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Help needed with graphing exponential functions

Hi

In the question I have been wondering how to do which is on a calculus paper, first it shows this table of values:

x 0 0.5 1 1.5 2 2.5 3
y=3e^2x 3 8.15 22.17 60.26 163.79 445.24 1210.27

Then the question asks to draw the above graph, however I do not know where to start.

Do I need to differentiate the y=3e^2x?

Or do I plot it as it is?

Thanks
Reply 1
Avatar for davros
davros
16
Original post by dc2209
Hi

In the question I have been wondering how to do which is on a calculus paper, first it shows this table of values:

x 0 0.5 1 1.5 2 2.5 3
y=3e^2x 3 8.15 22.17 60.26 163.79 445.24 1210.27

Then the question asks to draw the above graph, however I do not know where to start.

Do I need to differentiate the y=3e^2x?

Or do I plot it as it is?

Thanks


Just plot it - you have the values!

There are no subtleties with the exponential function - no maxima or minima - and you should know the general shape of the curve. The only thing to mark is the y-intercept value.
Reply 2
Avatar for dc2209
dc2209
OP
2
Original post by davros
Just plot it - you have the values!

There are no subtleties with the exponential function - no maxima or minima - and you should know the general shape of the curve. The only thing to mark is the y-intercept value.


Ah thank you it seems I was over complicating it.

What do you mean mark the y=intercept value? I thought the y axis needed to follow the structure of y=mx+c and C is the intercept? I'm a bit confused

Thanks anyway
Reply 3
Avatar for dc2209
dc2209
OP
2
Original post by davros
Just plot it - you have the values!

There are no subtleties with the exponential function - no maxima or minima - and you should know the general shape of the curve. The only thing to mark is the y-intercept value.



Also, when I enter the question in wolfram alpha, I get this http://gyazo.com/dfef4ded2cf3f1b963be38ffae4e1b8e

however as you can see the question provides values on the y axis which go up to 1210.27?
Reply 4
Avatar for davros
davros
16
Original post by dc2209
Ah thank you it seems I was over complicating it.

What do you mean mark the y=intercept value? I thought the y axis needed to follow the structure of y=mx+c and C is the intercept? I'm a bit confused

Thanks anyway


The y-intercept is just where the graph crosses the y-axis i.e. the value when x = 0.

Original post by dc2209
Also, when I enter the question in wolfram alpha, I get this http://gyazo.com/dfef4ded2cf3f1b963be38ffae4e1b8e

however as you can see the question provides values on the y axis which go up to 1210.27?


If you look at what it's plotted, it thinks you've entered y=(3e2)xy = (3e^2) x not y=3e2xy = 3e^{2x} so it's given you a straight line! I don't know wolfram alpha but I suspect you need to enter some brackets around the 2x.
Reply 5
Avatar for dc2209
dc2209
OP
2
Original post by davros
The y-intercept is just where the graph crosses the y-axis i.e. the value when x = 0.



If you look at what it's plotted, it thinks you've entered y=(3e2)xy = (3e^2) x not y=3e2xy = 3e^{2x} so it's given you a straight line! I don't know wolfram alpha but I suspect you need to enter some brackets around the 2x.



Ah I see thank you it is figured out now and I have drawn the graph.

Btw if I wanted find the gradient when x = 2 from the graph would that be as simple as drawing a line from x = 2 to the curve and then reading the value on the y axis?

Thanks
Original post by dc2209
...


No. That is f(2) f(2) while the gradient is f′(2) f'(2) which requires differentiation. :K:
Reply 7
Avatar for dc2209
dc2209
OP
2
Original post by lazy_fish
No. That is f(2) f(2) while the gradient is f′(2) f'(2) which requires differentiation. :K:




so it isnt right?

this is the graph i have drawn

It should follow this general shape...except with differing values (as the equation is different)
inv_egraph.gif2.9KB

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