Problem 44 - Algebra and Trigonometry

Question

An equation of a quadratic function is given. a. Determine, without graphing, whether the function has a minimum value or a maximum value. b. Find the minimum or maximum value and determine where it occurs. c. Identify the function's domain and its range. $$ f(x)=6 x^{2}-6 x $$

Step-by-Step Solution

Answer

The function has a minimum value. The minimum value is -3/2 and it occurs at x = 1/2. The domain of the function is \(-\infty, \infty\) and its range is \([-3/2, \infty)\).

Step 1: Identify if the function has a minimum or maximum

In the function \(f(x) = 6x^{2} - 6x\), the leading coefficient is 6, which is greater than 0. Thus, the function opens upwards and it has a minimum value.

Step 2: Find the minimum value and where it occurs

The x-coordinate of the minimum value (vertex) can be found using the formula \(-b/(2a)\). In this function, \(a = 6\) and \(b = -6\). So, \(\(-b/(2a) = --6/(2*6) = 1/2\). Substituting \(x = 1/2\) into the function, the minimum value of the function is \(f(1/2) = 6*(1/2)^{2} - 6*(1/2) = -3/2\). Thus, the function has a minimum value of -3/2 at \(x = 1/2\).

Step 3: Identify the function's domain and its range

The domain of a quadratic function is all real numbers, so the domain of this function is \( (-\infty, \infty)\). Since the function has a minimum value and it opens upwards, the range is \([-3/2, \infty)\).

Key Concepts

Minimum and Maximum Values

In the world of quadratic functions, understanding whether a function possesses a minimum or maximum value is crucial. A quadratic function is typically written in the standard form as \(f(x) = ax^2 + bx + c\). The behavior of the parabola, which represents the quadratic function, can help determine these values.
When you examine the function \(f(x) = 6x^2 - 6x\), you will notice that the coefficient of \(x^2\) is positive ( plus greater than zero). This indicates that the parabola opens upwards like a cup, and, as a result, the function has a minimum value.

• If the coefficient were negative, the parabola would open downwards, resembling a frown, and the function would have a maximum value instead.
• The simplest way to predict whether the function reaches a minimum or maximum point is by considering the sign of this crucial coefficient.

Domain and Range

Two essential components of understanding quadratic functions are the domain and range. The domain refers to all possible inputs (or values of \(x\)) that the function can accept. For any quadratic function, especially in the standard form, the domain is always all real numbers. This means you can substitute any real number for \(x\), and the function will provide a valid output. Thus, for the function \(f(x) = 6x^2 - 6x\), the domain is \((-");infty, ");infty)\).
The range, however, differs based on the orientation of the parabola. Since this function opens upwards and has a minimum value at \(x = \frac{1}{2}\), the lowest value, \(-\frac{3}{2}\), forms the beginning of the range. Consequently, this function’s range is \([-\frac{3}{2}, ");infty)\).

• The range illustrates that the function's output values start from \(-\frac{3}{2}\) and extend to infinity.
• Bear in mind that the range will vary depending on whether the parabola has a minimum or maximum value.

Vertex of a Parabola

The vertex is known as the turning point of a quadratic function, reflecting the lowest or highest point on the parabola, depending on its direction. To find the vertex of the function \(f(x) = 6x^2 - 6x\), you can use the vertex formula for a quadratic function, which is given by the expression \(x = -\frac{b}{2a}\).
In this context, you will find that \(a = 6\) and \(b = -6\). Plugging these values into the formula will yield \(x = \frac{1}{2}\). This \(x\)-value represents the horizontal coordinate of the vertex.

• To uncover the complete vertex, you replace \(x\) in the function to find the \(y\)-coordinate or the minimum or maximum value of the function.

Here, substituting \(x = \frac{1}{2}\) into the equation \(6\left(\frac{1}{2}\right)^2 - 6\left(\frac{1}{2}\right)\) results in a \(y\)-coordinate of \(-\frac{3}{2}\). Therefore, the vertex of this parabola is \(\left(\frac{1}{2}, -\frac{3}{2}\right)\).
This crucial point helps understand the function’s behavior and placement on the graph. Regardless of the parabola's orientation, determining the vertex is a vital step in analyzing quadratic functions.