Problem 68 - College Algebra with Modeling and Visualization

Question

Determine the \(x\) - and \(y\) -intercepts on the graph of the equation. Graph the equation. $$ y=-1.5 x+15 $$

Step-by-Step Solution

Answer

The x-intercept is \( (10, 0) \) and the y-intercept is \( (0, 15) \).

Step 1: Determine the x-intercept

The x-intercept occurs where the value of y is zero. To find the x-intercept, set y to zero in the equation and solve for x.\[\begin{align*} 0 &= -1.5x + 15 \ 1.5x &= 15 \ x &= \frac{15}{1.5} \ x &= 10 \end{align*}\]Thus, the x-intercept is at \((10, 0)\).

Step 2: Determine the y-intercept

The y-intercept occurs where the value of x is zero. To find the y-intercept, set x to zero in the equation and solve for y.\[\begin{align*} y &= -1.5(0) + 15 \ y &= 15 \end{align*}\]Thus, the y-intercept is at \((0, 15)\).

Step 3: Graph the Equation

To graph the equation \(y = -1.5x + 15\), plot the intercepts found in the previous steps. Place a point at the x-intercept \((10, 0)\) and another at the y-intercept \((0, 15)\). Draw a straight line through these points, which represents the graph of the equation.

Key Concepts

Graphing

When we talk about graphing, we are referring to the process of plotting points on a coordinate plane to visually represent equations. In the case of our example, we are working with the linear equation \(y = -1.5x + 15\). Graphing this equation involves identifying key points—particularly the intercepts—and drawing a line through them.

Here's how it's done in simple terms:

• First, we find the intercepts of the line by setting either \(x\) or \(y\) to zero and solving for the other variable. These intercepts, specifically the \(x\)-intercept and \(y\)-intercept, are crucial points because they are where the line crosses the axes.
• Next, plot these points on the graph. This provides two key coordinates that define the line.
• Finally, the line of the equation is drawn by connecting these points. Since linear equations form straight lines, only two points are necessary to draw the full line accurately.

Graphing helps us understand the relationship between variables and how they change in response to each other. It's a visual tool that brings algebraic equations to life, making them easier to comprehend.

x-intercept

The \(x\)-intercept is the point where the graph of an equation crosses the \(x\)-axis of a coordinate plane, meaning the \(y\) value at this point is zero. To find the \(x\)-intercept, follow a simple procedure:

• Take the original equation and substitute zero in place of \(y\).
• Solve for \(x\) to find out where the line touches the \(x\)-axis.
• For instance, with our equation \(y = -1.5x + 15\), setting \(y = 0\) gives us \(0 = -1.5x + 15\).
• Solving this, you get \(x = 10\), indicating that the \(x\)-intercept occurs at the point \((10, 0)\).

Understanding the \(x\)-intercept is crucial as it gives insight into the behavior of the graph along the \(x\)-axis. This point shows where the output (\(y\)) of the function is zero, providing valuable information in both mathematics and physical interpretations, such as determining where a particular value might be "experienced as zero" in real-world scenarios.

y-intercept

The \(y\)-intercept is found where the graph of an equation crosses the \(y\)-axis, which means the \(x\) value here is zero. Determining the \(y\)-intercept is a straightforward process:

• Replace \(x\) with zero in the original equation, since this reveals the \(y\) value at the point where the line crosses the \(y\)-axis.
• In our example equation \(y = -1.5x + 15\), substituting \(x = 0\) simplifies to \(y = 15\).
• This calculation shows that the \(y\)-intercept is located at \((0, 15)\).

Knowing the \(y\)-intercept is particularly useful because it often serves as the starting point for graphing a linear equation. Additionally, it has real-world applications, like understanding initial conditions in contexts where the graph models a situation with a natural starting point at \(x = 0\). The \(y\)-intercept primarily helps anchor the graph, setting one of its absolute points on the plane.