# If (-3,6) is a point on the graph of y= f(x), what point do you know is on the graph of y=f(x+3)?

*print*Print*list*Cite

### 3 Answers

If `x=-6,` we have

`f(-6+3)=f(-3)=6,`

**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).`

` `

**Sources:**

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.

6=f(-3) (i)

what information do we have about

y=f(x+3) ?

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.

y=f(x)= ax^2+b

6=f(-3)=9a+b

6=f(x+3)=a(x^2+9+6x)+b=9a+b+ax(x+6)

ax(x+6)=0

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.

i.e y=f(x)=b

let b=3

6=f(-3)=f(x+3)

(2) Let x=-6

then y=f(x)=-6a+b

-6a+b=6

9a+b=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.