Chapter 1.1, Problem 57PS

Solution

To find: The height $h$ of the cable above the road at its lowest point.

The height of the cable at the lowest point is $10$ .

Given information:

The function modeling the cable is given as $y=\frac{1}{9000}{x}^2-\frac{7}{15}x+500$ .

Concept used:

Vertex form a quadratic function is given as $y=a{\left(x-h\right)}^2+k$ , where $\left(h,k\right)$ is the vertex of the quadratic function.

The vertex of the quadratic function represents the maximum/minimum point of the function.

If $a$ positive, the vertex is the minimum point and if $a$ is negative, the vertex is the maximum point of the function.

Calculation:

Find the vertex of the given function by writing it into vertex form.

  $\begin{array}{c}y=\frac{1}{9000}{x}^2-\frac{7}{15}x+500\\ =\frac{1}{9000}\left(x^2-2\cdot t\cdot 2100+{\left(2100\right)}^2-{\left(2100\right)}^2\right)+500\\ =\frac{1}{9000}\left({\left(x-2100\right)}^2-{\left(2100\right)}^2\right)+500\\ =\frac{1}{9000}{\left(x-2100\right)}^2-\frac{1}{9000}{\left(2100\right)}^2+500\\ =\frac{1}{9000}{\left(x-2100\right)}^2+10\end{array}$

Here, $a=\frac{1}{9000}$ , which is positive. So, the vertex is the minimum point of the function.

The vertex is $\left(2100,10\right)$ . Therefore, the height of the cable at its lowest point is $10$ .

Conclusion:

The height of the cable at the lowest point is $10$ .