# Find the point on the graph y=x12 which is closest to the point (3,0).?

Update:

find the point on the graph y=x^1/2 which is closest to the point (3,0).

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• 2 weeks ago

x^(1/2) = √x

The perpendicular distance from a point to a line is the shortest distance from a fixed point to any point on a fixed infinite line in Euclidean geometry. It is the length of the line segment which joins the point to the line and is perpendicular to the line.

Let (x, √x) be a point on the graph of y = √x.

distance between point (x, √x) and point (3, 0) is

d = √[(3 - x)² + (0 - √x)²]

= √(9 - 6x + x² + x)

= √(x² - 5x + 9)

We want to minimize that distance. However, the algebra is easier if we minimize the square of the distance, which is justifiable because the square root function is strictly increasing.

So, want to minimize

d² = x² - 5x + 9

critical points are when the first derivative equals zero:

2x - 5 = 0

2x = 5

x = 5/2

the corresponding y-value using the original function is

y = √(5/2)

= (√5 / √2)

= (√5 / √2) * (√2 / √2) ……….. rationalizing the denominator

= √(10) / 2

The closest point on y = √x to point (3, 0) is (5/2 , √(10) / 2) .

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In the attached graph, I figured out the equation of the tangent line to y = √x at point (5/2 , √(10) / 2) in order to visually show that the segment AB in the graph is perpendicular to the tangent line.