Distance Formula and closest to point on graph.

Question

Find the points on the graph of the function that are closest to the given point; f(x) = x^2 − 5, (0, −2)

Answer 1

distance squared from (0,-2) to (x,y) on graph is:

d²=(x-0)²+(x²-5-(-2))²=x⁴-5x²+9

d(d²)/dx=4x³-10x

derivative is zero when x=0 and x=+/-(√10)/2

when x=0 ,,, d²=9

when x=+/-(√10)/2 ,,,d²=11/4

x=0, y=-5 is local maximum, x=+/- (√10)/2, y=5/2 are local minimums

Answer 2

Let point on the graph opf f(x) is (x, y) = (x, x² - 5). Then square of the distance is D(x) = (x - 0)² + (x² - 5 + 2)² = x² + x⁴ - 6x² + 9 = x⁴ - 5x² + 9; D'(x) = 4x³ - 10x = 0; x = 0 or x = ±√10/2

D''(x) = 12x²- 10; D''(0) = - 10 < 0 max D''(±√10/2) = 20 > 0 min

x = 0 y = - 5

x = ± √10/2 y = - - 5/2

Answer 3

The point on the function c;losest to (0, -2) shall be on a normal (perpendicular) to the function.

The instananeous slope of the function is given by the first derivative.

f'(x) = 2x

The slope of the perpendicular is -1/2

Use slope-intercept form to determine perpendicula2

y = (-x/2) - 2

Determine intersection of f(x) and y = (-x/2) - 2, that is closest point.