Question

1. the st. louis arch can be represented as a function on the coordinate plane. according to the model and the graph, at which points on the coordinate plane does the st. louis arch touch the ground? annotate the points on the coordinate plane and then mark your answers in the space provided.

Explanation:

Step1: Understand the problem

The ground - level corresponds to \(y = 0\) in the coordinate plane for the function representing the arch. We need to find the \(x\) - intercepts of the function.

Step2: Assume a general form

If the function of the arch is a quadratic function \(y=ax^{2}+bx + c\), we set \(y = 0\) and solve for \(x\) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). But without the actual function, if we assume symmetry about the y - axis (a reasonable assumption for the St. Louis Arch), and if the base of the arch has a certain width. Let's assume the arch is symmetric about \(x = 0\). If the width of the base of the arch is \(w\), the points where it touches the ground are \((-\frac{w}{2},0)\) and \((\frac{w}{2},0)\). In reality, the St. Louis Arch has a base width of 630 feet. So the points are \((- 315,0)\) and \((315,0)\) (assuming appropriate units where the width is measured).

Answer:

\((-315,0)\) and \((315,0)\)