Problem 17 - 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. \(f(\frac{1}{2}) = -6\), \(f(4) = -3\)

Step-by-Step Solution

Answer

The linear function is \(f(x) = x - 6.5\). The graph of the function is a straight line with a slope of 1 and y-intercept of -6.5, passing through the points \((\frac{1}{2}, -6)\) and \((4, -3)\).

Step 1: Calculate the Slope (m)

We have two points, \((x_1,y_1)\) as \((\frac{1}{2}, -6)\) and \((x_2,y_2)\) as \((4, -3)\). Use these to calculate the slope of the line, using the formula \(m = \frac{(y_2-y_1)}{(x_2-x_1)} = \frac{(-3--6)}{(4-\frac{1}{2})} = 1.\)

Step 2: Derive the Linear Function

Utilizing slope intercept form \(y = mx + b\), substitute \(m = 1\), \(x = \frac{1}{2}\), and \(y = -6\) into the formula to find \(b\), the y-intercept. Our function becomes \(-6 = 1*\frac{1}{2} + b\), which simplifies to \(b =-6-1*\frac{1}{2} = -6.5\).So the linear function is \(f(x) = x - 6.5\).

Step 3: Sketch the Graph

The function \(f(x) = x - 6.5\) is a straight line with a slope of \(1\), indicating it will be an upward slanting line. The y-intercept \(-6.5\) means the line will cross the y-axis at that point. Sketching the line means finding the point where \(x = 4\) corresponds to \(y = -3\), and also where \(x = \frac{1}{2}\) corresponds to \(y = -6\), ensuring that these points fall on the same straight line that crosses the y-axis at \(y = -6.5\).

Key Concepts

Slope Calculation

Understanding the slope of a linear function is crucial as it determines the direction and steepness of the line in the graph. The slope is calculated using two points on the line, with a common formula: \[ m = \frac{(y_2-y_1)}{(x_2-x_1)} \]In this exercise, we used the points \((\frac{1}{2}, -6)\) and \((4, -3)\) to find the slope (m). This resulted in a slope of 1, which tells us that for every unit we move horizontally to the right, the vertical change (rise over run) is also 1 unit upwards. This is a key concept in understanding how the line will look on a graph.

Slope-Intercept Form

The slope-intercept form of a linear equation is one of the most popular ways to express these functions. It's given by \[ y = mx + b \]where \(m\) is the slope of the line and \(b\) is the y-intercept, which is the y-coordinate where the line crosses the y-axis. When we determined the linear function in the exercise, we used this form with our calculated slope and one of the given points to find \(b\). Once \(b\) was identified as \(-6.5\), our equation took the final form of \(f(x) = x - 6.5\), showcasing the simplicity and utility of the slope-intercept form.

Graphing Linear Equations

To visualize a linear function, graphing it is essential. It involves plotting points and drawing a line through them. We start by finding the y-intercept and placing a point at \((-6.5, 0)\) on the y-axis. Then, we use the slope to determine how to move from this point to the next. With a slope of 1, from the y-intercept, for every step right (positive x-direction), we step one unit up (positive y-direction). We also plot the given points \((4, -3)\) and \((\frac{1}{2}, -6)\) on the graph, ensuring they lie on the straight line defined by our equation. The process reinforces the connection between algebraic equations and their graphical representations.

Function Values

Function values correspond to the output of the function for specific input values (x-coordinates). In our exercise, we were given function values to start with: \(f(\frac{1}{2}) = -6\) and \(f(4) = -3\). These values were used to help us find the slope and ultimately define the linear function. This illustrates how specific points can give us a wealth of information about the linear function and how we can use these values to graph the function or solve for other unknowns within the function's equation.