Problem 55 - College Algebra and Calculus: An Applied Approach

Question

Maximum Height of a Diver The path of a diver is given by \(y=-\frac{4}{9} x^{2}+\frac{24}{9} x+10\) where \(y\) is the height (in feet) and \(x\) is the horizontal distance from the end of the diving board (in feet) (see figure). Use a graphing utility and the trace or maximum feature to find the maximum height of the diver.

Step-by-Step Solution

Answer

The maximum height of the diver is \( \frac{1458}{16} feet.

Step 1: Identify the quadratic function

The equation given for the path of the diver is \(y=-\frac{4}{9} x^{2}+\frac{24}{9} x+10\). This is already in the form \(y=ax^{2}+bx+c\), where \(a, b,\) and \(c\) are coefficients.

Step 2: Converting the standard equation to vertex form

The vertex form of a quadratic equation is \(y=a(x-h)^2 + k\), where (h, k) is the vertex (maximum point in this case). By completing the square, the standard equation can be converted to vertex form. First, factor the quadratic term: \(y = -\frac{4}{9}(x^2 -\frac{54}{4}x)+10\). Then, complete the square and rewrite the equation by adding and subtracting \(\left(\frac{54}{8}\right)^2\) inside parentheses: \(y = -\frac{4}{9}(x^2 -\frac{54}{4}x + \left(\frac{54}{8}\right)^2 - \left(\frac{54}{8}\right)^2)+10\). This can simplify to: \(y = -\frac{4}{9}(x - \frac{27}{4})^2 + \frac{1458}{16}\).

Step 3: Identify the maximum height

The vertex form of the equation is now \(y = -\frac{4}{9} (x - \frac{27}{4}) ^ 2 + \frac{1458}{16}\). The vertex (h, k) is the point (\(\frac{27}{4}, \frac{1458}{16}\)). Hence, the maximum height of the diver, which is the y-coordinate of the vertex (k), is \( \frac{1458}{16} feet\).

Key Concepts

Graphing Utility

When dealing with quadratic functions, a graphing utility is an invaluable tool for visualizing the shape and key characteristics of the parabola. A quadratic function typically creates a U-shaped curve called a parabola, which can either open upwards or downwards depending on the sign of the leading coefficient. In our exercise, students are instructed to use a graphing utility's trace or maximum feature to locate the maximum point of the diver's path, represented by a quadratic function.

Using the graphing utility simplifies the process of understanding complex functions by providing a visual context. For the maxima or minima, many graphing utilities offer features like 'Trace', 'Max', or 'Min' which allow users to directly identify the highest or lowest point on the graph— in this exercise, that would be the maximum height of the diver. This is crucial for students who might struggle with the algebraic manipulation involved in converting standard quadratic equations to their vertex forms.

Vertex Form of a Quadratic

The vertex form of a quadratic equation is an expression written as \(y = a(x-h)^2 + k\). This form is beneficial because it easily reveals the vertex of the parabola, which corresponds to the maximum or minimum value of the function. In the vertex form, \((h, k)\) represents the vertex's coordinates. Here's why this is important:

When given a quadratic equation in standard form, as in our exercise, converting it to vertex form enables students to immediately identify the maximum height of the parabola without extensively analyzing the properties of the quadratic equation. Moreover, this form makes the direction of the parabola's opening apparent: if 'a' is positive, the parabola opens upward; if 'a' is negative, as it is in our diver's path problem, the parabola opens downward, indicating that the vertex gives the maximum point.

Completing the Square

The method of completing the square allows us to convert a quadratic function from standard form \(y = ax^2 + bx + c\) to vertex form. This algebraic technique involves the following steps:

• Rearrange the equation so that the terms with \(x\) are together, and the constant term is on the other side.
• Factor out the coefficient of the \(x^2\) term if it is not 1.
• Find the value needed to complete the square, which is \(\frac{b}{2a}\), and add and subtract it inside the parentheses squared.
• Add the same value to balance the equation, keeping in mind the factored out coefficient.

This process transforms a seemingly complicated standard quadratic equation into a form that directly shows the vertex thus facilitating the immediate calculation of the maximum or minimum value of a quadratic function. For our diver's problem, completing the square simplifies finding the maximum height.

Coordinates of the Vertex

Understanding the coordinates of the vertex is essential in analyzing quadratic functions. Specifically, in a real-world scenario like our diver's path, the vertex represents the peak of the diver's trajectory. The y-coordinate of the vertex signifies the maximum height reached, which is what we are searching for in the exercise.

In our case, the coordinates of the vertex are derived algebraically by completing the square. Once the quadratic equation is transformed into vertex form, the vertex coordinates \((h, k)\) can be directly read off as (\(\frac{27}{4}, \frac{1458}{16}\)). For the given exercise, \(h\) and \(k\) explicitly indicate the horizontal distance from the diving board and the peak height in feet, respectively, providing the student with the specific information required to solve the problem.