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If `x=-6,` we have
so `(-6,6)` is on the graph of `f(x+3).`
You might also know that the graph of `f(x+3)` is the same as the graph of `f(x)` shifted to the left by 3 units. Thus if `(-3,6)` is on the graph of `f(x),` `(-6,6)` must be on the graph of `f(x+3).`
Here are the graphs of two functions to serve as an example. I picked `f(x)=-x+3`. The black graph is the graph of `f(x)` and the red graph is the graph of `f(x+3)=-(x+3)+3=-x.` Notice that `(-3,6)` is on the graph of `f(x)` and `(-6,6)` is on the graph of `f(x+3).`
You should know that a point `(x_1,y_1) ` is on a graph of a function `y = f(x) if y_1 = f(x_1).`
Since the problem provides the information that (-3,6) is on the graph of the function `y = f(x)` , yields that `6 = f(-3)` .
Since the problem provides the information that `y = f(x+3)` and `f(x) = y` , yields:
`f(x + 3) = f(x) = y`
Substituting -3 for x and 6 for y, yields:
`f(-3 + 3) = f(-3) = 6 => f(0) = f(-3) = 6`
Hence, evaluating the coordinates of the point that belongs to the graph `y = f(x + 3)` , under the given conditions, yields `x = 0` and `y = 6` , thus, the point (0,6) is on the graph `y = f(x + 3).`
Ans. Answer to this question is not simple as it appears ,many informations missing.
But let us try.
y=f(x) , (-3,6) is on this graph.
what information do we have about
In question it is not given if function f(x) is linear or nonlinear.
If f(x) is linear then it may be invertible , a non linear fuction may be invertible.
If a is not zero ,then bis also not zero.
then either x=0 or x=-6
(1) If x=0 then f is constant fuction.
(2) Let x=-6
gives a=0 ,which is contradicting fact a is not zero.
so what possible now x=0
and y=f(x) is constant fuction.
A line parallel to x axis at distance of +6 unit.
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