=36)
• In this situation, b = -85 because she must pay $85 and is NOT making $85.
• Thus, the situation can be represented by the equation y = 12x - 85
• As a function, if we let:
a = the number of pieces of art
P(a) = the profit made
...then our function would be P(a) = 12a - 85
• To extend this problem a bit, Ms. L would have to sell at least 8 pieces of art to make a profit.
P(7) = -1 means that selling 7 pieces of art would yield a "profit" of $-1...which means she actually lost money :(
P(8) = 11 means that selling 8 pieces of art would yield a profit of $11
• Mathematically, the domain would be (-∞,+∞) because you can put any real number in for x
• Contextually the domain would be [0,n] where n = the total number of pieces available to sell. The lower bound is 0 because she can't sell negative pieces and the upper bound is n because she can't sell more pieces than what she has.
• Contextually, the lower bound of the range would be [-85] because she can't LOSE any money after paying the $85 tent fee. The upper bound of the range would be dependent on the amount of pieces she has available to sell.
• We could set up an inequality to represent the question, "How many pieces of art would Ms. L have to sell in order to at least break even?
• If we define breaking even as having a "profit" of $0 or more, then we establish the inequality 12a - 85 ≥ 0
• Solving the inequality, we could get a ≥ 7.083, but since a can only be an integer (she can't sell partial pieces), we would have to round up to 8.
Average Rate of Change
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If a video references a "secant" line, it's talking about a straight line that connect two points on a curve.
Tangent lines aren't brought up in Integrated I, II, or III, but are a fundamental part of pre-calculus and calculus.
Tangent lines aren't brought up in Integrated I, II, or III, but are a fundamental part of pre-calculus and calculus.
Scatterplots
Correlation vs. Causation
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The Correlation Coefficient
Rearranging Formulas: Standard to Slope-Intercept
2 things to check when you get to Standard Form:
- A (the coefficient of x) cannot be negative
- There cannot be fractions (for A, B, or C)
- A (the coefficient of x) cannot be negative
- There cannot be fractions (for A, B, or C)
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Equations to Graphs
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(For the second video, start at 10:00 to see the "weird" cases of 0 and undefined slopes...)