Solving Exercise 5 Algebra II Honors

  • Thread starter Thread starter darshanpatel
  • Start date Start date
  • Tags Tags
    Algebra Honors
Click For Summary

Homework Help Overview

The discussion revolves around finding the equation of a secant line through specified points based on a given function, f(x) = sqrt(x-1). Participants are exploring the requirements of the problem and the necessary steps to derive the equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants express confusion about the problem's wording and the process of finding the secant line's equation. Questions arise regarding the specific function values at given points and the correct form of the line equation. Some participants attempt to apply point-slope form and seek validation of their approaches.

Discussion Status

The discussion is active, with participants sharing their attempts and seeking clarification on mathematical notation and processes. Some guidance has been provided regarding the use of point-slope form and the importance of verifying calculations against initial data.

Contextual Notes

Participants are working under the constraints of a homework exercise that requires them to derive an equation based on a specific function and given points. There is an emphasis on understanding the mathematical concepts rather than simply arriving at a solution.

darshanpatel
Messages
139
Reaction score
0
I don't get the wording of the problem. What does it mean find the equation?

Problem: Find the equation of the secant line through the points (x, f(x)) and (x2, f(x2)) for parts (a)-(e) in Exercise 5.

There is no work because I don't know where to begin.

Just as an example, (a) in Exercise 5 is x=2 and x2=3...
What do I do?

They also give me m(sec)= f(x2)-f(x)/x2-x
 
Physics news on Phys.org
darshanpatel said:
I don't get the wording of the problem. What does it mean find the equation?

Problem: Find the equation of the secant line through the points (x, f(x)) and (x2, f(x2)) for parts (a)-(e) in Exercise 5.

There is no work because I don't know where to begin.

Just as an example, (a) in Exercise 5 is x=2 and x2=3...
What do I do?

They also give me m(sec)= f(x2)-f(x)/x2-x

The equation of a line is often of the form
y = mx + b

for instance, y = 3x - 7​
 
I know it is y=mx+b but how would i find that? Can you please show using the example numbers?
 
darshanpatel said:
I know it is y=mx+b but how would i find that? Can you please show using the example numbers?

Were you given a specific function for this problem?
 
Yes for exercise 5 it said consider the function given by f(x)=sqrt(x-1)
 
What are f(2) and f(3) ?
 
what do you mean?
 
SammyS said:
What are f(2) and f(3) ?

darshanpatel said:
what do you mean?

You are given the function [itex]f(x)=\sqrt{x-1}\,.[/itex]

So I asked, "What are f(2) and f(3) ?" . That's a very basic question.
 
oh, f(2)=1 and f(3)=sqrt2
 
  • #10
I tried putting it into point-slope form and got y-1=(sqrt2-1)(x-2), is that a right start?

Reduced that down to y=(sqrt2x -x)-2sqrt2 +3
 
Last edited:
  • #11
SammyS u still want to help me?
 
  • #12
darshanpatel said:
I tried putting it into point-slope form and got y-1=(sqrt2-1)(x-2), is that a right start?
Yes
Reduced that down to y=(sqrt2x -x)-2sqrt2 +3
Right, though it is usually better to write it with just one x,
viz., y = (sqrt2 - 1)x -2sqrt2 + 3

But you are not finished yet. The final step is to check that this equation fits your initial data, to avoid the embarrassment of scoring some red crosses when your work is marked.

When x=2, does this produce a y value of 1?
when x=3, does this give y = sqrt2?

If this all tallies, then it must be right. :smile:
 
  • #13
darshanpatel said:
I tried putting it into point-slope form and got y-1=(sqrt2-1)(x-2), is that a right start?

Reduced that down to y=(sqrt2x -x)-2sqrt2 +3
This -- sqrt2x -- is terrible notation because it is ambiguous. Does it mean [itex]\sqrt{2x}[/itex] or does it mean [itex]\sqrt{2}\cdot x[/itex]? Because this expression came from (=(√2-1)(x-2), you apparently intend for what you wrote to mean [itex]\sqrt{2}\cdot x[/itex]. A better way to write that is to put x in front of the radical, as x√2, which is clear and unambiguous.
 
  • #14
Thank you, sorry, i see it written a lot like sqrt2 or what ever, but how do you do the symbols?
 
  • #15
darshanpatel said:
Thank you, sorry, i see it written a lot like sqrt2 or what ever, but how do you do the symbols?
To use the typesetting fonts and symbols, you have to invest time in learning Latex-family formatting. This site makes it as effortless as possible: http://www.codecogs.com/latex/eqneditor.php

You can construct your itex formatting on that site, then cut and paste it into your posts.
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
6
Views
5K
  • · Replies 23 ·
Replies
23
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
9
Views
2K
Replies
2
Views
2K