+56 Determine the equation of the radical function below in the form of \(y = \sqrt b(x-p)+q\)
Link to picture: https://drive.google.com/file/d/1zNitd9lgemypcl2cDNwPf9flWtQdBbaz/view?usp=sharing
+56 I still haven't figured this one out yet unfortunately. A bit of assistance would be appreciated, review is a pain in the butt.
+118608 This is a sideways parabola
it has a vertex of (5,-1)
so the formula is
\((y+1)^2=4a(x-5)\)
and it goes through (1,3)
Sub that in to find a
\((y+1)^2=4a(x-5)\\ (3+1)^2=4a(1-5)\\ 16=4a(-4)\\ a=-1\\ (y+1)^2=-4(x-5)\\ \)
Now you want it in the given form
\((y+1)^2=4(5-x)\\ \text{It is the top half so I am only going to worry about the pos squareroot}\\ y+1=2\sqrt{5-x}\\ y=2\sqrt{5-x}-1\\ or\\ y=\sqrt{4(5-x)}-1\\ y=\sqrt{-4(x-5)}-1\\\)
Mmmm it seems your question was not correct I think it should have been
Determine the equation of the radical function below in the form of
\(y = \sqrt{ b(x-p)}+q\)
