Problem 4 - 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 \(f(x)=5 x^{2}+30 x+17\).

Step-by-Step Solution

Answer

The minimum function value is -28.

Step 1: Identify the Type of Function

Recognize that the given function \( f(x) = 5x^2 + 30x + 17 \) is a quadratic function. The standard form of a quadratic function is \( ax^2 + bx + c \), where in this case, \( a = 5 \), \( b = 30 \), and \( c = 17 \). Since \( a > 0 \), the parabola opens upwards, and the vertex represents the minimum point of the function.

Step 2: Determine the Vertex Using the Formula

To find the vertex of the parabola, use the formula for the x-coordinate of the vertex, \( x = -\frac{b}{2a} \). Substitute \( b = 30 \) and \( a = 5 \) into the formula:\[x = -\frac{30}{2 \times 5} = -\frac{30}{10} = -3\]

Step 3: Calculate the Minimum Function Value

Now that we have the x-coordinate of the vertex, \( x = -3 \), substitute it back into the function \( f(x) = 5x^2 + 30x + 17 \) to find the minimum value:\[f(-3) = 5(-3)^2 + 30(-3) + 17\]Calculate each term:\[5(-3)^2 = 5(9) = 45\]\[30(-3) = -90\]Now compute the entire expression:\[f(-3) = 45 - 90 + 17 = -28\]

Step 4: State the Minimum Value

The minimum value of the function \( f(x) = 5x^2 + 30x + 17 \) occurs at the vertex, and the minimum function value is \(-28\).

Key Concepts

Vertex

The vertex of a quadratic function is a significant point that offers much insight into the function's characteristics. In the context of a parabola, the vertex is the highest or lowest point on the graph, depending on the direction in which the parabola opens. For the function given in the exercise, the quadratic equation is formulated as \(f(x) = 5x^2 + 30x + 17\). Here, the vertex is where the function reaches its minimum point, because the leading coefficient \(a = 5\) is positive, indicating that the parabola opens upwards. The x-coordinate of the vertex can be found with the formula \(x = -\frac{b}{2a}\), a handy tool for pinpointing this crucial location. Substituting the specific values from our function—\(b = 30\) and \(a = 5\)—we find that \(x = -3\). This calculation reveals the "center" of the parabola in terms of its symmetry and height.

Parabola

The shape of the quadratic function graph is called a parabola. In mathematics, a parabola is a U-shaped curve that can open up or down, depending on the coefficient of the squared term. A positive \(a\) value, like \(5\) in our example, suggests an upward opening parabola. This indicates that the graph makes a U-shape, and the vertex at \(x = -3\) is its lowest point.

• The parabola is symmetrical about the vertical line, or the axis of symmetry, which passes through the vertex.
• Every parabola has specific properties such as a focal point and a directrix, though they are not necessary to find the minimum value.

Understanding the nature of a parabola helps visualize how the function behaves across its domain, which in this case extends indefinitely on either side of the vertex while maintaining the same upward curve.

Minimum Function Value

The minimum value of a quadratic function is vital as it represents the smallest output \(f(x)\) the function can produce. This value is located at the vertex when dealing with upward opening parabolas like in our function \(f(x) = 5x^2 + 30x + 17\). After determining that the x-coordinate of the vertex is \(-3\), it's straightforward to calculate this minimum value by substituting \(-3\) back into the original function.Calculating \(f(-3)\), we find:\[f(-3) = 5(-3)^2 + 30(-3) + 17 = 45 - 90 + 17 = -28 \]This calculation shows that the smallest value the function can achieve is \(-28\), and it occurs precisely at \(x = -3\). Knowing the minimum function value allows us to understand not just the function's behavior but also its constraints within defined conditions.