Exponential functions and graphing them.

**Rumor** · September 22nd 2009, 07:17 AM

I'll just get right to the point and give the problem.

"If a 100 mg (milligram) tablet of an asthma drug is taken orally and if none of the drug is present in the body when the tablet is first taken, the total amount of A in the bloodstream after t minutes is modeled by the function:

A=100(1-e^(-0.05t)).

a) How many minutes are needed for 70mg of the drug to enter the bloodstream? Find the answer in decimal form.

b) Draw the graph of the function for the first two hours (120 minutes). Label the axes and indicate the value of any asymptotes for your graph."

Any help, please?

**Prove It** · September 22nd 2009, 07:22 AM

Originally Posted by Rumor

I'll just get right to the point and give the problem.

"If a 100 mg (milligram) tablet of an asthma drug is taken orally and if none of the drug is present in the body when the tablet is first taken, the total amount of A in the bloodstream after t minutes is modeled by the function:

A=100(1-e^(-0.05t)).

a) How many minutes are needed for 70mg of the drug to enter the bloodstream? Find the answer in decimal form.

b) Draw the graph of the function for the first two hours (120 minutes). Label the axes and indicate the value of any asymptotes for your graph."

Any help, please?

Let $A = 70$ .

$70 = 100\left(1 - e^{-0.05t}\right)$

$0.7 = 1 - e^{-0.05t}$

$e^{-0.05t} = 0.3$

$-0.05t = \ln{0.3}$

$t = -20\ln{0.3}$

$t = -20\left(\ln{\frac{3}{10}}\right)$

$t = -20\left(\ln{3} - \ln{10}\right)$

$t = 20\ln{10} - 20\ln{3}$

$t \approx 24.0795\,\textrm{sec}$ .

As for the graphing, do you own a graphing calculator? Do you understand how to apply transformations to the graph of $y = e^x$ ?

**Rumor** · September 22nd 2009, 07:31 AM

Originally Posted by Prove It

Let $A = 70$ .

$70 = 100\left(1 - e^{-0.05t}\right)$

$0.7 = 1 - e^{-0.05t}$

$e^{-0.05t} = 0.3$

$-0.05t = \ln{0.3}$

$t = -20\ln{0.3}$

$t = -20\left(\ln{\frac{3}{10}}\right)$

$t = -20\left(\ln{3} - \ln{10}\right)$

$t = 20\ln{10} - 20\ln{3}$

$t \approx 24.0795\,\textrm{sec}$ .

As for the graphing, do you own a graphing calculator? Do you understand how to apply transformations to the graph of $y = e^x$ ?

I do own one. When I put the original equation in, it shows up fine on the graph. But when I attempt to plug the number in for t, it gives me an error. So I must not understand it completely.

**Prove It** · September 22nd 2009, 07:43 AM

Originally Posted by Rumor

I do own one. When I put the original equation in, it shows up fine on the graph. But when I attempt to plug the number in for t, it gives me an error. So I must not understand it completely.

Remember to put in the multiplication symbol.

Also, it's possible that your window needs adjusting.

**Rumor** · September 22nd 2009, 07:54 AM

Originally Posted by Prove It

Remember to put in the multiplication symbol.

Also, it's possible that your window needs adjusting.

When entering it into y= on the calculator, it's supposed to look like this:

100(1-e^(-0.05*24.0795)) , right?

Whenever I try to graph it, it gives me a window error because when I press ZoomFit, it places the Ymin and Ymax at the same value. But even when I change the values manually, all the graph shows me is a straight line.

**Prove It** · September 22nd 2009, 07:57 AM

Originally Posted by Rumor

When entering it into y= on the calculator, it's supposed to look like this:

100(1-e^(-0.05*24.0795)) , right?

Whenever I try to graph it, it gives me a window error because when I press ZoomFit, it places the Ymin and Ymax at the same value. But even when I change the values manually, all the graph shows me is a straight line.

No.

You need to enter it as:

y = 100*(1 - e^(-0.05*x))

**Rumor** · September 22nd 2009, 08:12 AM

Originally Posted by Prove It

No.

You need to enter it as:

y = 100*(1 - e^(-0.05*x))

Ah, okay. I apologize for being so slow with this. D: But thank you for your patience! :]

**pacman** · September 22nd 2009, 08:25 AM

A=100(1-e^(-0.05t)). for t = -30 to 30

**pacman** · September 22nd 2009, 08:29 AM

A=100(1-e^(-0.05t)). for t = -30 to 150

**Prove It** · September 22nd 2009, 08:40 AM

Originally Posted by pacman

A=100(1-e^(-0.05t)). for t = -30 to 150

Note that $t \geq 0$ as it is impossible to have negative time.

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Exponential functions and graphing them.

graph of A=100(1-e^(-0.05t)).

t = -30 to 150

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