Problem 48 - Introductory Algebra for College Students

Question

A ball is thrown upward and outward from a height of 6 feet. The height of the ball, \(y,\) in feet, can be modeled by $$y=-0.8 x^{2}+2.4 x+6$$ where \(x\) is the ball's horizontal distance, in feet, from where it was thrown. a. What is the maximum height of the ball and how far from where it was thrown does this occur? b. How far does the ball travel horizontally before hitting the ground? Round to the nearest tenth of a foot. c. Graph the equation that models the ball's parabolic path.

Step-by-Step Solution

Answer

a. The maximum height of the ball is the y-coordinate of the vertex and this occurs at the x-coordinate of the vertex. b. The ball travels horizontally until it hits the ground, this corresponding to the positive root of the equation when \(y = 0\). c. The graph will be a downward opening parabola, with the vertex as the maximum point and y-intercept being 6.

Step 1: Calculate the Maximum Height

First, determine the x-coordinate of the vertex, which represents the distance from the throw point to the ball's highest point. Use the formula for the x-coordinate of the vertex of a parabola, \(x=-\frac{b}{2a}\). For given equation \(y = -0.8x^{2}+2.4x+6\), the values of a, b, and c are -0.8, 2.4, and 6 respectively. So input these values, and calculate \(x = -\frac{2.4}{2*-0.8}\). Next, substitute x in the equation to find the maximum height.

Step 2: Calculate Horizontal Distance

Set y=0 in the equation \(y = -0.8x^{2}+2.4x+6\) and solve the resulting quadratic equation for x. This will give two solutions. Discard any negative solution as distance cannot be negative.

Step 3: Graph of the Equation

Plot the vertex, which is the maximum height calculated in step 1. The y-intercept (the value of y when x=0) is 6. Then, plot a few more points on either side of the vertex by choosing x values and calculating corresponding y values. Then, draw a smooth curve through all these points to form the graph.

Key Concepts

Parabolas

Parabolas are beautiful curves that are the graphs of quadratic functions. They either open upwards or downwards. These curves have symmetric properties, and their shapes are determined by the quadratic equation, which is typically in the form \( y = ax^2 + bx + c \). In this form, \( a \), \( b \), and \( c \) are constants, with \( a \) determining the direction of the parabola. If \( a \) is positive, the parabola opens upwards like a U. If it is negative, the parabola opens downwards like an upside-down U.

A key feature of parabolas is their symmetry about a vertical line called the axis of symmetry. This line always passes through the vertex of the parabola. Understanding parabolas involves recognizing these characteristics to predict their shape and the path they describe, such as in projectile motions like the ball in the exercise. Parabolas describe real-world scenarios where an object follows a curvilinear path under the influence of gravity, like the ball's movement in the given equation. The parabola here is downward-opening due to the negative coefficient of \( x^2 \).

Vertex of a Parabola

The vertex of a parabola is a point where it turns. In a real-world example like the one in the exercise about a ball's path, the vertex represents the maximum height point of the ball. It is a crucial point because it gives important information on the behavior of the parabolic path.

To find the vertex, we use a formula involving the coefficients \( a \) and \( b \) from the quadratic equation \( y = ax^2 + bx + c \). Specifically, the x-coordinate of the vertex is given by \( x = -\frac{b}{2a} \). Once you have the x-coordinate, you substitute it back into the original equation to find the y-coordinate, which gives the maximum height in our example.

The vertex tells us not just about the height, but also how long the object took to reach that height. In the example, using the values \( a = -0.8 \) and \( b = 2.4 \), we find that the maximum height occurs at a certain horizontal distance from the starting point. This is where the parabola peaks, and understanding it aids in graphing and analyzing the motion described by the quadratic equation.

Graphing Quadratic Functions

Graphing quadratic functions involves plotting points and drawing a curve that passes through these points, maintaining the characteristic U shape of the parabola. The graph of a quadratic function is symmetric at the axis of the vertex. For the given equation \( y = -0.8x^2 + 2.4x + 6 \), graphing it is essential to visualize the ball's path.

Start by identifying key points:

• The vertex, which is the highest point of this downward-opening parabola.
• The y-intercept, which you find by setting \( x = 0 \) and solving for \( y \). Here, the y-intercept is 6.
• Other points can be determined by choosing x-values to the left and right of the vertex and computing corresponding y-values.

Once you have these points, plot them on a coordinate plane. Drawing a curve through the points results in the parabola representing the quadratic function.

This visual representation helps to better understand the range and domain of the function, and in real-life applications, predict where the object will land—in the case of the exercise, where the ball hits the ground.