maths help

[mufc802]

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 ?

 

[carvegio]

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

 

[mufc802]

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.

 

[Dr Delphi]

(x + 2) - 2 = x + 2 - 2 = x

(As nothing is multiplying the brackets they can be ignored and it can be considered one expression).

 

[icecold]

2-2=0

 

[cheechm]

 

[Reaper 392]

Shouldn't it be (x+2)(-2x-4)^2 ?

it would be if the question was:

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

 

[Garee]

( x - 2 + 2 ) = x

 

[mufc802]

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?

 

[vertica]

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

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

 

[SideWinder]

Is it due in tomorrow?

 

[Dr Delphi]

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.

 

[Six7th]

Just to add to what's been said, if its a quadratic you'll usually get it to factorise into something like (x-2)(x+3)=0. For this to be true then x=2 or x=-3 and these are the points where it crosses the x-axis. If it can't factorise then you would have to use the quadratic formula, http://mathworld.wolfram.com/QuadraticFormula.html

 

[mufc802]

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]

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..

 

[Dr Delphi]

f(x) = y

therefore:

y = x(x-2)^2 (for the original function) and
y = x(x-2)^2 + 2 (for the shifted function)

You need to expand out the brackets, set y=0 and then solve for x! It always simpler to expand the brackets and that leaves you with what you have to work with.

To solve the quadratic equation remember:

SQRT(b^2 - 4ac) / 2a

gives you the solution where:

y =ax^2 + bx + c

 

[Reaper 392]

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..

you dont even need to know it for FP1, 2 or 3 (further maths)

thank god