Isn’t the dy/dx of the second graph 3(-2e^(-2(x-1) +2e^( 1(x-1))? since de^u/dx= e^u *du/dx and d-2(x-1) /dx =-2?
**3(-2e^(-2(x-1) +2e^( 1(x-1))
Oops, yes of course you are right. Thanks very much for pointing it out. Not sure how I wrote what I did — definite brain-fade. (Of course, the shape of the graph is unchanged).
You got your x and y co ordinate the wrong way round on the second graph
Doesn’t look like it to me. Can you explain what you mean in more detail?
yea, its completely wrong, wrong way round
No it isn’t. Try typing it in to your graph sketching calculator or sketching it on graphsketch.com
I am grateful for everything you are doing with your website, but its just that your x-axis intercept has the coordinates interchanged, thats all.
Ah, that’s what you mean! Yeah you’re right, thanks for pointing it out.
Hi, I was just wondering how do u know that the y is less than 0 when when x tends to infinity? is it based on observation or is there some formula? here we rely too much on GC, our graph isnt that strong thanks!!
As x tends to infinity, e^-(x-1) is bigger than e^-2(x-1) (because the second term is the square of the first term and they are both small). So e^-(x-1) – 2e^-2(x-1) must be negative.
The other way that you could see this is to see that dy/dx is only zero once, and this is a minimum with y < 0. So when x tends to infinity y must tend to zero from below.
So is GC a graphing package? If so, I think you are right to be concerned, and it is really worth practising graph sketching yourself. It’s really not that hard to do with a bit of practice and is usually worth well over 5% of the marks.
Thanks for making this.
Would e^(-2)(x-1) not equal (2/3)e^(-1)(x-1), rather than (2)e^(-1)(x-1)?
Yes, you’re absolutely right. I carelessly left out the 3.
Actually looking at this again my original answer is right. The factor of 3 is applied to both the terms, so y = 0 when e^(-2)(x-1) equals (2)e^(-1)(x-1), not (2/3)e^(-1)(x-1).
Hey thank you you for the clearn working as usual, was just wondering for exponential functions the asymtote like here would always be zero without working because there is no vertical translation going on right? otherwise if i added like a +4 after the asymtote would be y=4
Thank you for the work and the motivation to keep working you told me in the comments of another question, do you have any tips for the PAT.
Also i was sick for one of my math module exams and was going through some heartbreaking parent issues so i might get a B for AS math which is well below me its a slight possiblity i might see b on results day but im predicted A* for math at a2, if i do get a B should i still attempt the pat? Ive been told that oxford will forgive my AS b so long as im predicted a* at a2 and get a good PAT score.
Thank you for the great work you do
Hi — yes you’re right. You can see that both of the terms are of the form exp(-nx) so they both tend to zero as x gets large.
I think you should certainly attempt the PAT if that is what you want to do. As I understand it, if you get a good enough mark in the PAT then you will get an interview and if you do well enough in the interview you will get an offer, based on your A2 scores, so your AS scores don’t matter that much.
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