maths help

mufc802 · 30 May 2011 at 16:13

y=f(x) = x(x-2)^2

f(x+2)

(x+2)[(x+2)-2]^2
(x+2)(x)^2

Can someone explain how you get from [(x+2)-2]^2 to (x)^2 please. Shouldn't it be (x+2)(-2x-4)^2 ?

mufc802 · 30 May 2011 at 16:23

carvegio said:
Only the expression in the square bracket is being squared and (x + 2) - 2 = x.

How does it equal x? Please do step by step.

mufc802 · 30 May 2011 at 17:19

Oops.

Another thing I'm trying to work out is this:

f(x) = x(x-2)^2

y = f(x) + 2

So it moves up the vertical +2. How do I work out the place where it crosses the x?

mufc802 · 30 May 2011 at 17:47

Delphi-uk said:
To find out where something crosses the x-axis you have to set y = 0 and solve the equation, conversely to find where the y-intercept can be found (the technical name for a crossing point) then set x = 0 and solve.

Yes I know but I've got f(x) = x(x-2)^2 and y = f(x) + 2 so I don't know how to make y=0 before I've put the function into it which I don't know how to do.

I can do it when y=f(x+2) which makes (x+2)[(x+2)-2]^2 = (x+2)(x)^2 which means it crosses at x=-2 and x=0.

But when y=f(x)+2 when the 2 is outside the bracket I don't know how to substitute it into x(x-2)^2.

And no this isn't in for tomorrow, I'm trying to teach myself and just need some help.

mufc802 · 30 May 2011 at 17:50

vertica said:
sub in y=0 as that is the x axis.

f(x)=x(x-2)²

y = f(x)+2
y = x(x-2)² + 2
y = x(x²-4x+4)+2
y = x³ - 4x² +4x +2

where y=0
0 = x³ - 4x² +4x +2
solve to get x= -0.359304 (with the other two imaginary roots of 2.17965±0.903013i)

Oh and for future reference "²" = Alt0178 on num pad and "³" = Alt0179

That's complicated.. explains why they didn't ask to work out the x then.

Got 0 = x³ - 4x² +4x +2 but not got a clue what to do next, guess I don't need to know for C1..