Thread: A line perpindicular to an equation

1. A line perpindicular to an equation

Hi everone,

I'm currently reviewing for a Calculas exam I have soon, and although there are a few review questions I can't seem to get, I'll stick with this one. (By the way, if it is super simple, excuse me, I really CANT do calculas.)

Question: Determine the equation of the line perpindicular to f(x)=1/2e^x+1 at its y-intercept.

What I did was found the dervitive, which I got to be the same thing (f'(x)=1/2e^x+1) although it could be wrong. Only problem is that I have no clue what to do after that. The answer is suppose to be y=e/2 x+e/2

Any help is appreciated!

2. Originally Posted by memoIhad
Hi everone,

I'm currently reviewing for a Calculas exam I have soon, and although there are a few review questions I can't seem to get, I'll stick with this one. (By the way, if it is super simple, excuse me, I really CANT do calculas.)

Question: Determine the equation of the line perpindicular to f(x)=1/2e^x+1 at its y-intercept.

What I did was found the dervitive, which I got to be the same thing (f'(x)=1/2e^x+1) although it could be wrong. Only problem is that I have no clue what to do after that. The answer is suppose to be y=e/2 x+e/2

Any help is appreciated!
Well, since it requires a line through the y-intercept, it would be a good idea to find that! The y-intercept of a graph is where it crosses the y-axis: where x= 0. Now assuming you meant $\displaystyle f(x)= \frac{1}{2}e^{x+1}$ which is NOT what you wrote (what you wrote was $\displaystyle f(x)= \frac{1}{2}e^x+ 1$) but does have the property that its derivative is still $\displaystyle \frac{1}{2}e^{x+1}$, then since $\displaystyle f(0)= \frac{1}{2}e$, the y-intercept is $\displaystyle (0, \frac{1}{2}e)$. The slope of the tangent line to the graph is the derivative evaluated at x= 0 which is again $\displaystyle \frac{1}{2}e$. Do you know how to find the slope and equation of the perpendicular line from that?

3. Originally Posted by HallsofIvy
Well, since it requires a line through the y-intercept, it would be a good idea to find that! The y-intercept of a graph is where it crosses the y-axis: where x= 0. Now assuming you meant $\displaystyle f(x)= \frac{1}{2}e^{x+1}$ which is NOT what you wrote (what you wrote was $\displaystyle f(x)= \frac{1}{2}e^x+ 1$) but does have the property that its derivative is still $\displaystyle \frac{1}{2}e^{x+1}$, then since $f(0)= \frac{1}{2}e$, the y-intercept is $\displaystyle (0, \frac{1}{2}e)$. The slope of the tangent line to the graph is the derivative evaluated at x= 0 which is again [tex]\frac{1}{2}e[/itex]. Do you know how to find the slope and equation of the perpendicular line from that?
Thanks for the quick reply. Yes, I did mean the equation you wrote . .....EDIT: nvm, I understand how you found the y-intrecept....As for the equation of the line perpindicular to that, I would assume that it would a slope that is the negative reciprocal of the original equation, so -2? So do I just plug that into the equation of the line? Which would be y-y1=m(x-x1)?

4. Hi

As written by HallsofIvy the slope of the tangent line to the graph is the derivative evaluated at x= 0 which is $\displaystyle \frac{1}{2}e$.

This means that a direction vector of the tangent has coordinates (1,e/2).

You have to find the coordinates of a vector perpendicular to this one.

5. Originally Posted by memoIhad
Thanks for the quick reply. Yes, I did mean the equation you wrote . .....EDIT: nvm, I understand how you found the y-intrecept....As for the equation of the line perpindicular to that, I would assume that it would a slope that is the negative reciprocal of the original equation, so -2? So do I just plug that into the equation of the line? Which would be y-y1=m(x-x1)?
The slope of the tangent line is NOT 1/2. It is, as I said, $\displaystyle \frac{1}{2}e$ or $\displaystyle \frac{e}{2}$. So the slope of the perpendicular line is $\displaystyle -\frac{2}{e}$.

Now, what is the equation of a line with slope $\displaystyle -\frac{2}{e}$ passing through $\displaystyle (0, \frac{e}{2})$?

6. So, would the answer just be y=-2/e+e/2 ? that's the answer I got by plugging in the points (m=-2/e, y=e/2) The answer though, is different...am I missing something?

Thanks a lot for being patient with me

7. Originally Posted by memoIhad
So, would the answer just be y=-2/e+e/2 ? that's the answer I got by plugging in the points (m=-2/e, y=e/2) The answer though, is different...am I missing something?

Thanks a lot for being patient with me
Your answer is correct (apart from the observation made in the next post!)

The answer is suppose to be y=e/2 x+e/2
This is the equation of the tangent at the y-intercept.

8. Originally Posted by memoIhad
So, would the answer just be y=-2/e+e/2 ? that's the answer I got by plugging in the points (m=-2/e, y=e/2) The answer though, is different...am I missing something?
Well, yes, you are missing something! There is no "x"!
The equation of the line with slope -2/e, passing through (0, e/2) is
y= (-2/e)x+ e/2.

Thanks a lot for being patient with me