Creating Functions off of Coordinates

[Wister]

I am in Precalculus, and I was wondering how one would one determine a and b in this square root function, where the graph of that function passes through ( -4, 6 ) and ( 1, 4 )?
y = (ax + b)^(1/2)

I just don't know where to start on this. Thanks. ^-^' I understand how to shift functions "normally", when there aren't restrictions as to what can be added, but not when there are only two positions that I am allowed to put in values for.

 

[Deleted member 4993]

I am in Precalculus, and I was wondering how one would one determine a and b in this square root function, where the graph of that function passes through ( -4, 6 ) and ( 1, 4 )?
y = (ax + b)^(1/2)

I just don't know where to start on this. Thanks. ^-^' I understand how to shift functions "normally", when there aren't restrictions as to what can be added, but not when there are only two positions that I am allowed to put in values for.

You have two unknowns - a & b - so you need two equations to solve those.

The first equation is derived from the fact that the curve is described by y = (ax + b)^(1/2) and it passes through (-4,6)

Then:

6 = (-4a + b)^1/2 →

36 = -4a + b ..............................(1)

Now derive the second equation and solve for a & b.

 

[Wister]

Thank you~!
(Wow, I'm sorry that I didn't see that. ._.')