Problem 82 - Precalculus

Question

Determine the domain and range of the quadratic function. \(f(x)=2 x^{2}-4 x+2\)

Step-by-Step Solution

Answer

Domain is \((-\infty, \infty)\), range is \([0, \infty)\).

Step 1: Understanding the function

The given function is a quadratic function of the form \(f(x) = ax^2 + bx + c\) where \(a = 2\), \(b = -4\), and \(c = 2\). Quadratic functions have parabolic graphs that either open upwards (\(a > 0\)) or downwards (\(a < 0\)). In this case, since \(a = 2\), the parabola opens upwards.

Step 2: Determine the domain

For any polynomial function, including quadratic functions, the domain is all real numbers because there are no restrictions on \(x\). This means the function can accept any real number as \(x\). Thus, the domain of \(f(x) = 2x^2 - 4x + 2\) is \((-\infty, \infty) \).

Step 3: Find the vertex of the parabola

The vertex form of a quadratic function is instrumental in determining the range. The vertex \((h, k)\) can be found using the formulas: \[h = \frac{-b}{2a} \quad \text{and} \quad k = f(h)\] For this function, \(h = \frac{-(-4)}{2 \times 2} = 1\). Calculate \(k\) by substituting \(h = 1\) back into the function: \(f(1) = 2(1)^2 - 4(1) + 2 = 0\). So, the vertex is \((1, 0)\).

Step 4: Determine the range using the vertex

Since the parabola opens upwards, the \(y\)-coordinate of the vertex is the minimum value of the function. Thus, the range of the function is all real numbers greater than or equal to 0 (\(k\) value). Therefore, the range of \(f(x) = 2x^2 - 4x + 2\) is \([0, \infty)\).

Key Concepts

Domain of a Function

The domain of a function refers to all the possible input values (typically represented by \(x\)) for which the function is defined. For a quadratic function like \(f(x) = 2x^2 - 4x + 2\), there are no restrictions or boundaries limiting the values of \(x\). This is because quadratic functions are polynomials, and polynomials are always defined for all real numbers.

• Simply put, you can plug any real number into \(x\), and the function will produce a result.
• There are no divisions by zero or square roots of negative numbers to worry about here.

Therefore, the domain of this quadratic function is all real numbers, which can be written as \((-\infty, \infty)\).

Range of a Function

The range of a function consists of all the possible output values (typically represented by \(f(x)\)) that the function can produce. To find the range of a quadratic function like \(f(x) = 2x^2 - 4x + 2\), it's crucial to understand the shape and direction of its graph.

• The graph of a quadratic function forms a parabola, which can open upwards or downwards based on the sign of \(a\) in \(ax^2 + bx + c\).
• In our exercise, \(a = 2\), meaning the parabola opens upwards, forming a "U" shape.
• The minimum point of the parabola when it opens upwards is at its vertex.

For \(f(x) = 2x^2 - 4x + 2\), the vertex is \((1, 0)\). Since the parabola opens upwards from this vertex, the minimum value of the function, or the smallest \(y\)-value it can have, is 0. Consequently, the range of this quadratic function is all real numbers greater than or equal to 0, expressed as \([0, \infty)\).

Parabola Vertex

The vertex of a parabola in a quadratic function is a significant point because it represents the peak or the lowest point of the function, depending on its direction. For a quadratic function in the form \(ax^2 + bx + c\), the vertex can be found using the vertex formula:

• \(h = \frac{-b}{2a}\)
• Once you have \(h\), find \(k\) by plugging \(h\) back into the function: \(k = f(h)\)

For the function \(f(x) = 2x^2 - 4x + 2\), we calculate:

• \(h = \frac{-(-4)}{2 \times 2} = 1\)
• \(k = f(1) = 2(1)^2 - 4(1) + 2 = 0\)

Thus, the vertex is \((1, 0)\). This is where the parabola changes direction, and for an upward-opening parabola, it marks the minimum point. Recognizing the vertex aids in determining both the range of the function and key attributes of its graph.