How can I accurately sketch a complex graph with functions like 2x-⅜+¾e^-2x?

[Kajan thana]

Hi guys,

I need some help on sketching graph complex functions such as ( 2x-⅜+¾e^-2x).
Can someone please help me on sketching a graph like the one that I mentioned above. Is there any useful videos or website I can use. And please let me know if there are any good tips to get accurate graph.

I know the basics such find the y and x intercepts and the turning points. How would I able to recognise the general shape and if there is any asymptotes.

Thank you so much.

 

[FactChecker]

In examples like yours, look for divisions by zero and exponents of negative numbers, where the function would be undefined. The function may be asymptotic as x approaches those places. In general, you may be able to piece together the behavior of simple parts and terms of the equation (exponentials, periodic trig functions, low order polynomials, etc.)

There is a limit of what you can do without plugging in numbers. If you want any accuracy at all, you will ultimately have to plug in numbers.

EDIT: changed "negative exponents" to "exponents of negative numbers"

 

[Kajan thana]

In examples like yours, look for divisions by zero and negative exponents, where the function would be undefined. The function may be asymptotic as x approaches those places. In general, you may be able to piece together the behavior of simple parts and terms of the equation (exponentials, periodic trig functions, low order polynomials, etc.)

There is a limit of what you can do without plugging in numbers. If you want any accuracy at all, you will ultimately have to plug in numbers.

So for the example above,( 2x-⅜+¾e^-2x), there is no asymptotes, am I right ?

 

[mfb]

$2x-\frac 3 8 + \frac 3 4 e^{-2x}$? If that is the correct interpretation, it doesn't have vertical asymptotes.

 

[Kajan thana]

$2x-\frac 3 8 + \frac 3 4 e^{-2x}$? If that is the correct interpretation, it doesn't have vertical asymptotes.

but if the value of x increase then e^-2x eventually be zero, but the ( 2x-3/8) will be there still[/QUOTE]

 

[mfb]

2x - 3/8 is a linear function, it does not have vertical asymptotes. It has a different asymptote, sure.

 

[Kajan thana]

2x - 3/8 is a linear function, it does not have vertical asymptotes. It has a different asymptote, sure.

So how do we determine if a function have a astmptotes

 

[mfb]

There is no general set of rules that works for every function (at least not until you get to Laurent series). Look at its components, see if some things converge to a fixed value, see what happens to the rest.