SOLUTION: Given the graph below, write the radical function for f(x). These are the coordinates that I took from my homework: (-2,3)(-1,2)(2,1)(7,0) I am sorry if I don't have a graph t

Algebra ->  Functions -> SOLUTION: Given the graph below, write the radical function for f(x). These are the coordinates that I took from my homework: (-2,3)(-1,2)(2,1)(7,0) I am sorry if I don't have a graph t      Log On


   



Question 1075366: Given the graph below, write the radical function for f(x).
These are the coordinates that I took from my homework:
(-2,3)(-1,2)(2,1)(7,0)
I am sorry if I don't have a graph to place on this website, but radical function means A function that contains a radical expression with the independent variable in the radicand. Also a radicand is the number or expression inside a radical symbol.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If your graph has only those 4 points,
it is a cumbersome problem.
I would start with x=ay%5E2%2Bby%2Bc ,
would solve for a, b, and c,
and then would solve for y.
Substituting the coordinates of (7,0),
you get 7=c .
Point (2,1) gives you
2=a%2Bb%2Bc=a%2Bb%2B7 ---> a%2Bb=-5 .
Point (-1,2) gives you
-1=4a%2B2b%2B7--->4a%2B2b=-8
Solving system%28a%2Bb=-5%2C4a%2B2b=-8%29 , you get
a=1 and b=-6 .
So, x=y%5E2-6y%2B7 fits 3 of the points.
The fourth point confirms it.
Solving y%5E2-6y%2B7-x=0 for y, you get
y=%286+%2B-+sqrt%2836%2B4%28x-7%29%29%29%2F2 ,
which simplifies to y=3+%2B-+sqrt%28x%2B2%29 .
That is 2 choices.
Your function increases as x increases, so it is
f%28x%29=3%2Bsqrt%28x%2B2%29 .

If your graph is a continuous curve that looks like this:

you would think that it looks just like g%28x%29=sqrt%28x%29 ,
but translated 2 units to the left and 3 units up.
That would make it f%28x%29=3%2Bsqrt%28x%2B2%29 .
I would bet that is what the teacher wants, and if so,
there should be something to tell you that the function was a transformed y=sqrt%28x%29 .

At least you should think there is
an x%2B2 inside the squate root,
because the function is not defined for x%3C-2 <---> x%2B2%3C0 .
Then you would also think there is a positive coefficient
multiplying the square root, because y increases with x,
and a 3 added to the square root term,
because the graph starts at y=3.