Ok... thnx for that... i do like differentiation but i just looked at the book and all the weird symbols are scaring me but i guess ill take it as it goes... i have 93 average ums for my as level and i think *hope* i did well in m1 so i can afford to get slightly lower... ok i know u r probs not very interested but this is my form of calming myself down
my teacher thinks that this would be a hard paper ? and he said there could be a big question like differentiation/trig when they c3 will be difficult like how cause all the papers seem to look same? which paper was the hardest till now?
my teacher thinks that this would be a hard paper ? and he said there could be a big question like differentiation/trig when they c3 will be difficult like how cause all the papers seem to look same? which paper was the hardest till now?
Trigonometry and Differentiation are both very 'big' chapters so we must expect 'big' questions involving these two chapters.
S2 didnt go too well either... I need an average of 83UMS on S2 C3 C4
Yeah I repeated M1 in early may hoping to have got at least 70 UMS. S1 went really well so hoping for85-90 UMS which would leave me needing about 75 in C3 and C4 but I'm terrible at both of them so
Yeah I repeated M1 in early may hoping to have got at least 70 UMS. S1 went really well so hoping for85-90 UMS which would leave me needing about 75 in C3 and C4 but I'm terrible at both of them so
I'm confused by this question in my textbook, I know it's just modulus but any help would be appreciated!
Sketch this graph: y = |x+4| + |x-1|
Also does anyone have any useful ways of remembering the vector transformations because I am not good at those!
I have no idea how to sketch y = |x+4| + |x-1| tbh, that's mad hard.
But if you're asking about graph transformations (I assumed that's what you meant by vector transformations?), then I remember how to do graph transformations using this:
If y=f(x) and 'a' is a number, then perform the following actions to the graph:
y = af(x) -----> Multiply all y-coordinates by 'a' y = f(x) + a -----> Add 'a' to all y-coordinates y = f(ax) -----> Multiply all x-coordinates by '1/a' y = f(x+a) -----> SUBTRACT 'a' from all x-coordinates.
Also, for me, I do graph transformations step-by-step and do the things closest to the 'x' first / using BIDMAS, so, if it gives you y = f(x) and says draw the graph of:
y = 2f(3x) - 5
I draw the graph of y=f(3x) first, by multiplying all x-coordinates by '1/3',
then I draw the graph of y = 2f(3x) by multiplying all y-coordinates by '2',
then I draw the final graph of y = 2f(3x) - 5 by subtracting '5' from all y-coordinates, which is the graph they're looking for.
When you apply a function to an equation, it must be applied to the whole of each individual side, as you would when dividing each side of an equation or multiplying each side of an equation. Hence you can't apply logarithms like that.
i.e. The second line should read ln(e^x + 3e^(-x)) = ln(4) which you can't do anything with.