Problem 25 - College Algebra and Calculus: An Applied Approach

Question

Complete the table. Use the resulting solution points to sketch the graph of the equation. \(y=\frac{3}{4} x-1\) $$\begin{array}{|l|l|l|l|l|l|}\hline x & -2 & 0 & 1 & \frac{4}{3} & 2 \\\\\hline y & & & & & \\ \hline\end{array}$$

Step-by-Step Solution

Answer

The completed table is: \[ \begin{array}{|l|l|l|l|l|l|}\hline x & -2 & 0 & 1 & \frac{4}{3} & 2 \\\hline y & -2.5 & -1 & -0.25 & 0 & 0.5 \ \hline\end{array} \]. The graph of the equation is a straight line passing through these points.

Step 1: Finding Y values

First, for each x value listed in the table, substitute it into the equation to find the corresponding y value. Do this by multiplying 3/4 by the x value, then subtract 1.

Step 2: Substitute x=-2 into equation

For \(x=-2\), the equation becomes \(y = \frac{3}{4}(-2) - 1\), which evaluates to \(y=-2.5\)

Step 3: Substitute x=0 into equation

For \(x=0\), the equation becomes \(y = \frac{3}{4}(0) - 1\), which evaluates to \(y=-1\)

Step 4: Substitute x=1 into equation

For \(x=1\), the equation becomes \(y = \frac{3}{4}(1) - 1\), which evaluates to \(y=-0.25\)

Step 5: Substitute x=4/3 into equation

For \(x=\frac{4}{3}\), the equation becomes \(y = \frac{3}{4}(\frac{4}{3}) - 1\), which evaluates to \(y=0\)

Step 6: Substitute x=2 into equation

For \(x=2\), the equation becomes \(y = \frac{3}{4}(2) - 1\), which evaluates to \(y=0.5\)

Step 7: Completing The Table

After these calculations, the table will be:\[\begin{array}{|l|l|l|l|l|l|}\hline x & -2 & 0 & 1 & \frac{4}{3} & 2\\hline y & -2.5 & -1 & -0.25 & 0 & 0.5\\hline\end{array}\]

Step 8: Sketch The Graph

Now, it's time to graph the equation. The obtained pairs of x and y values {[(-2,-2.5), (0,-1), (1,-0.25), (4/3,0), (2,0.5)]} from the table will serve as coordinate points to plot on a grid. Draw a line thorough these points to get the graph of the equation.

Key Concepts

Graphing Linear Functions

Graphing linear functions is an essential skill in mathematics to visualize equations. In simple terms, a linear function is any equation of the form: \[ y = mx + c \]where:

• \( m \) is the slope of the line, determining its steepness and direction.
• \( c \) is the y-intercept, the point where the line crosses the y-axis.

To accurately graph these functions, follow these easy steps:

• Identify the y-intercept \((c)\). Mark this point on the y-axis.
• Use the slope \((m)\). Start from the y-intercept and use the slope to mark a second point. For example, a slope of \(\frac{3}{4}\) moves right 4 units and up 3 units.
• Draw a straight line through your points. This will extend infinitely in both directions.

These steps help translate algebraic expressions into visual representations. It’s a crucial process for visualizing relationships between variables.

Coordinate Geometry

In coordinate geometry, we use the Cartesian plane, a unique tool for visualizing mathematical equations. It consists of two perpendicular number lines (axes):

• The horizontal axis (x-axis).
• The vertical axis (y-axis).

Each point on the plane is represented by a coordinate \((x, y)\). When graphing the equation \( y=\frac{3}{4}x-1 \), each solution pair \((x, y)\) tells us an exact position on this plane. Here's how it looks:

• Start with the coordinate (0, -1). This is where our line crosses the y-axis.
• Place the next point using a table of values and mark the points: (-2, -2.5), (0, -1), (1, -0.25), \(\frac{4}{3}, 0\), and (2, 0.5).

Using coordinate geometry, we don't just see numbers. We see the shape and slope of line. It creates a visual cross-check, ensuring equations balance perfectly.

Algebra

Algebra helps us manipulate and understand equations, including linear ones. This involves solving for unknowns and understanding relationships. With the example equation \(y=\frac{3}{4}x - 1\), algebra guides us to do the following:

• Solve for \(y\) as seen in the steps of substituting values for \(x\).
• Simplify expressions to get a clear answer. For instance, multiplying \(\frac{3}{4} \times (-2)\) gives a straightforward calculation: \(-2.5\).

Algebraic methods transform complex calculations into simple steps. Learning these principles aids in solving not just equations, but also function analysis, which is crucial in various fields such as science and engineering. Remember, practice makes perfect when learning how to effectively employ algebra.