Problem 37 - Algebra and Trigonometry

Question

Find the maximum or minimum value of the function. $$ f(s)=s^{2}-1.2 s+16 $$

Step-by-Step Solution

Answer

The minimum value of the function is 15.64.

Step 1: Identify the Type of Quadratic Function

The given function is \( f(s) = s^2 - 1.2s + 16 \). This is a quadratic function of the form \( ax^2 + bx + c \). Here, \( a = 1 \), \( b = -1.2 \), and \( c = 16 \). Since \( a > 0 \), the parabola opens upwards, indicating a minimum point.

Step 2: Determine the Vertex

The vertex of the quadratic function \( f(s) = as^2 + bs + c \) can be found using the formula for the \( s \)-coordinate of the vertex: \( s = -\frac{b}{2a} \). Substituting the values: \( s = -\frac{-1.2}{2 \times 1} = \frac{1.2}{2} = 0.6 \). Thus, the \( s \)-coordinate of the vertex is 0.6.

Step 3: Calculate the Minimum Value

Substitute the \( s \)-coordinate of the vertex back into the function to find the minimum value. \( f(0.6) = (0.6)^2 - 1.2 \times 0.6 + 16 \). Calculate each part: \((0.6)^2 = 0.36\), \(-1.2 \times 0.6 = -0.72\). Therefore, \( f(0.6) = 0.36 - 0.72 + 16 = 15.64 \).

Step 4: State the Maximum or Minimum Value

Since the parabola opens upwards, the minimum value of the function is at the vertex. Thus, the minimum value of \( f(s) \) is 15.64.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a critical point in understanding its shape and orientation. For any quadratic function in the form \( f(x) = ax^2 + bx + c \), the vertex can be regarded as either the maximum or minimum point of the graph. In our specific function \( f(s) = s^2 - 1.2s + 16 \), the coefficient \( a = 1 \) is positive.
This implies that the parabola opens upwards and features a minimum point. To find the vertex, we use the formula for the \( s \)-coordinate: \( s = -\frac{b}{2a} \). By substituting the given values \( b = -1.2 \) and \( a = 1 \) into the formula, we obtain:

• \( s = -\frac{-1.2}{2 \times 1} \)
• \( s = \frac{1.2}{2} = 0.6 \)

This gives us the \( s \)-coordinate of the vertex. The vertex is hence located at \((0.6, f(0.6))\).

Minimum Value of a Function

The minimum value of a quadratic function that opens upward is located at its vertex. In the function \( f(s) = s^2 - 1.2s + 16 \), since we know the vertex \( s \)-coordinate is 0.6, the task is to find the corresponding minimum value by substituting this back into the original function.
Perform the calculations carefully:

• First calculate \( (0.6)^2 = 0.36 \)
• Next, multiply \(-1.2 \times 0.6 = -0.72 \)
• Add all results: \( 0.36 - 0.72 + 16 = 15.64 \)

Therefore, the minimum value of \( f(s) \) is 15.64, which occurs at the vertex of the parabola.

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of a quadratic equation \( ax^2 + bx + c = 0 \). The formula is expressed as \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). However, in the context of finding the vertex and extremum points, the component \( -\frac{b}{2a} \) from the formula works as the \( x \)-coordinate of the vertex.For parabola-oriented problems like our exercise, it’s important to recognize:

• The formula provides crucial vertex information without computing roots.
• It's especially useful when calculating extremum points like minimum or maximum directly from the vertex.

While our exercise didn’t require finding roots, the quadratic formula remains essential in solving many quadratic-related problems, providing further insights into the behavior and solutions of quadratic functions.