How to find values in a square root graph given an equation with 2 variables I'm trying to figure out how to do this problem below:
Find the values alpha and beta that will make the function 
$g(x)= 2 \sqrt{x - \alpha} + \beta$
pass through the points ($3,-2$), ($4,0$), and ($7,2$).
The answer is: $g(x) = 2 \sqrt{x - 3} - 2$.
I can't figure out how to get to the answer.  This question comes directly from the book: No BS Guide to Math & Physics, problem E.21.  
 A: Plug in each of the known points to get:
$$-2 = 2\sqrt{3-\alpha}+\beta$$
$$0 = 2\sqrt{4-\alpha}+\beta$$
$$2 = 2\sqrt{7-\alpha}+\beta$$
Subtract the first equation from the second to get:
$$2=2\left(\sqrt{4-\alpha} - \sqrt{3-\alpha} \right)$$
Now you have one equations in one variables.  Rewrite the equation to isolate one square root on each side of the equation:
$$1+\sqrt{3-\alpha} = \sqrt{4-\alpha}$$
Now square both sides:
$$1+2\sqrt{3-\alpha} + 3-\alpha = 4 - \alpha$$
Now isolate the square root, square both sides of the equation again, and solve for $\alpha$.  Back-substitute into one of the original equations and solve for $\beta$.  When you are all done, go back and check to make sure that the values of $\alpha$ and $\beta$ you've found work in the third equation too (as only the first two equations were used to find them).  Can you finish it from here?
A: Here is another solution, based on recognizing $g(x)$ as a transformation of the toolkit function $y=\sqrt{x}$.  Suppose you happen to notice that


*

*$(4,0)$ is 1 unit (horizontally) from $(3,-2)$, and

*$(7,2)$ is 4 units (horizontally) from $(3, -2)$, and

*both $1$ and $4$ are perfect squares.


Then you might want to consider the function $h(x) = g(x-3)$, whose graph is the same as that of $g(x)$ but shifted $3$ units to the left.  This graph would pass through $(0, -2)$, $(1, 0)$ and $(4, 2)$.  Now these points also have some interesting properties:


*

*$(1, 0)$ is $2$ units (vertically) above $(0, -2)$, and

*$(4, 2)$ is $4$ units (vertically) above $(0, -2)$, and

*$2 = 2\sqrt{1}$ and $4 = 2\sqrt{4}$.


These observations might lead you to consider the function $k(x) = h(x) + 2$, which passes through the points $(0, 2)$, $(1, 2)$ and $(4, 4)$.  By inspection, these points all satisfy $y=2\sqrt{x}$.  So we have:


*

*$k(x) = 2\sqrt{x}$,  and therefore

*$h(x) = k(x)-2 = 2\sqrt{x} - 2$, so finally

*$g(x) = h(x - 3) = 2\sqrt{x-3} - 2$.


This method is rather ad hoc and does not necessarily generalize; it works in this instance because the given points correspond to transformations of $(x,y)$ for $x=0, 1, 2$ on the toolkit function, and are therefore not too hard to recognize.
