find the coordinates of the single stationary point of the curve with equation y=8x^2 + 2/x

rewrite the equation in a more helpful way:y=8x² + 2x^-1 differentiate:dy/dx = 16x -2x^-2set equal to 0 and solve for x16x -2x^-2 = 016x -2/ x²=016x= 2/ x²16x³=28x³=1x³=1/8x=1/2note: x cannot be -1/2sub x into original curve equation:y= 8(0.5)² + 2/0.5y= 2+ 4y=6