Problem 67 - Precalculus : Building Concepts and Connections

Question

Graph each quadratic function by finding a suitable viewing window with the help of the TABLE feature of a graphing utility. Also find the vertex of the associated parabola using the graphing utility. $$s(t)=-16 t^{2}+40 t+120$$

Step-by-Step Solution

Answer

The vertex of the parabola \(s(t)=-16 t^{2}+40 t+120\) is (1.25, 130)

Step 1: Determine the Coefficients

Observe the quadratic equation \(s(t)=-16 t^{2}+40 t+120\). It follows the form \(ax^{2}+bx+c\), where \(a=-16\), \(b=40\), and \(c=120\)

Step 2: Graph the Function

Using a graphing tool, plot the equation \(s(t)=-16 t^{2}+40 t+120\). Find a suitable viewing window that clearly shows the arc of the parabola and where it intersects the y-axis.

Step 3: Determine the Vertex

To find the vertex of the parabola, use the graphing tool's vertex finding feature or the formula \(-b/2a\), which gives the x-coordinate of the vertex. Substitute \(a=-16\) and \(b=40\) into the formula to get \(x = -(-40)/(2*-16) = 1.25\). Substitute \(x = 1.25\) back into the equation to find the y-coordinate of the vertex: \(s(1.25)=-16*(1.25)^{2}+40 *1.25 +120\)

Key Concepts

Vertex of Parabola

A vertex is a fundamental part of a parabola's structure. It is the point that represents either the highest or lowest point on the graph, depending on whether the parabola opens upwards or downwards. For a given quadratic function in the form \( ax^2 + bx + c \), the vertex can be determined using the formula \( -b/(2a) \) for the x-coordinate.
Consider the quadratic function \( s(t) = -16t^2 + 40t + 120 \):

• Here, \( a = -16 \) and \( b = 40 \).
• Using the formula, the x-coordinate of the vertex is \( x = -40/(2*(-16)) = 1.25 \).
• Substitute \( x = 1.25 \) back into the equation to find the y-coordinate:
• \[ s(1.25) = -16(1.25)^2 + 40*1.25 + 120 = 125 \]

The vertex of this parabola is \((1.25, 125)\), indicating the highest point since the parabola opens downwards (\(a<0\)).

Parabola Graphing

Graphing a parabola involves plotting the curve represented by a quadratic equation. The graph is a symmetrical U-shaped curve that extends infinitely. There are two possible orientations for a parabola, either opening upwards or downwards:

• Upwards when \( a > 0 \)
• Downwards when \( a < 0 \)

You begin by identifying key components such as the vertex and the y-intercept. In the case of \( s(t) = -16t^2 + 40t + 120 \), the graph will open downwards due to \( a = -16 \). The y-intercept is straightforward as it is the point where the graph intersects the y-axis, given by \( s(0) = 120 \).
To accurately display the complete curve of the parabola, use a graphing utility to visualize the full span of the parabola, ensuring a clear view of the vertex and intercepts.

Graphing Utilities

Graphing utilities are powerful tools for visualizing mathematical equations, specifically beneficial for quadratic functions and their parabolas. They help to find precise values like the vertex, zeros, and intersections.
Using a graphing calculator or software, you input the quadratic equation, such as \( s(t) = -16t^2 + 40t + 120 \). These utilities are equipped with features to find:

• The vertex
• Intersections with the axes
• Enhanced viewing options for better analysis

They provide a visual representation that shows how the coefficients affect the graph's shape. Adjust the window settings to get an optimal view, capturing crucial points like the vertex at \((1.25, 125)\).
Use a variety of settings to understand how the parabola expands or contracts and how it shifts along the axes.

Coefficients in Quadratic Equations

Quadratic equations appear in the standard form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are coefficients that determine the parabola's shape and position on a graph.

• The coefficient \( a \) influences the direction and width of the parabola. If \( a > 0 \), the parabola opens upwards. If \( a < 0 \), it opens downwards.
• The coefficient \( b \) affects the position of the vertex along the x-axis.
• The constant \( c \) gives the y-intercept of the graph, where the parabola crosses the y-axis.

For the equation \( s(t) = -16t^2 + 40t + 120 \), we identify that:

• \( a = -16 \), indicating a downward opening.
• \( b = 40 \), positions the vertex at \( x = 1.25 \).
• \( c = 120 \) sets the y-intercept at \( s(0) = 120 \).

The interplay of these coefficients defines the parabola's curvature and initial position, helping predict its behavior when graphed.