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

 
 Nov 24, 2022
 #1
avatar+36455 
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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 :

 

 

 
 Nov 24, 2022
 #2
avatar+36455 
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.....

 
 Nov 26, 2022

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