Problem 46 - Algebra and Trigonometry

Question

Find the domain and range of the function. $$ f(x)=x^{2}-2 x-3 $$

Step-by-Step Solution

Answer

Domain: \((-\infty, \infty)\); Range: \([-4, \infty)\).

Step 1: Identify the Type of Function

The function given is a quadratic function of the form \( f(x) = ax^2 + bx + c \). For this exercise, \( f(x) = x^2 - 2x - 3 \), where \( a = 1 \), \( b = -2 \), and \( c = -3 \). Quadratic functions are parabolas.

Step 2: Determine the Domain

The domain of a quadratic function is all real numbers because there are no restrictions on the values \( x \) can take. Thus, the domain is \( (-\infty, \infty) \).

Step 3: Determine the Vertex

Use the vertex formula for a quadratic function, \( x = -\frac{b}{2a} \). Substitute \( b = -2 \) and \( a = 1 \) to find \( x = -\frac{-2}{2\cdot1} = 1 \). Then, calculate \( f(1) = 1^2 - 2\cdot1 - 3 = -4 \). The vertex is \( (1, -4) \).

Step 4: Determine the Range

Since the parabola opens upwards (\( a > 0 \)), the range starts from the y-value of the vertex \(-4\) and goes to infinity. Thus, the range of \( f(x) = x^2 - 2x - 3 \) is \([-4, \infty)\).

Key Concepts

Function Domain

The concept of the function domain relates to all the possible input values that a function can accept. For quadratic functions like our given equation, \( f(x) = x^2 - 2x - 3 \), understanding the domain is straightforward.

Quadratic functions are polynomials, specifically second-degree polynomials. They're known to form parabolas, which open either upwards or downwards. A remarkable property of all polynomials, including quadratics, is that they are defined for every real number. That means you can plug any real number into \( x \) and get a corresponding value of \( f(x) \).

Therefore, the domain of any quadratic function is all real numbers, denoted in interval notation as \( (-\infty, \infty) \). This indicates that there are no breaks, gaps, or undefined points in the function for any real number value of \( x \).

Function Range

The function range refers to all the possible output values that a function can produce. In the context of quadratic functions like \( f(x) = x^2 - 2x - 3 \), the range is determined by the direction in which the parabola opens and the vertex's y-value.

Quicks facts to remember about quadratic ranges:

• If the parabola opens upwards (when \( a > 0 \)), the vertex represents the lowest point on the graph.
• Conversely, if the parabola opens downwards (when \( a < 0 \)), the vertex is the highest point.

In our given function, \( a = 1 \), meaning the parabola opens upwards. Therefore, the y-coordinate of the vertex forms the minimum value of the range. We calculated the vertex to be \( (1, -4) \); thus, the smallest value that \( f(x) \) can take is \(-4\).

Since there is no upper bound as the parabola extends infinitely upwards, the range of our function is expressed as \([-4, \infty)\), showcasing that it includes \(-4\) and every number greater than \(-4\).

Vertex of a Parabola

The vertex of a parabola is a crucial point that provides valuable insights into the function's graph. For quadratic functions, this point serves as either the maximum or minimum of the function, depending on the direction the parabola opens.

To locate the vertex for a quadratic function of the form \( ax^2 + bx + c \), use the formula:
\[ x = -\frac{b}{2a} \]
Applying it to our function \( f(x) = x^2 - 2x - 3 \) with \( a = 1 \) and \( b = -2 \), we find:
\[ x = -\frac{-2}{2 \cdot 1} = 1 \]

Once we have the x-coordinate of the vertex, we substitute it back into the function to find the y-coordinate:
\( f(1) = 1^2 - 2 \cdot 1 - 3 = -4 \)
So, the vertex is located at \( (1, -4) \).

The vertex not only provides a reference point on the graph but also helps to define the function's range and the parabola's axis of symmetry.