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

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### 4 Answers

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` **

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

`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:

Called:

`f^(+)=-2+sqrt(x-1)`

and:

`f^(-)=-2-sqrt(x-1)`

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.

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`

`y^2+4y+5-x=0`

Now se solve quadratic equation:

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

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

TO BE CONTINUE........