Problem 3 - A Graphical Approach to Precalculus with Limits

Question

These exercises review topics covered in earlier sections. The concepts are used in solving the applications that follow in this exercise set. Do not use a calculator. Find the minimum function value of \(f(x)=3 x^{2}-24 x+50\).

Step-by-Step Solution

Answer

The minimum function value is 2 at \( x = 4 \).

Step 1: Identify the Form of the Quadratic Equation

The given function \( f(x) = 3x^2 - 24x + 50 \) is a quadratic equation of the standard form \( ax^2 + bx + c \), where \(a = 3\), \(b = -24\), and \(c = 50\). A quadratic function creates a parabola, which opens upwards if \(a > 0\). Thus, it has a minimum value at the vertex of the parabola.

Step 2: Find the Vertex using the Vertex Formula

The x-coordinate of the vertex for a quadratic equation \( ax^2 + bx + c \) is given by \( x = -\frac{b}{2a} \). Substituting \( a = 3 \) and \( b = -24 \) into the formula, we get: \[x = -\frac{-24}{2 \times 3} = \frac{24}{6} = 4.\]

Step 3: Substitute the x-coordinate back into the Function

Now that we know the x-coordinate of the vertex is \( x = 4 \), substitute \( x = 4 \) back into the function to find the minimum function value:\[f(4) = 3(4)^2 - 24(4) + 50.\]Calculate:\[f(4) = 3(16) - 96 + 50 = 48 - 96 + 50 = 2.\]Thus, the minimum value is \( f(4) = 2 \).

Key Concepts

Vertex Formula

The vertex formula is a crucial tool in dealing with quadratic functions in the form of \( ax^2 + bx + c \). A quadratic function graph is a parabola, and the vertex is its highest or lowest point, depending on whether it opens upwards or downwards. To find this vertex, we use the formula for the x-coordinate:

• \( x = -\frac{b}{2a} \)

For our given function \( f(x) = 3x^2 - 24x + 50 \), we identify \( a = 3 \) and \( b = -24 \). By substituting these values into the formula, we compute:

• \( x = -\frac{-24}{2 \times 3} = 4 \)

This calculation provides the x-coordinate of the vertex, which is paramount for finding the extremum of the function. Once we know \( x \), we can further assess the behavior of the parabola by determining its minimum or maximum value at this point.

Parabola

A parabola is the U-shaped graph formed by a quadratic function like \( f(x) = 3x^2 - 24x + 50 \). Its general properties make it a fascinating subject in algebra. A parabola can open either up or down. In our case:

• If \( a > 0 \), it opens upwards.
• If \( a < 0 \), it opens downwards.

Since our function's coefficient \( a = 3 \) is positive, the parabola opens upwards. This particular orientation indicates the vertex as the point where the minimum value of \( f(x) \) occurs. Each parabola is symmetric about its vertex, meaning if you fold it along the vertex line, both sides will match perfectly. This symmetry is helpful when analyzing and predicting values on either side of the vertex.

Minimum Value

The minimum value is a key concept while studying parabolas that open upwards, like the one from our function \( f(x) = 3x^2 - 24x + 50 \). Because the parabola opens upwards, the vertex represents the lowest point, or the minimum value of the function. To find this minimum value, once you have the x-coordinate of the vertex (which is \( x = 4 \) for our example), you substitute it back into the original function:

• \( f(4) = 3(4)^2 - 24(4) + 50 \)

After performing the arithmetic operations, you find:

• \( f(4) = 2 \)

This tells us that the minimum value is \( 2 \) when \( x = 4 \). Understanding how to find the minimum or maximum values through the vertex is critical in solving various real-world problems, such as optimizing resources or minimizing costs.