Unit 7. Radical & Rational Functions

Hey guys, it's Manpreet.

I'll be blogging about what we did in your class today. First we finished the Radical functions using transformation part of the unit, where we learned how to determine a radical function from a graph.

With only a graph of a trasformated radical fuction, we can determine the traformation by first drawing the original function (y=√x), then using one of the two methods, comparing vertical distance (vd), or horizontal distance (hs). Vd is used to determine the value of (a) in the function y=a√(bx-h)+k, and hd is used to determine the value of (b).

Then we went to the next part of the unit, which is The Square Root of a Function. This is when you convert a any function to a radical function. 

There are some rules that are very importent when converting and graphing square root function:

          1) f(x)<0  √f(x) does not exist

          2) f(x)=0 

          3) 0<=f(x)<=1  √f(x) will be above f(x)

          4) f(x)>1  √f(x) will be below f(x)

This one is fairly straight forword, there are a couple of rules the you should follow. You should, when mapping the number for the original function, try to find the value of x where y will equal 1 and/or 0 so it will make it a lot easier in find the invariant points.

This one deals with quadratic functions, which as first might seem difficult to understand but practicing this will help in using certain methods to complete it. The most importent thing to do first is to find the x and y intercepts. This will help you in graphing the square root function.

There is one more thing, Mr. P told us to find the reason why we can't use f(x)>0 to get the restrictions, and I couldn't find anything on it but I tried to solve it anyways. So, I found a way to get the answer but in a really weird way, atleast to me.

   

First I took the equation and arranged it in the proper order, then I factored it out. I wrote the equation twice, one to solve (-1/2x+1)>=0 and the other to solve (x+2)>=0. The reason that you can solve for either one, is because if 0 is divide by any value you get 0, so then you can solve for either. For (-1/2x+1)>=0 I got x<=2 and for (x+2)>=0 I got x>=-2. Then putting them together I got -2<=x<=2. 

*I don't know if this works on any other equation.. So far worked on this. If you are having trouble graphing the function try it.

shameless plug: https://www.youtube.com/user/themanpreet23/videos

The End.