SOLUTION: Determine the point {{{P(x, y) }}} on the graph of the equation {{{y= sqrt (x+1) }}} such that the slope of the line through the point (3, 2) and P is 3/8

Question 1065873: Determine the point on the graph of the equation such that the slope of the line through the point (3, 2) and P is 3/8
Found 2 solutions by josgarithmetic, Boreal:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
using formula for slope of a line

Can you continue from here for the x you need, and find y?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!

The point is (x,y). The slope is (2-y)/(3-x)=(3/8)
That is 16-8y=9-3x
or 3x+7=8y; but y = sqrt (x+1)
right side is 8 sqrt (x+1)
square both sides
9x^2+42x+49=64(x+1)=64x+64
9x^2-22x-15=0
(9x+5)(x-3)=0
x=3 (doesn't help) and x=-5/9
y is sqrt ( -5/9 +1)= sqrt (4/9) =2/3
the slope between (-5/9,2/3) and (3,2)=2-(2/3) or 12/9 divided by 3-(-5/9)=32/9
12/9 divided by 32/9 is 3/8.
The point has to be to the left of (3,2), because the derivative is 1/2 sqrt (x+1), which at x=3 gives a slope of 1/(2*2)=1/4, and the slope gets less further to the right.
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