Problem 93 - College Algebra with Modeling and Visualization

Question

Use transformations of graphs to model the table of data with the formula \(f(x)=a(x-h)^{2}+k .\) (Answers may vary.) Average sales price of a home in thousands of dollars $$ \begin{array}{|cccccc} \hline \text { Year } & 1970 & 1980 & 1990 & 2000 & 2005 \\ \hline \text { Price } & 30 & 80 & 150 & 210 & 300 \end{array} $$

Step-by-Step Solution

Answer

Model: \( f(x) = \frac{2}{3}(x - 1990)^2 + 150 \). Check additional data points for accuracy.

Step 1: Identify the Function Type

The problem asks to model the data using a quadratic function in the form \( f(x) = a(x-h)^2 + k \). Quadratic functions can be used to model a wide variety of data sets that seem to have a parabolic shape.

Step 2: Choose Two Points

To determine a reasonable model, we need to select two pairs from the data to calculate the parameters \( a, h, \) and \( k \). Let's choose the years 1970 and 2005 with prices 30 and 300.

Step 3: Determine the Vertex

The vertex of the parabola appears midway between the highest and lowest points. Consider using \((1990, 150)\) as the vertex point, giving us \( h = 1990 \) and \( k = 150 \). This means that when \( x = 1990 \), \( f(x) = 150 \).

Step 4: Establish the Equation

We start with the equation \( f(x) = a(x - 1990)^2 + 150 \). We will use the other point, such as \((2005, 300)\), to solve for \( a \).

Step 5: Solve for 'a' Using Another Data Point

Using the point (2005, 300), substitute into the equation: \( 300 = a(2005 - 1990)^2 + 150 \). Simplify to solve for \( a \):\[300 = a(15)^2 + 150\]\[300 = 225a + 150\]\[225a = 150\]\[a = \frac{150}{225}\]\[a = \frac{2}{3}\].

Step 6: Construct the Final Equation

Substituting the values of \( a \), \( h \), and \( k \) back into the general form, we obtain:\[ f(x) = \frac{2}{3}(x - 1990)^2 + 150 \].

Step 7: Verify the Model

Finally, use other data points (e.g., 1970 and 1980) to check if the model approximates well. For instance, substitute 1980 into the function to check if the outcome approximates the price 80. Adjust \( a \), \( h \), or \( k \) if necessary to improve the fit.

Key Concepts

Transformations of Graphs

Transformations of graphs are vital in understanding how to shift and modify the shape and position of a function on a coordinate plane. When dealing with functions in vertex form, such transformations become clearer and more applicable. In our quadratic function, we see that each component, like \(a\), \(h\), and \(k\), plays a specific role:

• \(a\) determines the width and direction of the parabola. If \(a\) is positive, the parabola opens upwards; if negative, downwards. A larger absolute value of \(a\) makes the parabola thinner, while a smaller absolute value makes it wider.
• \(h\) moves the parabola left or right along the x-axis. This horizontal shift means the parabola's vertex is not at the origin but at \((h, k)\).
• \(k\) moves the graph up or down along the y-axis. This vertical shift sets the base level from which the parabola rises or falls.

Considering these transformations is crucial when modeling real-world data, as it allows us to accurately reflect changes and movements seen in the data on a graph.

Modeling Data

Modeling data involves creating mathematical representations that simplify and describe real-world phenomena. Quadratic functions are particularly useful because their parabolic shape can fit many naturally occurring patterns, such as the trajectory of objects under gravity or economic trends. When we model the given data using a quadratic function, the main goal is to find a function that closely approaches each data point. Start by choosing an appropriate vertex, which aligns with a significant middle point in the data—in this case, the year 1990 with a price of 150. This serves as our baseline when developing the model. By selecting the years 1970 and 2005, we establish points that help determine the shape and curvature essential in making accurate predictions. Through this methodical approach, one can ensure the resulting graph aligns closely with actual data, providing insights and predictions about future trends.

Vertex Form of a Quadratic Function

The vertex form of a quadratic function is expressed as \(f(x) = a(x-h)^2 + k\). This form is powerful because it directly reveals the vertex, \((h, k)\), of the parabola, allowing for easy translation and transformation of the graph.This structure highlights the vertex

• The vertex \((h, k)\) is the highest or lowest point of the parabola, depending on the sign of \(a\).
• The parameter \(a\) affects the openness and direction of the parabola.

To find the vertex form in practical applications, identifying a straightforward point such as a vertex within a data set helps set the stage for further modeling. The vertex essentially divides the parabola symmetrically, making it easier to predict and understand the path it follows.

Solving Quadratic Equations

Solving quadratic equations is a central aspect of working with quadratic functions, especially when developing models from data. Our task involves deriving and confirming the values of the parameters \(a\), \(h\), and \(k\) that bring a quadratic model closer to the given data points.To solve for \(a\) in the equation \(f(x) = a(x-h)^2 + k\), select a specific point, plug it into the equation, and isolate \(a\). For our data set, we've identified the point (2005, 300), which after substituting and simplifying, helps determine \(a = \frac{2}{3}\).Once \(a\) is found, it's essential to verify the quadratic model by comparing calculated outputs with known data points. This validation process can involve plugging in different data points to see if the resulting values reasonably approximate the expected results. Fine-tuning these parameters is often necessary to ensure that the model captures the essence of the given data accurately.