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

A) 2 inches

Explanation:

All credits go to guy above me

just making it simple

Answer 2

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=(-(-1)\pm √((-1)^2-4* (0.01) * 27))/(2* 0.01)\\\\x=(1\pm √(1-1.08))/(2* 1)`

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

Roots can be obtained by the formula

`x=(-b \pm √(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.