# Thread: Cubic Transformations - Graph shown is best represented by the equation:

1. I am confused about using horizontal transformations such as

f(x+a) and f(x-a) to interpret these graphs.

2. Originally Posted by confusedatmath
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I am confused about using horizontal transformations such as

f(x+a) and f(x-a) to interpret these graphs.
A rule of thumb I use to work out horizontal translations is that the graph moves along the x-axis in the opposite direction to the sign in the function. That is $\displaystyle f(x+a)$ moves $\displaystyle a$ units to the left (-ve) and $\displaystyle f(x-a)$ moves $\displaystyle a$ units to the right (+ve).

You can verify the direction by plugging values in and seeing what happens. Your example is a cubic so suppose we have the "base" function $\displaystyle f(x) = x^3$. It is pretty clear that $\displaystyle f(x) = 0 \text{ when } x = 0$. Now suppose we have $\displaystyle f(x-4)$ (where a=4). This translation is shifted 4 units to the right according to the previous paragraph and $\displaystyle f(x-4) = 0 \text{ when } x-4 = 0 \therefore x=4$ which is 4 units to the right of 0.

Let me know if you meant something else

edit: If I take your first example the point (a,b) is to the left of 0 on the x-axis so it'll be which sign inside the function

But the answer is f(x)=-(x-a)^3 +b .....

4. Originally Posted by confusedatmath
But the answer is f(x)=-(x-a)^3 +b .....
That doesn't make sense to me. I tried it with a graph in showing the graphs of $\displaystyle f(x) = -(x+5)^2 \text{ with } g(x) = -x^3$ for comparison and the graph of f(x) is shifted 5 units left compared to g(x).

5. The first thing you should notice is that when x= a, y= b. Since all of the options have "+ b", the cubic portion must be 0 when x= a so those that have "x+ a" are impossible. That eliminates D and E.

The second thing you should notice is that the usual $\displaystyle x^3$ is reversed- this graph rises to the left, not the right. That means x is swapped for -x. Since we are using "x- a" instead of x, we must have $\displaystyle -(x- a)^3$ which is the same as (a- x)^3. That eliminates A leaving B and C which are identical.