Problem 40 - A Graphical Approach to College Algebra

Question

Graph each function in a viewing window that will allow you to use your calculator to approximate (a) the coordinates of the vertex and (b) the \(x\) -intercepts. Give values to the nearest hundredth. $$y=2.96 x^{2}+1.31$$

Step-by-Step Solution

Answer

Vertex is at \((0, 1.31)\). No real \(x\)-intercepts.

Step 1: Identify the Function Type

The function given is a quadratic function of the form \( y = ax^2 + bx + c \). For the function \( y = 2.96x^2 + 1.31 \), it clearly has no \( x \)-term, so it's a special case of \( ax^2 + c \).

Step 2: Recognize the Characteristics

For a quadratic function \( y = ax^2 + bx + c \), the vertex is at \( x = -\frac{b}{2a} \). Since there is no \(x\)-term, \( b = 0 \) and therefore \( x = 0 \) for the vertex. The vertex is at \( (0, c) \).

Step 3: Determine the Vertex

Having identified that \( x = 0 \) for the vertex, substitute \( x = 0 \) into the equation to find \( y = 2.96(0)^2 + 1.31 = 1.31 \). So, the vertex is \( (0, 1.31) \).

Step 4: Setting Up the Viewing Window

Adjust your viewing window on the calculator to clearly see the vertex and potential \( x \)-intercepts. A suggested window setting is from \(-3 \) to \( 3 \) on the \( x \)-axis and \(-2 \) to \( 5 \) on the \( y \)-axis.

Step 5: Graph the Function

Enter the equation \( y = 2.96x^2 + 1.31 \) in the calculator and graph it using the set window to view its shape and key points.

Step 6: Find the X-Intercepts

To find the \( x \)-intercepts, set \( y = 0 \) in \( 2.96x^2 + 1.31 = 0 \). Solve for \( x \) using \( x^2 = -\frac{1.31}{2.96} \). Since the right side is negative, there are no real solutions; therefore, there are no real \( x \)-intercepts.

Key Concepts

Vertex calculation

Vertex calculation is an essential part of understanding quadratic functions. For any quadratic equation of the form \( y = ax^2 + bx + c \), the vertex is found using the formula \( x = -\frac{b}{2a} \). This formula provides the \( x \)-coordinate of the vertex.
In our specific case, the equation is \( y = 2.96x^2 + 1.31 \). Notice that it doesn't have a \( b \) term, which is typical in the expression, meaning \( b = 0 \).

• Substituting into the vertex formula gives \( x = -\frac{0}{2 \times 2.96} = 0 \)

To find the \( y \)-coordinate of the vertex, plug \( x = 0 \) back into the equation:

• \( y = 2.96(0)^2 + 1.31 = 1.31 \)

So, the vertex of the function is at the point \( (0, 1.31) \). This tells us that this quadratic opens upward from the point \( (0, 1.31) \), which is its minimum point.

Graphing calculator

Using a graphing calculator can greatly assist in visualizing quadratic functions. By entering the equation directly into the calculator, students can adjust the view to better comprehend the parabola's shape. This involves setting a suitable viewing window.
In this scenario, our equation is \( y = 2.96x^2 + 1.31 \). To ensure you see key characteristics like the vertex, adjust the window settings. A suggested range could be:

• \( x \)-axis: from \( -3 \) to \( 3 \)
• \( y \)-axis: from \( -2 \) to \( 5 \)

These settings should include the vertex and allow for a clear picture of how the graph behaves around it. Once entered, the calculator will display the parabola, making it easier to understand the function's characteristics, especially when coupled with numerical approximations of critical points.

Benefits of Graphing Calculators

• They offer a visual representation of equations.
• Key points like intercepts and vertices can be quickly identified.
• They help compare multiple functions or changes in parameters effortlessly.

X-intercepts

The \( x \)-intercepts are the points where the graph crosses the \( x \)-axis. For quadratic functions, finding the \( x \)-intercepts involves setting \( y = 0 \) and solving the resulting equation \( ax^2 + bx + c = 0 \).
In our specific function, \( y = 2.96x^2 + 1.31 \), set \( 0 = 2.96x^2 + 1.31 \) to find the \( x \)-intercepts. Rearrange the equation to solve for \( x \):

• \( x^2 = -\frac{1.31}{2.96} \)

This yields a negative value under the square root, indicating no real number solutions. Thus, there are no real \( x \)-intercepts in this case.
This scenario, where no real \( x \)-intercepts are found, indicates that the entire function is "above" the \( x \)-axis, reflecting a parabola that doesn't touch or cross the \( x \)-axis. Understanding these intercepts helps in comprehending the entire behavior of quadratic functions.