Problem 64 - Precalculus : Building Concepts and Connections

Question

Graph each quadratic function by finding a suitable viewing window with the help of the TABLE feature of a graphing utility. Also find the vertex of the associated parabola using the graphing utility. $$g(s)=-s^{2}-15$$

Step-by-Step Solution

Answer

The vertex of the quadratic function \(g(s) = -s^{2} - 15\) is at (0, -15).

Step 1: Format the function

Rewrite the function in the standard form of a quadratic equation \(y = ax^{2} + bx + c\). In this case, the function is already in the required form: \(g(s) = -s^{2} - 15\).

Step 2: Find the coordinates of the vertex

The 'x'-coordinate of the vertex can be found using the equation \(-\frac{b}{2a}\). Substituting \(a = -1\) and \(b = 0\), we get \(s = \frac{0}{2*-1} = 0\). The 'y'-coordinate of the vertex can be found by substituting this 's' value into the equation: \(g(0) = -0^{2} - 15 = -15\). So, the vertex of the function is at (0, -15).

Step 3: Graph the function

Using a graphing utility, plot the function by choosing suitable 's' values and calculating the corresponding 'g(s)' values. Mark the vertex on the graph. The graph will open downwards (since 'a' is negative) and has its vertex at the point (0, -15). Remember to interpret the graph correctly and ensure it matches the expected shape and position of a quadratic function.

Key Concepts

Finding the Vertex

When graphing quadratic functions, identifying the vertex is crucial as it represents the highest or lowest point on the parabola, depending on whether the parabola opens upwards or downwards.

The vertex (\textbf{h, k}) of a standard quadratic equation of the form \(y = ax^2 + bx + c\) can be found using the formula for the 'x'-coordinate: \(h = -\frac{b}{2a}\). Once you determine the 'x'-coordinate, substitute it back into the original equation to find the 'y'-coordinate, \(k\).

For example, if you have the function \(g(s) = -s^2 - 15\), the coefficient \(a\) is -1, and \(b\) is 0. Plugging these into the vertex formula gives us an 's' coordinate of 0. Plugging \(s = 0\) back into \(g(s)\) yields \(k = -15\), making the vertex (0, -15). This point is crucial in understanding the function's shape and behavior.

Quadratic Equation Standard Form

The standard form of a quadratic equation is given by \(y = ax^2 + bx + c\). Each letter represents a specific element: \(a\) affects the parabola's direction and width, \(b\) controls the axis of symmetry and vertex location, and \(c\) determines the y-intercept.

Understanding the standard form is key to graphing quadratics because it directly tells us if the parabola opens upwards (\(a > 0\)) or downwards (\(a < 0\)). For the provided exercise, we know that our given function is \(g(s) = -s^2 - 15\), which is already in the standard form and indicates a parabola that opens downwards because the leading coefficient (\(a = -1\)) is negative.

The standard form also allows us to simplify the process of finding the vertex and plotting the parabola on a graph.

Graphing Utility Usage

Graphing utilities are an essential tool in visualizing quadratic functions. They allow students to translate a quadratic equation into a visual parabolic curve that provides deeper insight into the function's characteristics.

To use one effectively, input the quadratic function into the graphing utility, then adjust the viewing window to encompass the significant features of the graph, particularly the vertex and the intercepts. For example, for the function \(g(s) = -s^2 - 15\), you would plot points by calculating \(g(s)\) at various 's' values and then use the TABLE feature to organize these values and select the best viewing window.

It's also possible to use a graphing utility to directly find the vertex. Utilize the utility's built-in functions to navigate to the vertex or set it to automatically identify key features of the graph, such as intercepts, maximums, and minimums.