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Determine the equation of the radical function below in the form of $$y = \sqrt b(x-p)+q$$

May 9, 2021

#1
+56
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I still haven't figured this one out yet unfortunately. A bit of assistance would be appreciated, review is a pain in the butt.

May 10, 2021
#2
+117872
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May 10, 2021
#3
+117872
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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

$$y = \sqrt{ b(x-p)}+q$$

May 10, 2021