Problem 38 - A Graphical Approach to College Algebra

Question

Solve each problem. Path of a Frog's Leap Refer to Exercise \(37 .\) Suppose that the initial position of the frog is \((0,4)\) and its landing position is (6, 0). The height of the frog in feet is given by $$h(x)=-\frac{1}{3} x^{2}+\frac{4}{3} x+4$$ (a) What was the horizontal distance \(x\) from the base of the stump when the frog reached maximum height? (b) What was the maximum height?

Step-by-Step Solution

Answer

(a) 2 feet; (b) 5.33 feet.

Step 1: Understanding the Quadratic Equation

The given quadratic function for the height of the frog, \( h(x) = -\frac{1}{3}x^2 + \frac{4}{3}x + 4 \), represents a parabola. The parabola opens downwards because the coefficient of \( x^2 \) is negative.

Step 2: Finding the Vertex Formula for Maximum Height

The maximum height of a parabola given by a quadratic equation in the form \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). For our function, \( a = -\frac{1}{3} \) and \( b = \frac{4}{3} \).

Step 3: Calculate the Horizontal Distance for Maximum Height

Substitute \( a \) and \( b \) into the vertex formula:\[x = -\frac{\frac{4}{3}}{2(-\frac{1}{3})} = -\frac{\frac{4}{3}}{-\frac{2}{3}} = 2\]So, the frog reaches its maximum height when the horizontal distance \( x \) from the base of the stump is 2 feet.

Step 4: Calculate the Maximum Height

Substitute \( x = 2 \) back into the height equation to find the maximum height:\[h(2) = -\frac{1}{3}(2)^2 + \frac{4}{3}(2) + 4 = -\frac{1}{3} \times 4 + \frac{8}{3} + 4 = -\frac{4}{3} + \frac{8}{3} + 4\]Simplify the expression:\[= \frac{4}{3} + 4 = \frac{4}{3} + \frac{12}{3} = \frac{16}{3} = 5.33\]Thus, the maximum height of the frog is approximately 5.33 feet.

Key Concepts

Parabola

A parabola is a U-shaped curve that can open upwards or downwards. When dealing with quadratic functions, the shape of the graph is always a parabola. This comes from the squared term. For the quadratic function given in this exercise, \( h(x) = -\frac{1}{3}x^2 + \frac{4}{3}x + 4 \), the parabola opens downwards. This is indicated by the negative coefficient of \( x^2 \), which is \( -\frac{1}{3} \).

A downward-opening parabola represents scenarios where there's a maximum point, like the highest point a frog can jump. The vertex of the parabola marks this highest point, known as the maximum height, when dealing with projectile motions or jumps. Understanding how a parabola corresponds to real-life situations like a frog's leap helps to interpret the mathematics behind it.

Vertex Formula

The vertex formula is essential in finding the maximum or minimum value of a quadratic function. When working with a quadratic equation in the standard form \( ax^2 + bx + c \), the x-coordinate of the vertex is determined using the formula \( x = -\frac{b}{2a} \). This x-value tells you where on the x-axis the vertex, or the peak and lowest point of the parabola, is located.

For our given function \( h(x) = -\frac{1}{3}x^2 + \frac{4}{3}x + 4 \):

• \( a = -\frac{1}{3} \)
• \( b = \frac{4}{3} \)

Substituting these values into the vertex formula, we find:\[x = -\frac{\frac{4}{3}}{2(-\frac{1}{3})} = 2\\]This calculation tells us that the maximum height of the frog's leap occurs when the frog is 2 feet away from the base in terms of horizontal distance. This spot marks where the frog's energy reaches its peak height before descending.

Maximum Height

The maximum height of a object or projectile, like our frog here, is calculated by substituting the vertex x-coordinate back into the original quadratic equation. This gives the y-coordinate of the vertex, representing the highest point reached. In our example, after using the vertex formula, we found the horizontal distance to be \( x = 2 \).

We substitute \( x = 2 \) back into the function \( h(x) = -\frac{1}{3}x^2 + \frac{4}{3}x + 4 \) to find the height:

• \( h(2) = -\frac{1}{3}(2)^2 + \frac{4}{3}(2) + 4 \)
• \( = -\frac{4}{3} + \frac{8}{3} + 4 \)
• \( = \frac{4}{3} + 4 \)
• \( = \frac{16}{3} \approx 5.33 \) feet

Thus, the frog reaches a maximum height of approximately 5.33 feet. This is the highest point on the parabolic path of the leap. Understanding and calculating this maximum height can be very helpful in various physics and engineering problems, especially those involving projectile motion.