For question 7, shouldn't the vertex be (10,8) instead of (-10, 8) because with 8 - abs(x+10) wouldn't the minus sign turn the +10 to -10 and then to get the vertex points you reverse the sign on the x value giving +10? Thanks
This is the same as asking where the vertex for the parabola y = 8 - (x + 10)^2 is, which is (-10,8). The minus sign in front of the modulus sign, or in front of the ( )^2 doesn't affect the location of the vertex. You can play around with the numbers here: www.desmos.com/calculator/tnzbvrflr8
The graphs of y = |1 + x| and x = |1 + y| are reflections of one another in the line y = x as you can see here on Desmos: www.desmos.com/calculator/8ynskmkhnc (this is done by swapping the xs and ys)
Well let's say you have y = |x - 3| and y = |2x + 4|. The first is a V shape with vertex (3,0) and the second is a V shape with vertex (-2,0). The 2nd one has steeper sides due to the 2x.
Oh I see - when it gets to that stage, plotting points may be your best bet. Alternatively, you can see that if you remove the modulus signs you would get y = 2x, so that would give you part of the graph. It would probably be easier to explore these via Desmos: www.desmos.com/calculator/gkgr8awv51
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You deserve way more subs than u have atm. Thanks for making a fellow CIE A level student less stressed! Bless u
Thanks! Have you seen my CIE exam paper walkthroughs? sites.google.com/view/tlmaths/home/cie-a-level-maths?authuser=0
Thank you so much, I never knew u could complete the square with these to find the vertex
🥳🥳🥳best teacher ever!!!!
Thank you very much sir for this spectacular way of making maths easy, you deserve more subs!
Thank you!
Youve just made these totally trivial. Nice!
You are saving me so much, deserve way more subs!
thank you! i finally understand these now! :)
This is the epitome of hero does't ware cape
Who knew math could be so easy. Thank you
Thx, helps a lot for my act algebra
can you make a video about writing modulus inequalities in a required form please (I got taught the rule |x-a|
I've just uploaded B7-11 to B7-15 for you - hope they help!
Hi teacher how do i sketch the graph for Y=|1/2x-1| the completing square pov cant seems to work on this equation.
Yes it can if you factor the 1/2 out of the modulus bracket:
y = (1/2)*|x-2|
It therefore has its vertex at (2,0)
@@TLMaths owhhhh, so we must try to always factories the equation? And make the value infrony of x=1?
Yes
@@TLMaths ok thanks a lot teacher now i can sketch the graph in a blick of an eye😆😆😆
Sir thank you so much made my life simple
For question 4 and 8 why does it turn into 5/8 and 3/5?
You have to divide the whole bracket by the number before the x in order to move it to the outside
amazing thank you!
for question eight , why cant the modulus be open downwards? As it is originally -5 at the very first
As in the video, |3 - 5x| is the same as |5x - 3|
Great video!
For question 7, shouldn't the vertex be (10,8) instead of (-10, 8) because with 8 - abs(x+10) wouldn't the minus sign turn the +10 to -10 and then to get the vertex points you reverse the sign on the x value giving +10? Thanks
This is the same as asking where the vertex for the parabola y = 8 - (x + 10)^2 is, which is (-10,8). The minus sign in front of the modulus sign, or in front of the ( )^2 doesn't affect the location of the vertex. You can play around with the numbers here: www.desmos.com/calculator/tnzbvrflr8
@@TLMaths Got it, thanks a lot
thanks a lot for teaching me
Good job
is it the same process if the question requires you to sketch a modulus of a y function?
Can you give me an example?
@@TLMaths x=|1+y| - although it appears online that no such equation existed when there is a modulus of a y term
The graphs of y = |1 + x| and x = |1 + y| are reflections of one another in the line y = x as you can see here on Desmos: www.desmos.com/calculator/8ynskmkhnc (this is done by swapping the xs and ys)
Can you put factors back into the modulus like when you took them out?
Yes so 2|x+3| = |2x+6|
@@TLMaths Does this apply with negative factors too?
The negative would have to stay outside of the modulus
-2|x+3| = -|2x+6|
how did you know the gradient in q6 was 1?
thank you :)
what does b707 mean? I just came for OCR a2! Thanks for the vid anyway
It's the coding system I use: sites.google.com/site/tlmaths314/home/a-level-maths-2017/full-a-level
How would you sketch a graph consisting of two modulus
Well let's say you have y = |x - 3| and y = |2x + 4|. The first is a V shape with vertex (3,0) and the second is a V shape with vertex (-2,0). The 2nd one has steeper sides due to the 2x.
@@TLMaths I mean say something like y=|x+1|+|x-1|
Oh I see - when it gets to that stage, plotting points may be your best bet. Alternatively, you can see that if you remove the modulus signs you would get y = 2x, so that would give you part of the graph. It would probably be easier to explore these via Desmos: www.desmos.com/calculator/gkgr8awv51
@@TLMaths thnks a lot man!!
Kya bat hai
I'm Indian🇮🇳 so i could not understand ur language
Not sure how I can help change that 🤷
I’m Norwegian 🇳🇴 so i could not understand the point of stating your nationality
@@TLMaths sir, I always speak in hindi so i can not understand your language
Yep, so I'm not sure what you expect me to do about it?