• Support PF! Buy your school textbooks, materials and every day products Here!

Graphing functions and identifying features

  • Thread starter zebra1707
  • Start date
  • #1
107
0

Homework Statement



Hi there

I need assistance with two graphs that are causing me some problems

y=e(^-2x)(-1) and y=ln(x-2)+1

Homework Equations



I just need some guidence as to where to start - starting with a table - what range is approriate?

The Attempt at a Solution



Stuck
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,770
911
[itex]e^x[/itex] goes to 0 very rapidly for x< 0 and goes up very rapidly for x> 0. I would recommend taking x from -1 or -2 to +4 or +5.

ln(x) is only defined for x> 0 so ln(x- 2) is only defined for x> 2. I would recommend taking x from 2 up to, say 10.

Wouldn't it have been faster to just play around with some numbers rather than wait for someone to respond here? Do you not have a graphing calculator? It would have taken only a few seconds to try various value on a calculator.
 
  • #3
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,219
945

Homework Statement



Hi there

I need assistance with two graphs that are causing me some problems

y=e(^-2x)(-1) and y=ln(x-2)+1

Homework Equations



I just need some guidance as to where to start - starting with a table - what range is appropriate?

The Attempt at a Solution



Stuck
I assume you mean [itex]\displaystyle y=(-1)e^{-2x}\,,[/itex] which you could write as y = (e^(-2x))(-1), (the location of parentheses is important) or y = -e^(-2x), or better yet, y = -e-2x,

and y = ln(x-2) + 1 .

Are you familiar with the graphs of:
[itex]\displaystyle y=e^{x}\,,[/itex]​
and
[itex]\displaystyle y=\ln(x)\ ?[/itex]​
That's the place to start.

Then use what you've (hopefully) been learning about shifting, stretching, shrinking, flipping, etc. graphs.
 
  • #4
107
0
[itex]e^x[/itex] goes to 0 very rapidly for x< 0 and goes up very rapidly for x> 0. I would recommend taking x from -1 or -2 to +4 or +5.

ln(x) is only defined for x> 0 so ln(x- 2) is only defined for x> 2. I would recommend taking x from 2 up to, say 10.

Wouldn't it have been faster to just play around with some numbers rather than wait for someone to respond here? Do you not have a graphing calculator? It would have taken only a few seconds to try various value on a calculator.
Thank you, yes, you are right. I think just a lack of confidence in this aspect of Maths. Sorry to be a bother. Regards
 
  • #5
107
0
I assume you mean [itex]\displaystyle y=(-1)e^{-2x}\,,[/itex] which you could write as y = (e^(-2x))(-1), (the location of parentheses is important) or y = -e^(-2x), or better yet, y = -e-2x,

and y = ln(x-2) + 1 .

Are you familiar with the graphs of:
[itex]\displaystyle y=e^{x}\,,[/itex]​
and
[itex]\displaystyle y=\ln(x)\ ?[/itex]​
That's the place to start.

Then use what you've (hopefully) been learning about shifting, stretching, shrinking, flipping, etc. graphs.
Many thanks, I appreciate the guidance.
 

Related Threads for: Graphing functions and identifying features

  • Last Post
Replies
4
Views
1K
Replies
2
Views
5K
  • Last Post
Replies
1
Views
1K
Replies
6
Views
1K
  • Last Post
Replies
7
Views
832
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
7
Views
548
Top