Graphing functions and identifying features

[zebra1707]

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

 

[HallsofIvy]

e^x 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.

 

[SammyS]

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 \displaystyle y=(-1)e^{-2x}\,, 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:

\displaystyle y=e^{x}\,,​

and

\displaystyle y=\ln(x)\ ?​

That's the place to start.

Then use what you've (hopefully) been learning about shifting, stretching, shrinking, flipping, etc. graphs.

 

[zebra1707]

e^x 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

 

[zebra1707]

I assume you mean \displaystyle y=(-1)e^{-2x}\,, 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:

\displaystyle y=e^{x}\,,​

and

\displaystyle y=\ln(x)\ ?​

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.