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

AI Thread Summary
The discussion centers on understanding horizontal transformations of cubic functions, specifically how to interpret equations like f(x+a) and f(x-a). It emphasizes that the graph shifts in the opposite direction to the sign in the function, with f(x+a) moving left and f(x-a) moving right. A specific example using the cubic function f(x) = x^3 illustrates this concept, showing how f(x-4) translates the graph 4 units to the right. The conversation also highlights the importance of analyzing points on the graph, particularly when determining the correct transformation based on given coordinates. Overall, the participants are working through the complexities of cubic transformations and their graphical representations.
confusedatmath
Messages
14
Reaction score
0
View attachment 1901
View attachment 1902

I am confused about using horizontal transformations such as

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

Attachments

  • Screen Shot 2014-02-02 at 2.18.37 pm.png
    Screen Shot 2014-02-02 at 2.18.37 pm.png
    8.2 KB · Views: 94
  • Screen Shot 2014-02-02 at 2.18.43 pm.png
    Screen Shot 2014-02-02 at 2.18.43 pm.png
    11.3 KB · Views: 97
Mathematics news on Phys.org
confusedatmath said:
<snip>

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 $$f(x+a)$$ moves $$a$$ units to the left (-ve) and $$f(x-a)$$ moves $$a$$ 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 $$f(x) = x^3$$. It is pretty clear that $$f(x) = 0 \text{ when } x = 0$$. Now suppose we have $$f(x-4)$$ (where a=4). This translation is shifted 4 units to the right according to the previous paragraph and $$f(x-4) = 0 \text{ when } x-4 = 0 \therefore x=4$$ which is 4 units to the right of 0.

Let me know if you meant something elseedit: 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
+. Giving us (x+a)^3
 
But the answer is f(x)=-(x-a)^3 +b ...
 
confusedatmath said:
But the answer is f(x)=-(x-a)^3 +b ...

That doesn't make sense to me. I tried it with a graph in wolfram showing the graphs of $$f(x) = -(x+5)^2 \text{ with } g(x) = -x^3$$ for comparison and the graph of f(x) is shifted 5 units left compared to g(x).
 
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 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 -(x- a)^3 which is the same as (a- x)^3. That eliminates A leaving B and C which are identical.
 
Last edited by a moderator:
so we sub x=a because in the graph it says (a,b)

what if the question said (-a,b) ??
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Back
Top