TMUA - Graphs by Jacqueline Tyler

[@Paige-ez5g]

What a way to start my Sunday afternoon

[@filipmartinez6774]

for real

[@BinyIII]

at 7:48 just do f(-x)=f(x+3) and then compare coefficients to get b=3

[@RajSandhu-gm8iz]

In Q2: A few other things that can help work out what a graph looks like which are helpful related to symmetry include, as mentioned symmetry about x-axis the y powers are even, symmetry about y-axis the powers of x are even. Symmetry about line y=x is tested by swapping x with y and y with x, the equation stays the same, for symmetry about line y=-x, swap x with -y and y with -x, the equations stays the same. Might be useful to know as well.

[@neochen580]

Another way for solving Q10 is by saying that f(-x) = f(x+3), then applying the transformations to the equation and equationg x coefficients.

[@sd20052]

fro question 12, if k=0, isn't the x axis an asymptote?

[@AB-nf8li]

a gift which keeps giving

[@IlterP]

question 12 i dont think the 0 should be included in the range for k in the answer, if k = 0 then y=x/(x^2-4x), which is not defined for y=0. that kinda threw me off

[@adityashethwala8936]

For Q15, when you square the roots, wouldn't you get 0, pi^2, 4pi^2, 9pi^2 etc... and not just pi,4pi,9pi .....? @R2Drew2

[@taxrate]

drew are you single by any chance

[@Mustafa_Shahzad]

Aw hell nah 😂

[@abdeazbaid1194]

[@jagzey]

ain’t no way nga😂

[@wdlwunde2120]

🤣🤣🤣🤣🤣

[@cicyove1227]

He is double

[@skmaths-help]

For question 5 I don't understand how you just conclude that y = x must be an asymptote from the (x-y) bracket being zero? Doesn't that just mean there are no solutions there not necessarily that this is an asymptote?

[@ashleypatel4489]

If you sub in an x value equal to the y value, the question would read as 0=1, which isn’t possible, which would make it an asymptote I think?

[@mg9432]

Great video! Could you do one on circles if you haven't already

[@periodic5866]

For Q12, can you say that f(x) having a range of all real numbers means that at x = k, x^2 - 4x - k has to not be equal to 0 (so that f(x) can be equal to 0), meaning k^2-5k is not equal to zero? Not exactly sure where you go from there but I think if that is correct then it already narrows your choices down to either C or D, where you might be able to take a guess even if you don't have enough time to fully work it out.

[@shreyanpankhania2203]

someone correct me but for Q12 i found that all options but C were eliminated after you find k > -4 so it makes it much shorter

[@jetblue2758]

I think this is true but it doesn't necessarily rule out option E. I mean, there could be more boundaries for k which you haven't found out through just that inequality, so E could still be a possibility. That being said C is definitely a good option to pick based off your logic, as it's much more likely to be correct than E is, though I don't think it's necessarily correct

[@RS-mw9vg]

can you make a video with all important formulas and tips and tricks for each topic. thanks a lot.

[@1nf3rno_06]

After having done all past papers and doing all the questions I got wrong without looking at the mark scheme, would it make sense to go through them again or should I move onto something like STEP 1 or MAT?

[@OrtancaCOCUK-qe4dv]

Redo em, Ive realised Im still struggling after going over them. Make a note of tips, tricks and mistakes you’ve made to revise on day of the exam

[@hokagedattebayo7623]

i love you r2drew2

[@poghuman8676]

Awesome channel

[@jamiehenshaw8419]

Surely it can't just be me counting 8 regions there for Q4? 3 regions above the quadratic in the top left quarter, the massive region on the bottom left makes the 4th region, the 2 finite regions in the middle makes 6, and the 2 regions in the right hand side make 8. Someone tell me if im being completely stupid plz lol

[@dontspam7186]

remember to exclude the coordinate axis

[@jamiehenshaw8419]

@@dontspam7186 perfect cheers for that

[@Molar7335]

For Q16, I managed to get it solely from the fact that 2/x cannot be 0 therefore the only possible graph must be A

[@VybesLOL]

can u make a vid on congruency boss?

[@rtwodrew2]

kzbin.info/www/bejne/n6rPeWund6-Va80 kzbin.info/www/bejne/nqKcn2t-eth-qMk kzbin.info/www/bejne/pHmogXaVZp6ngpo