Problem 18 - Precalculus with Limits

Question

In Exercises 11-18, (a) write the linear function \(f\) such that it has the indicated function values and (b) sketch the graph of the function. \(f(\frac{2}{3}) = -\frac{15}{2}\), \(f(-4) = -11\)

Step-by-Step Solution

Answer

The linear function that satisfies the given conditions is \(f(x) = x - 8\). The graph of this function starts at the point (0, -8) and slopes upward at a 45-degree angle, passing through the points (\(\frac{2}{3}, -\frac{15}{2}\)) and (-4, -11).

Step 1: Determine the Slope of the Linear Function

The slope measures the vertical change for each unit of horizontal change. As such, the slope is given by the equation \(m = \frac{y_2 - y_1}{x_2 - x_1}.\) Using the given points, \((\frac{2}{3}, -\frac{15}{2})\) and \((-4, -11)\), the slope can be calculated: \(m = \frac{-11--\frac{15}{2}}{-4-\frac{2}{3}}\) which simplifies to \(m = 1\).

Step 2: Determine the Y-Intercept of the Linear Function

Now, using the slope and one of the points, the y-intercept can be calculated. Let's use the first point, \((\frac{2}{3}, -\frac{15}{2})\):Substitute into \(f(x)=mx+c\), here \(f(x)\) is \(y\), \(m\) is the slope (which is 1), and \(x\) is \(\frac{2}{3}\)So, \(-\frac{15}{2} = (\frac{2}{3}*1) +c\)Solving for \(c\) yields \(c = -\frac{15}{2}-\frac{2}{3} = -8\).Therefore, the linear function is \(f(x) = x - 8\).

Step 3: Graphing the Linear Function

To graph the linear function, start by drawing the y-intercept on the y-axis at -8. Then, knowing the slope is 1, go one unit up for every one unit moved to the right. Also, to reaffirm the accuracy of the graph, place the points (\(\frac{2}{3}, -\frac{15}{2}\)) and (-4, -11) on the graph.

Key Concepts

Slope of a Linear Function

The slope of a linear function is a measure of how steep the line is. It is calculated as the ratio of the vertical change (the rise) to the horizontal change (the run) between two points on a line.

The formula to find the slope, designated as m, is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). If the slope is positive, the line slants upward to the right. If it's negative, the line slants downward to the right. A slope of zero means the line is horizontal, and an undefined slope (division by zero in the formula) corresponds to a vertical line.

To illustrate, given two points \((\frac{2}{3}, -\frac{15}{2})\) and \((-4, -11)\), the slope can be calculated as follows: \(m = \frac{-11 - (-\frac{15}{2})}{-4 - (\frac{2}{3})} = 1\). Thus, the line increases by 1 unit of the y-coordinate for every 1 unit increase in the x-coordinate.

Graphing Linear Equations

Graphing linear equations involves plotting points and drawing a line that connects them. To start, you need two pieces of information – the slope and a point, or the slope and the y-intercept.

Using the slope-intercept form of a line, \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept, is extremely helpful for graphing. Once the y-intercept is plotted, you can use the slope to determine the direction and angle of the line. For every rise over run as indicated by the slope, a point is placed corresponding to that movement. Then, draw a line through these points to represent the equation.

In the given example, with a slope of 1 and the y-intercept at -8, start plotting at the point \((0, -8)\) on the y-axis. Then, move up 1 unit and right 1 unit repeatedly to plot additional points. This slope of 1 indicates a 45-degree angle from the x-axis, making it easy to draw a straight line through the points.

Y-Intercept

The y-intercept of a linear function is where the line crosses the y-axis on a coordinate graph. It is represented by the letter \(b\) in the slope-intercept form \(y = mx + b\). This value tells us the point at which the input, or x-value, is zero.

For the exercise at hand, the y-intercept can be found by using one of the points given, say \((\frac{2}{3}, -\frac{15}{2})\) along with the slope. Substitute the slope and the x-value from the point into the equation and solve for \(b\): \(-\frac{15}{2} = 1 * \frac{2}{3} + b\). After rearranging, we get \(b = -8\), which means our line crosses the y-axis at \((0, -8)\).

This point remains constant for any value of x and serves as a starting point for graphing the entire line. The y-intercept is crucial because it gives a fixed location to begin the construction of the graph of the linear function.