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 + 27, 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

Answer with explanation:

To find the height of rope from other end of the equation,

       h=0.01 x² - x + 27, 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.

We will Substitute,

  h=0,and get ,two roots of the Quadratic function.

$\rightarrow 0.01 x^2- x + 27=0\\\\ x=\frac{-(-1)\pm \sqrt{(-1)^2-4\times (0.01) \times 27}}{2\times 0.01}\\\\x=\frac{1\pm \sqrt{1-1.08}}{2\times 1}$

→→For a quadratic function of the type, ax² +b x +c=0

Roots can be obtained by the formula

    $x=\frac{-b \pm \sqrt{D}}{2a},{\text{where}}, D=b^2-4 a c$

As the two roots obtained are not real.

It will not cut the x axis.

Now, finding the vertex of the parabola,which has coordinate, (50,2).

→Lowest part of the rope is 2 inches above the ground.

For,the quadratic function

  →y=0.01 x² - x + 27

 → y =0.01×(x² -100 x  + 2700)

 →y = 0. 01 ×[(x-50)²-2500+2700]

→y -2 = 0.01 (x-50)²

⇒Vertex = (50,2)

h=2 inches→→Lowest part  of the rope above the ground.

Answer 2

Answer:

A) 2 inches

Step-by-step explanation:

All credits go to guy above me

just making it simple