Solve for Linear Function: Algebra 2 Help | Homework Statement

In summary, the conversation discusses a problem with understanding linear functions and slope. The speaker explains their confusion with the process of finding the slope and asks for clarification. They also mention their struggles with ADHD and difficulty focusing. The expert provides a summary of the conversation and offers advice for better understanding the material.
  • #1
offtheleft
131
1

Homework Statement



having a little bit of trouble. some of the information is new to me. ill point out because I am copying this down from my notes.

let A & B define some linear function, [tex]y=f(x)[/tex].
A = (1, 2), B = (6, 6)
C = (x, y), D = (x, y)

let (x, y) represent some arbitrary points on the line defined by f but, not on points A or B.

C is between A and B and, D is somewhere past B.

so far, i know what's goin on. i have the slope which is [tex]\frac{4}{5}[/tex].

here's where I'm lost, i might have zoned out because of my ADHD but this is what's going on

1, from A [tex]\rightarrow[/tex] C, [tex]\frac{y-2}{x-1}[/tex]

2, from B [tex]\rightarrow[/tex] D, [tex]\frac{y-6}{x-6}[/tex]

3, from B [tex]\rightarrow[/tex] C, [tex]\frac{-(6-y)}{-(6-x)}[/tex]

4, and from A [tex]\rightarrow[/tex] D is the same as A [tex]\rightarrow[/tex] C


I have no idea how all that was figured out ( from 1 - 4 ) someone please explain that to me?

and it continues into the next part

[tex]\frac{y-2}{x-1}[/tex] = [tex]\frac{y-6}{x-6}[/tex]

i can comprehend that.

[tex](x-1)(x-6)[/tex] [tex]\frac{y-2}{x-1}[/tex] = [tex](x-1)(x-6)[/tex] [tex]\frac{y-6}{x-6}[/tex]

not sure how that all happened.

than it goes too..

[tex]\frac{x-1}{x-1}[/tex] [tex][(x-6)(y-2)][/tex] = [tex]\frac{x-6}{x-6}[/tex] [tex][(x-1)(y-6)][/tex]

(this is where i started paying attention again)
cancel stuff out and do some fun algebraic gymnastics and i came up with this;

[tex]xy-2x-6y+12=xy-6x-y+6[/tex]

more fun algebra stuff and i came up with;

[tex]y=\frac{4}{5}x+\frac{6}{5}[/tex]







i have no problem doing certain things. but that whole middle piece just lost me until it ws put into some fun polynomial expression. i need to know how i got that
 
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  • #2
offtheleft said:

Homework Statement



having a little bit of trouble. some of the information is new to me. ill point out because I am copying this down from my notes.

let A & B define some linear function, [tex]y=f(x)[/tex].
A = (1, 2), B = (6, 6)
C = (x, y), D = (x, y)

let (x, y) represent some arbitrary points on the line defined by f but, not on points A or B.

C is between A and B and, D is somewhere past B.

so far, i know what's goin on. i have the slope which is [tex]\frac{4}{5}[/tex].

here's where I'm lost, i might have zoned out because of my ADHD but this is what's going on

1, from A [tex]\rightarrow[/tex] C, [tex]\frac{y-2}{x-1}[/tex]

2, from B [tex]\rightarrow[/tex] D, [tex]\frac{y-6}{x-6}[/tex]

3, from B [tex]\rightarrow[/tex] C, [tex]\frac{-(6-y)}{-(6-x)}[/tex]
Yes, and each of those is equal to 4/5. you don't really need the "-" in (3) because -1/-1= 1. [tex]\frac{6- y}{6- x}[/itex] is sufficent. Of course that is exactly the same as [tex]\frac{y- 6}{x- 6}[/tex].

4, and from A [tex]\rightarrow[/tex] D is the same as A [tex]\rightarrow[/tex] C
Well yes, because C= D= (x,y)!


[/quote]I have no idea how all that was figured out ( from 1 - 4 ) someone please explain that to me?[/quote]
There wasn't anything "figured out" until you add the "= 4/5" part! What you wrote in 1- 4 is just [tex]\frac{y_1- y_0}{x_1- x_0}[/tex] for two points [tex](x_0, y_0)[/tex] and [tex](x_1, y_1)[/tex]. Those are all from the definition of "slope" of a line which you had already figured out was 4/5.

and it continues into the next part

[tex]\frac{y-2}{x-1}[/tex] = [tex]\frac{y-6}{x-6}[/tex]

i can comprehend that.
Do you know the definition of "slope" of a line? Do you know how [tex]tan(\theta)[/tex] is defined where [itex]\theta[/tex] is an angle of right triangle? Without knowing those, you can't comprehend it.

[tex](x-1)(x-6)[/tex] [tex]\frac{y-2}{x-1}[/tex] = [tex](x-1)(x-6)[/tex] [tex]\frac{y-6}{x-6}[/tex]

not sure how that all happened.
You are clearing the fractions by multiplying both sides by the denominators of the fractions.

than it goes too..

[tex]\frac{x-1}{x-1}[/tex] [tex][(x-6)(y-2)][/tex] = [tex]\frac{x-6}{x-6}[/tex] [tex][(x-1)(y-6)][/tex]
Okay, just to make it clear what is happening they move the corresponding numerator and denominator together so you can see they cancel. That was what I said before: they are clearing the fractions by multiplying by the denominators so the denominators cancel.

(this is where i started paying attention again)
So you weren't paying attention before? No wonder you didn't understand!

cancel stuff out and do some fun algebraic gymnastics and i came up with this;

[tex]xy-2x-6y+12=xy-6x-y+6[/tex]

more fun algebra stuff and i came up with;

[tex]y=\frac{4}{5}x+\frac{6}{5}[/tex]







i have no problem doing certain things. but that whole middle piece just lost me until it ws put into some fun polynomial expression. i need to know how i got that
Sometimes you can do mathematics by memorizing formulas. Sometimes you actually have to think. If you get to a part you don't understand, DON'T stop "paying attention". Concentrate on it and look at every detail. There was nothing there that you hadn't seen before.
 
  • #3
i didnt mean to stop paying attention. i have REALLY bad ADHD.

so that whole section where i had 1 - 4 was just a more complicated or elaborate way to find the slope?

after you clarified a few things i understand it more except for the [tex]
tan(\theta)[tex]. when it comes to sin, cos, tan, etc... I am lost. i had geom in high school, three years later i went from algebra 1 straight to algebra 2.
 

1. What is a linear function?

A linear function is a mathematical function that can be represented by a straight line on a graph. It follows the form y = mx + b, where m is the slope of the line and b is the y-intercept.

2. How do I solve for a linear function?

To solve for a linear function, you need to have two points on the line. You can then use the slope formula (m = (y2 - y1) / (x2 - x1)) to find the slope, and then plug the slope and one of the points into the slope-intercept form (y = mx + b) to find the y-intercept. This will give you the equation of the linear function.

3. What are some real-life applications of linear functions?

Linear functions are commonly used in areas such as economics, physics, and engineering. They can be used to model relationships between variables, such as the relationship between cost and quantity produced in economics, or the relationship between distance and time in physics.

4. How can I check if a set of points represents a linear function?

To check if a set of points represents a linear function, you can plot the points on a graph and see if they form a straight line. You can also calculate the slope between each pair of points and see if they are all equal, which is a property of linear functions.

5. Can linear functions have negative slopes?

Yes, linear functions can have negative slopes. This means that the line will be decreasing from left to right on a graph. The sign of the slope is determined by the relationship between the x and y values of the points on the line, not by the direction of the line on the graph.

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