f(x)=2x^2+4x+2 where XER and x>or equal to -1
(a) Express f(x) in the form a(x+b)^2+c is 2(x+1)^2+2 correct?!
(b) Describe fully two transformations which would map the graph of y=x^2, x>or equal to 0 onto the graph of y=f(x) (as stated above)
(c) Find an expression for the inverse of f(x)
Any help appreciated. Cheers !
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- Thread Starter
- 07-04-2006 00:13
- 07-04-2006 00:25
f(x)=2x^2 + 4x + 2
=2(x^2 +2x) +2
=2(x^2 + 2x + 1 ) +2 -2 (since adding 2 in the brackets leads to removing two here)
=2(x+1)^2 (so c=0)
- 07-04-2006 00:35
b) 1: Move the graph to the left parallel to the x-axis by 1 unit
2: Stretch parallel to the y axis by factor 2
- 07-04-2006 00:39
You translate y=x^2 left one unit, then stretch scale factor 2 parallel to the y-axis to get y=f(x).
y=2(x+1)^2 Rearrange this to make x the subject of the formula.
y/2 = (x+1)^2
root(y/2) = x+1
-1 + root(y/2) = x
so inverse isf^-1(x)= -1 + root(x/2)
- 07-04-2006 03:32
Excellent. Each of us conquered one part of the question.
(Was too tired to do the 3rd part earlier, heh )