Problem 29 - College Algebra with Corequisite Support

Question

For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible. \(f(-5)=-4,\) and \(f(5)=2\)

Step-by-Step Solution

Answer

The equation is \(y = \frac{3}{5}x - 1\).

Step 1: Determine the Slope

To find the equation of a line given two points, we start by calculating the slope (m). The formula for the slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For the points \((-5, -4)\) and \((5, 2)\), the slope calculation is: \[ m = \frac{2 - (-4)}{5 - (-5)} = \frac{6}{10} = \frac{3}{5}. \]

Step 2: Use Point-Slope Form

Now that we have the slope \(m = \frac{3}{5}\), we can use the point-slope form of a linear equation, which is \(y - y_1 = m(x - x_1)\). Using the point \((-5, -4)\), the equation becomes: \[ y + 4 = \frac{3}{5}(x + 5). \]

Step 3: Simplify to Slope-Intercept Form

Next, we simplify the equation to the slope-intercept form \(y = mx + b\). Distribute the slope \(\frac{3}{5}\): \[ y + 4 = \frac{3}{5}x + \frac{3}{5} \times 5. \] This results in: \[ y + 4 = \frac{3}{5}x + 3. \] Solving for \(y\), we get: \[ y = \frac{3}{5}x + 3 - 4. \] Hence, \[ y = \frac{3}{5}x - 1. \]

Step 4: Verify the Linear Equation

To ensure our equation is correct, we can check it against the given points. For \(x = -5\), \(f(-5) = \frac{3}{5}(-5) - 1 = -3 - 1 = -4\). For \(x = 5\), \(f(5) = \frac{3}{5}\times 5 - 1 = 3 - 1 = 2\). Since both points satisfy the equation, it is correct.

Key Concepts

Slope Calculation

Calculating the slope is the first step to finding the equation of a line between two points. The slope (\(m\)) represents how steep the line is as it moves from one point to another.
The formula we use to calculate the slope is:

• \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

This formula calculates the change in the y-values, or "rise," divided by the change in the x-values, or "run."
In this exercise, the points (-5, -4) and (5, 2) are given.
By substituting these values into the formula, we find:

• \(m = \frac{2 - (-4)}{5 - (-5)} = \frac{6}{10} = \frac{3}{5}\)

Thus, the slope of the line is \(\frac{3}{5}\). This tells us that for every 5 units you move right along the x-axis, the line moves 3 units up on the y-axis.

Point-Slope Form

The point-slope form is a foundational concept in linear equations, especially useful when a point on the line and the slope are known. The formula for a line in point-slope form is:

• \(y - y_1 = m(x - x_1)\)

Here, \((x_1, y_1)\) is a point on the line, and \(m\) stands for the slope.
Using this form with a point from our exercise (-5, -4) and the slope \(m = \frac{3}{5}\), we substitute these into the formula:

• \(y + 4 = \frac{3}{5}(x + 5)\)

This gives us a partial equation of the line, not fully simplified, but perfect for describing the line that passes through a given point with a specific slope.
The point-slope form is especially handy before converting an equation into a more universal format, like slope-intercept form, as it clearly showcases the specific point and slope.

Slope-Intercept Form

The slope-intercept form is one of the most common ways to express a linear equation. This form is denoted as:

• \(y = mx + b\)

In this equation, \(m\) is the slope, and \(b\) is the y-intercept, where the line crosses the y-axis.
To convert the equation from point-slope form to slope-intercept form, some simplifications are necessary.
Starting from:

• \(y + 4 = \frac{3}{5}x + 3\)

We simplify to isolate \(y\):

• \(y = \frac{3}{5}x + 3 - 4\)
• Thus, the slope-intercept form becomes \(y = \frac{3}{5}x - 1\)

This equation provides a direct way to understand both the line's direction and where it intersects the y-axis.
The slope-intercept form is valuable for quickly graphing or identifying these features of the line.

Verifying Solutions

Verifying solutions ensures the accuracy of your linear equation with respect to the points given.
To verify, substitute the x-values from the given points back into your simplified linear equation and check if the corresponding y-values match.
For our equation \(y = \frac{3}{5}x - 1\), let's verify both points (-5, -4) and (5, 2):

• Substitute \(x = -5\):
  • \(f(-5) = \frac{3}{5}(-5) - 1 = -3 - 1 = -4\)
  Correct, as it matches the y-value of the point (-5, -4).
• Substitute \(x = 5\):
  • \(f(5) = \frac{3}{5}\times 5 - 1 = 3 - 1 = 2\)
  Correct, as it matches the y-value of the point (5, 2).

Thus, both points verify the accuracy of the linear equation. Verifying solutions is critical, as it confirms that the equation represents the relationship between given data points accurately.
Always double-check your work by plugging in initial conditions to ensure they satisfy the final equation.