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#### confusedatmath

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- Jan 2, 2014

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- Jan 2, 2014

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- Mar 1, 2012

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A rule of thumb I use to work out horizontal translations is that the graph moves along the x-axis in the<snip>

I am confused about using horizontal transformations such as

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

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

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- Jan 2, 2014

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

- Mar 1, 2012

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That doesn't make sense to me. I tried it with a graph in wolfram 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).But the answer is f(x)=-(x-a)^3 +b .....

- Jan 29, 2012

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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 [tex]x^3[/tex] 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 [tex]-(x- a)^3[/tex] which is the same as (a- x)^3. That eliminates A leaving B and C which are identical.

The second thing you should notice is that the usual [tex]x^3[/tex] is

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- Jan 2, 2014

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so we sub x=a because in the graph it says (a,b)

what if the question said (-a,b) ??

what if the question said (-a,b) ??