Problem 37 - A Graphical Approach to College Algebra

Question

Solve each problem. Path of a Frog's Leap A frog leaps from a stump 3 feet high and lands 4 feet from the base of the stump. We can consider the initial position of the frog to be at \((0,3)\) and its landing position to be at \((4,0)\) CAN'T COPY THE IMAGE It is determined that the height \(h\) in feet of the frog as a function of its distance \(x\) from the base of the stump is given by $$h(x)=-0.5 x^{2}+1.25 x+3$$ (a) How high was the frog when its horizontal distance \(x\) from the base of the stump was 2 feet? (b) What was the horizontal distance from the base of the stump when the frog was 3.25 feet above the ground? (c) At what horizontal distance from the base of the stump did the frog reach its highest point? (d) What was the maximum height reached by the frog?

Step-by-Step Solution

Answer

(a) 3.5 feet, (b) about 2.28 feet, (c) 1.25 feet, (d) 3.78125 feet.

Step 1: Determine the Height at x = 2 Feet

To find how high the frog was when its horizontal distance was 2 feet, substitute \( x = 2 \) into the given function \( h(x) = -0.5x^2 + 1.25x + 3 \). Compute the height:\[ h(2) = -0.5(2)^2 + 1.25(2) + 3 \] Simplify the expression: \[ h(2) = -0.5(4) + 2.5 + 3 \] \[ h(2) = -2 + 2.5 + 3 \] \[ h(2) = 3.5 \] Therefore, the frog was 3.5 feet high when \( x = 2 \) feet.

Step 2: Find the Horizontal Distance for a Specific Height

We will use the quadratic formula to find \( x \) when \( h(x) = 3.25 \). Set the equation equal to 3.25 and solve for \( x \): \[ 3.25 = -0.5x^2 + 1.25x + 3 \] Rearrange the equation: \[ -0.5x^2 + 1.25x + 3 - 3.25 = 0 \] \[ -0.5x^2 + 1.25x - 0.25 = 0 \] Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = -0.5 \), \( b = 1.25 \), \( c = -0.25 \). Substitute these values to find \( x \). \[ x = \frac{-1.25 \pm \sqrt{1.25^2 - 4(-0.5)(-0.25)}}{2(-0.5)} \] Compute the discriminant: \[ 1.25^2 - 4(-0.5)(-0.25) = 1.5625 - 0.5 = 1.0625 \] Solve for \( x \): \[ x = \frac{-1.25 \pm \sqrt{1.0625}}{-1} \] \[ x = \frac{-1.25 \pm 1.03}{-1} \] \[ x_1 = \frac{-1.25 + 1.03}{-1} \approx -0.22 \] \[ x_2 = \frac{-1.25 - 1.03}{-1} \approx 2.28 \] Only \( x_2 = 2.28 \) is applicable for a horizontal distance since it is positive. Therefore, the frog was 3.25 feet above the ground at this horizontal distance (approximately).

Step 3: Determine the Horizontal Distance of Maximum Height

The vertex of a quadratic function given by \( ax^2 + bx + c \) is at \( x = -\frac{b}{2a} \). Here \( a = -0.5 \) and \( b = 1.25 \). Calculate: \[ x = -\frac{1.25}{2(-0.5)} \] \[ x = \frac{1.25}{1} = 1.25 \] Therefore, the horizontal distance from the base of the stump where the frog reaches its highest point is 1.25 feet.

Step 4: Find the Maximum Height

Substitute \( x = 1.25 \) into the height function to find the maximum height: \[ h(1.25) = -0.5(1.25)^2 + 1.25(1.25) + 3 \] \[ h(1.25) = -0.5(1.5625) + 1.5625 + 3 \] \[ h(1.25) = -0.78125 + 1.5625 + 3 \] \[ h(1.25) = 3.78125 \] The maximum height reached by the frog is 3.78125 feet.

Key Concepts

Function Evaluation

Function evaluation is a fundamental aspect of understanding quadratic functions and their applications. When we are given a function like the frog's height equation, evaluating it means finding the function's output, or height, at a specific input value — such as a particular horizontal distance, \( x \). To evaluate a function at a specific point, simply substitute that point into the function. For example, in part (a) of the exercise, the function \( h(x) = -0.5x^2 + 1.25x + 3 \) is evaluated at \( x = 2 \). Here's how it works:

• Substitute \( x = 2 \) into the function: \( h(2) = -0.5(2)^2 + 1.25(2) + 3 \).
• Calculate each part of the equation separately: \( -0.5(4) = -2 \), \( 1.25(2) = 2.5 \), and add the constant \( 3 \).
• Simplify to find \( h(2) = -2 + 2.5 + 3 = 3.5 \).

This tells us that the frog's height was 3.5 feet above the ground at a horizontal distance of 2 feet.

Vertex of a Parabola

The vertex of a parabola is a key feature when analyzing quadratic functions. The vertex represents either the maximum or minimum point of the parabola, depending on the orientation. For a downward-opening parabola like the frog's leap, the vertex is the highest point.To find the vertex of a quadratic function, such as \( y = ax^2 + bx + c \), use the formula \( x = -\frac{b}{2a} \). This formula calculates the x-coordinate of the vertex. In the frog's leap equation, \( h(x) = -0.5x^2 + 1.25x + 3 \), we determine:

• \( a = -0.5 \)
• \( b = 1.25 \)
• Calculate \( x = -\frac{1.25}{2(-0.5)} = 1.25 \)

This means the frog reaches its peak height at a distance of 1.25 feet horizontally from the base of the stump, making the x-coordinate of the vertex 1.25.

Quadratic Formula

The quadratic formula is a reliable technique for solving equations that cannot be factored easily. It provides solutions for any quadratic equation of the form \( ax^2 + bx + c = 0 \), with the formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula finds the solutions, or roots, of the quadratic equation. In the frog leap problem, we used the quadratic formula to determine the frog’s height of 3.25 feet to solve for \( x \):

• Set up the equation \( -0.5x^2 + 1.25x + 3 - 3.25 = 0 \), simplifying to \( -0.5x^2 + 1.25x - 0.25 = 0 \).
• Identify constants: \( a = -0.5 \), \( b = 1.25 \), \( c = -0.25 \).
• Plug into the quadratic formula: \( x = \frac{-1.25 \pm \sqrt{1.25^2 - 4(-0.5)(-0.25)}}{2(-0.5)} \).
• Compute the discriminant: \( 1.25^2 - 0.5 = 1.0625 \).
• Solve: \( x = \frac{-1.25 \pm 1.03}{-1} \) to find possible distances \( x_1 \approx -0.22 \) and \( x_2 \approx 2.28 \).

Since negative distances don't make sense here, the frog was 3.25 feet above ground at approximately 2.28 feet from the base.