let's say that
f(x) = a√x + b
Then we know that
a√1 + b = -2
a√5 + b = 0
a(√5-1) = 2
a = 2/(√5-1) = (√5+1)/2
b = -2 - (√5+1)/2 = -(√5+5)/2
f(x) = (√5+1)/2 √x - (√5+5)/2
Somehow I don't think that's what you are after.
How about
f(x) = √(ax+b)
then
√(a+b) = -2
√(5a+b) = 0
b = -5a
a = 1
f(x) = -√(5-x)
Okay, I have a graph of a sq root function. I'm told to write the radical function for f(x). Some points on the line are (1,-2) and (5,0). I know you probably can't help me a whole lot, but if you can give me any insight into how to figure this problem out, I'd really appreciate it.
1 answer
1 answer