Problem 9 - Calculus for Biology and Medicine

Question

Determine the equation of the line that satisfies the stated requirements. Put the equation in standard form. The line passing through \((0,-2)\) with slope \(-3\)

Step-by-Step Solution

Answer

The equation in standard form is \( 3x + y = -2 \).

Step 1: Use point-slope form

To find the equation of a line, we can start with the point-slope form of a line's equation, which is \( y - y_1 = m(x - x_1) \). Here, \( m \) is the slope and \( (x_1, y_1) \) is a point on the line. We are given the slope \( m = -3 \) and the point \( (0, -2) \).

Step 2: Plug values into point-slope form

Substitute the slope and point into the point-slope form: \( y - (-2) = -3(x - 0) \). This simplifies to \( y + 2 = -3x \).

Step 3: Simplify to slope-intercept form

Rearrange the equation to isolate \( y \): \( y = -3x - 2 \). This is the slope-intercept form of the equation, where \( y = mx + b \).

Step 4: Convert to standard form

The standard form of a linear equation is \( Ax + By = C \). To convert \( y = -3x - 2 \) to standard form, we rearrange it to \( 3x + y = -2 \).

Step 5: Verify standard form

Check to ensure that the standard form equation \( 3x + y = -2 \) satisfies the original point and slope requirements. Substitute \( x = 0 \) into the equation to find \( y \) and confirm the slope. This shows the equation is correct.

Key Concepts

Point-Slope Form

The point-slope form of a linear equation is a flexible way to describe a line using the slope and a single point on the line. The formula is written as:\[ y - y_1 = m(x - x_1) \]where:

• \( m \) is the slope of the line.
• \( (x_1, y_1) \) is a known point on the line.

This form is particularly useful when you know a point that the line passes through and the slope of the line. Using these values, you can easily construct the equation of the line by plugging the slope and point coordinates into the formula. Once these values are substituted in, you can solve for \( y \) to find the line’s equation in slope-intercept form.

This form highlights the rate of change (the slope) directly, so it's very informative for understanding how steep or shallow a line is relative to the axes.

Slope-Intercept Form

The slope-intercept form is the most straightforward way to express a linear equation. It is written as:\[ y = mx + b \]where:

• \( m \) represents the slope of the line.
• \( b \) is the y-intercept, where the line cuts the y-axis.

This form lets you quickly determine the slope and intercept of the line, making it easy to graph. To convert from point-slope form to slope-intercept form, solve the point-slope form equation for \( y \), which directly gives you the equation in slope-intercept form.

This method is most useful for understanding linear behavior. It shows how fast \( y \) changes with \( x \) (the slope), and where the line crosses the y-axis (the intercept). Using this form, you can swiftly sketch a graph since it directly gives two critical components: the slope and the intercept.

Standard Form

The standard form of a linear equation is another method to express linear relationships. The general look is:\[ Ax + By = C \]where:

• \( A \), \( B \), and \( C \) are integers.
• \( A \) should be a non-negative integer.

Standard form is especially beneficial because it's easy to compare two or more linear equations and determine parallel, perpendicular, or intersecting lines. It organizes the equation clearly, positioning variables on one side and the constant on the other.

To convert a slope-intercept form equation to standard form, rearrange the terms to place all variables on one side. This transformation often involves ensuring that the leading coefficient \( A \) is positive and that all coefficients are integers. This form is particularly useful for solving systems of equations. It provides a neat format for equation manipulation, making it easier to substitute and solve.