ok i see how that happens, but why is it that? is there proof?
because (3x-3)^2 can because 3(x-1)^2 is the dilation 3 now? do i get the dilation in its expanded form or factorised form e.g. a(x-b)^n +c where a is dilation,??
firstly, nice answer sir.jonse, i just thought i'd add to it;
^2)
is not the same as
^2)
^2 = [(3)(x-1)]^2 = 3^2 (x-1)^2 = 9(x-1)^2)
(not the same as
)
this does imply that;
a value of 9 outside the bracket has
the same impact on the shape of a parabola (a dilation by a factor of 9 parallel to y axis/away from x axis) as a coefficient of 3 on the x term does (a dilation by a factor of 1/3 parallel to x / away from y)
hence, dilation by 1/3 from y is the same as dilation by 9 from x
try this out and you'll see that it works (
for parabolas)
as for why it's a dilation of 1/3 when the 3 is on the x, well, it's kinda the same for y except we usually get y by itself. let me explain. a more native way of describing this dilation from the x axis (parallel to y axis) is by saying y_new = 9 * y_old (normally we say y_new is y' and y_old is just y)
so y' = 9y
and our equation is
^2)
for example
make the substitution,
^2))
is the new formula describing the dilation
for dilations in x it's the same story, x_new = 1/3 x_old describes a dialtion by a factor of 1/3 from the x axis
so x' = 1/3 x
and our equation is
but this time, to make the substitution and get a new equation with x' in it, we need to change x' = 1/3 x to 3x' = x
we then substitute 3x' for x, and get
^2)
as our new equation
hope that example helps a bit, this is one of the trickyest things to get your head around in methods!
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