Question

A jump rope held stationary by two children, one at each end, hangs in a shape that can be modeled by the equation h=0.01x^2 - 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?

Answer 1

The lowest height of a parabola ax²+bx+c is when x=-b/2a
in this case, b=-1, a=0.01, so x=1/0.02=50
when x=50, h=0.01*50²-50+28=3
so the lowest part of the rope is 3 feet inches above the ground. 

Answer 2

Answer:

3 inches.

Step-by-step explanation:

As the equation has a positive coefficient of $x^{2}$, the graph is a parabola that open up. So, the minimun height value will be the y value of the vertex. Now, the vertex is

$(\frac{-b}{2a},h(\frac{-b}{2a}) )$

where b=-1 and a=0.01. Then,

$\frac{-b}{2a}=\frac{-(-1)}{2*0.01}=50$

$h(50)= 0.01(50)^{2}-(50)+28=0.01(2500)-50+28=3.$

Then, the lowest part of the rope (50) is 3 inches close to the ground.