What is the smallest distance between the origin and a point on the graph of y = 1/2*(x^2 - 8)?

Guest Nov 24, 2022

#1**+1 **

Use the distance formula to find the disatnce from the point (x,y) to the origin....where y = -1/2x^2 - 4

d^2 = (x-0)^2 + ( -1/2 x^2 -4 -0)^2

= 1/4 x^4 -3x^2 + 16

Now take the derivative and set = 0 to find the local maxima/minma points ( this is where the graph of the function has a slope = 0)

x^3 -6x = 0

x ( x^2 -6) = 0 shows the 'x' coordinate possibilities to be ± √6 and 0 BUT we need to jnow if these points are maxima or minima

...... to determine this, take the derivative AGAIN

3x - 6 now sub in the found 'x' values .... a positive result means the 'x' point is a minima

when x = 0 the value is -6 <==== this would be a maxima

when x = ± √6 the result is positive <==== so either of these 'x' s are the coordinte for the nearest point

find the 'y' coordinate by sustituting in the 'x ' into the original equation to find y = - 1

__so the nearest point would be ( √6 , -1) __

Here is a graph to SEE this :

ElectricPavlov Nov 24, 2022

#2**0 **

I posted the nearest point to the origin.....this would be a distance of sqrt(7) units from the origin....as shown by the circle on the graph....sorry I left that part out.....

ElectricPavlov Nov 26, 2022