Question

find the point on the graph of the function that is closest to the given point:
f(x)=(x+1)^2 (5,3)

Answer 1

let the given point be A(5,3)
Let the closest point be P(x,y)

slope AP = (y-3)/(x-5)

f ' (x) = 2(x+1) = 2x + 2 which is the slope of the tangent at P
But at the closest point, AP must be perpendicular to that tangent
so 2x + 2 = -(x-5)/(y-3)
2xy+2y-6x-6 = -x+5
y(2x+2) = 5x+11
y = (5x+11)/(2x+2)

but y = (x+1)^2 = x^2 + 2x + 1

so x^2 + 2x + 1 = (5x+11)/(2x+2)
2x^3 + 2x^2 + 4x^2 + 4x + 2x + 2 = 5x+11
2x^3 + 6x^2 + x - 9 = 0
try x = 1
LS = 2 + 6 + 1 - 9 = 0 = RS
(How lucky was that ?)
so if x = 1
y = (2^2 = 4

the closest point is (1,4)