- 46 Views
- Uploaded on
- Presentation posted in: General

By: Steven Jacob. Algebra 1 Functions. Problem.

Algebra 1 Functions

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

By: Steven Jacob

Algebra 1 Functions

- Clark was calling on his cell-phone. Each minute costs 72 cents each. If he wants to know how much it would cost for each minute he uses it, what equation would you use to solve it? If a deal said that for every 10 minutes of the use of the phone costs $7.00, and Clark wanted to use his phone for 15 minutes of his phone, which minute plan would be cheaper? What is the Independent and Dependent Variables?

- Solving this type of equation is very simple.
- The first step to solving this algebraic equation is to write down a reasonable formula.
- The formula to this equation is C=x(.72)

- C represents the cost while x is the variable that represents the number of minutes that Clark uses his phone. .72 represents the cost of each minute.
- The second question states that if Clark received a deal for every 10 minutes of the use of his cell phone, it would cost him $7.00. He wants to use 15 minutes of his cell phone.

- To Solve this equation, you can cut $7.00 in half and get $3.50. Then you add $3.50 to $7.00 and get $10.50 for 15 minutes. The other deal was, for every minute on the cell phone. It would cost them 72 cents. You would multiply C=15(.72) and get $10.80.
- As a result, the 10 minute deal was better.

- The Cost of the minutes depends on how long Clark is using his phone.
- Tip( Y Depends on X!)
- It makes a positive trend because, the more minutes Clark uses his phone, the more expensive is the cost.

- The domain of an Algebraic expression is all of your X quadrant values. The range of an expression is all of your Y quadrant values.
- Let’s say my domain of an equation is (5) (10) (15) (20) (25)
- Let’s say my range of an equation is (-5) (-10) (-15) (-20) (-25)

- If I were to write my Domain and Range values in Set Builder Notation, it would look like this
- 5 ≤ x ≥ 25
- -25 ≤ y ≥ -5

- The Domain is Discrete.
- This is because, when you graph the function, the points are plotted as separate coordinates and cannot have points that represent part of a whole number.

- There are 3 types of functions.
- Linear
- Quadratic
- V shaped

- Linear Equations form a diagonal line with plot points that correspond with each other. To find if an equation is linear, you would normally look for x=y.

- These equations have a U shape known as a parabola to the equations. Whenever you want to know if an equation is quadratic, you would normally look for y=x .

2

- Absolute Value equations normally have a pattern with integers. They form a V shape on a graph. Suppose, you have a set of domains such as D:(-2,-1,0,1,2) To know if an equation is V Shaped, you would look for the absolute value, or y= x .

- If I had the equation y=2x + -.75, I could graph it with several variables.
- I could choose my domain which could be, D:(-2,-1,0,1,2)
- To find my Range values, I would have to input my Domain into the equations.

2

- I could rewrite my equations in function notation form.
- Writing a function notation form is very simple. It just requires you to replace Y as f(x).
- F(x) stands for the “function of x.”

- If I were to solve f(x)= 2x -.75, I would have to input my domains first!
- My domain would be D:(-2,-1,0,1,2)
- To find My Range Values, I can input my domain to my equation.
- An example of this is…

2

2

2

2

2

2

2

- By: Steven Jacob