Given the graph with the points: (1,-2)(5,0)(10,1),write the radical function for  f(x).

Expert Answers
ishpiro eNotes educator| Certified Educator

Consider the x-coordinates of the given points:``

x = 1, x = 5 and x = 10

Since we need to right a radical function, and the y-values of the given points are integers, we can assume that the number under the radical will be a perfect square.

Note that both 5 and 10 can be expressed as 4 + 1 and 9 + 1, and 4 and 9 are perfect squares. So the radical function can include `sqrt(x-1)` .

If we plug in x = 1 into this expression, we will need to obtain the -2 as a value of y. This can be accomplished by subtracting 2 from the radical:

`y = sqrt(x-1) - 2`

x = 1 results in `y = sqrt(1-1) - 2 = -2`

Plugging in x = 5 results in `y = sqrt(5 -1) - 2 = 0`

Pluggin in x = 10 results in `y=sqrt(10-1) - 2 = 1`

These values of y are the same as in the given point, so the guess is correct.

Answer: `y=sqrt(x-1) - 2`

oldnick | Student

Last part is the graph (I'm sorry , but have broken solution for editor problems)


Let's see that f(x) is a branch of a parbole lied down X axis

oldnick | Student

`x= (-4+-2sqrt(x-1))/2=-2+-sqrt(x-1)`

(We have double branch for have "broken" a parabole)

Now we have to choice wich of two branches is needed:





 we have:

`f^(+)(1)=-2`     `f^(-)(-1)=-2`

`f^(+)(5)=0`         `f^(-)(5)=-4`

`f^(+)(10)=1`       `f^(-)(10)=-5`

So it shows  `f(x)=f^(+)=sqrt(x-1)-2`

Is the function we'r searching for.



oldnick | Student

Setting the points  in a parabole of kind `g(x)=ay^2+by+c`

we have a system with three unknows: (the unlnows are the coefficents of parabole)

      `4a-2b+c=1`        `(y=-2; x=1)`

                      `c=5`        `(y=0; x=5)`

       `a+b+c=10`          `(y=1; x=10)` ``


That has solution:     `a=1;b=4;c=5` 

              `g(y)= y^2+4y+5`

Now we have to found inverse function: `f(x)` , that means find found function hold:    `g(y)-x=0`


Now se solve quadratic  equation:

`Delta= 16-4(5-x)=4(x-1)`

since  `x>=1`  equation  has almost one solution: