MathsM.Omar, what in the world have you just written there?
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MathsM.Omar, what in the world have you just written there?
D4 stuffM.Omar, what in the world have you just written there?
BB Z10Allah khair,tha konsa mobile
Well bro, i got most of that figured out myself, h and k are turning points and they're 2,8 so i got that right.My main problem was in the g(x) functions.Could you solve that out on a page and post?
for the first image...for g(x) to be an iverse we need to define domain for either x>=2 or x=< 2 ...from the range in question in question i.e k=<x=<4 we see tht x=>2 nd thus k=2
...well the second image is self explanatory
What part dont you understand?
In part i i dont understand how you got to "a-b=7" when it should be "a+b=7"I didn't attempt to do the sketch, though.
In part i i dont understand how you got to "a-b=7" when it should be "a+b=7"
explain please?
Thanks for replying btw,
SalmanPakRocks any new guideI am sorry, I didn't see that you already explained it. Whatever you wrote was correct and that's how argand diagrams are sketched. However, this guide is just for elaboration.
As Xtremepaper server isn't letting me upload the file in here, I've uploaded it on my dropbox account.
I hope this link works.
https://www.dropbox.com/s/9pdws1gqxo1b3yd/scan0007.pdf?m=
At x = 3 there is line of symmetry so A = x value so 1 is given zero and from 2(x-3)^2-5 we get A as 3 but we have 2 out the bracket hence A=6f(x) = 2x^2 - 12x + 13
f (x) = 2(x-3)^2 -5
for 0=< x =< A
state the value of A for which the graph of y=f(x) has a line of symmetry
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