Question

Which points on the graph of y=4-x² are closest to the point (0,2)?

Answer 1

The distance between the point (0,2) and any an arbitrary point on the graph of y = 4-x²can be given by the formula D=√(x²+ (2-y)²) where x is the horizontal distance to the y axis, and (2-y) is the vertical distance to the y axis.See the diagram:

Thisequation can be written as:

But since y =4-x^2,

From the Chain Rule:

S04x³ -10x = 0. Then solve for x:

x(4x² - 5) = 0

So x = 0, +√(5/4) and -√(5/4). Plugging these back into y, we get y = 4, 11/4 and 11/4 respectively.

So the points are:

(0,4)
(+sqrt(5/4), 11/4)
(-sqrt(5/4), 11/4)

Answer 2

Given:

Question Details

Which points on the graph of y=4-x² are closest to the point (0,2)?

We want the points on the graph of that areclosest to .

We can start by finding a function which is equal to the distance squared from to :

; or , we have:

.

Take the derivative of this , and we get :

.

This derivative is zero when --- and .

When ,. Wethus need to check the distance from to .

When , wehave .

We thus need to check the distance from to and from to

. Because of the symmetry of about the axis , these distances are equal ( remember that

is onthe axis ) . We have the distance from to is :

.

The distance from to the other two points is :

.

We therefore have the points and on are closest

to ,and the minimum distance is :

.