4 Answers | Add Yours
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
(We have double branch for have "broken" a parabole)
Now we have to choice wich of two branches is needed:
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`
Now we have to found inverse function: `f(x)` , that means find found function hold: `g(y)-x=0`
Now se solve quadratic equation:
since `x>=1` equation has almost one solution:
TO BE CONTINUE........
We’ve answered 318,944 questions. We can answer yours, too.Ask a question