Karl M. answered 08/24/22
Experienced, Highly Rated Undergraduate Math and Physics Tutor
The simplest way is to see what the value of function is when the argument of the square root (what's inside the square root) is equal to 1. Let's take a look at an example to see why that is:
y = A·sqrt(x) where A is some constant coefficient (a scale factor).
If x=1, then y=A since sqrt(1)=1. If we look at a graph of this function for a particular value of A, we would see the curve goes through the point (1, A) so whatever the value at x=1 is, is the value of A.
In cases where the argument is more than just x like:
y = A·sqrt(x + 4)
Then when x= -3, sqrt(x + 4)=1 and y=A and we know our curve goes through the point (-3, A).
The last piece that may crop up is if you have some vertical shift:
y = A·sqrt(x + 4) + 1
In this case, we should still go to x=-3 but since everything gets shifted up by 1, we should subtract 1 from the value at x=-3 to get A.
In short to wrap up, go to x coordinate that makes the inside of the square root become 1 and the value of the function there is the scale factor A plus whatever vertical shift there is in the function.
Karl M.
The -4 is a vertical shift, meaning that all the values will be shifted down by 4. From the last point in the answer above, the value (in your case, 7) is A + vertical shift. So: 7 = A + (-4) implies A=11.08/25/22
Kate H.
The equation I am comparing sqrtx to is y=Asqrt(x-2)-4, where the graph is very steep. I can see that sqrt(x-2)=1 when x=3. The point on the graph is (3,7). How do I find A from here? Thank you.08/25/22