Solve the problem.
A jump rope held stationary by two children, one at each end, hangs in a shape that can be modeled by the equation h = 0.01 x squared minus x + 28, where h is the height (in inches) above the ground and x is the distance (in inches) along the ground measured from the horizontal position of one end. How close to the ground is the lowest part of the rope?

Question

PWRDrEaMz · Answer

Maybe try to Solve x

surjithayer · Answer

$$h=0.01x^2-x+28$$
$$=\frac{ x^2 }{ 100 }-x+28$$
$$=\frac{ 1 }{100 }(x^2-100x+2800)$$
$$=\frac{ 1 }{ 100 }(x^2-100x+2500-2500+2800)$$
$$=\frac{ 1 }{ 100 }(x-50)^2+3$$
closest point to the ground is at 3 inches high from the ground.