Problem 81 - Precalculus

Question

Determine the domain and range of the quadratic function. \(f(x)=x^{2}+6 x+4\)

Step-by-Step Solution

Answer

Domain: \(\mathbb{R}\). Range: \([-5, \infty)\).

Step 1: Identify the Type of Function

The given function is quadratic, as it is of the form \(f(x) = ax^2 + bx + c\). Here, \(a = 1\), \(b = 6\), and \(c = 4\). Quadratic functions are parabolas, and their domain is always all real numbers since you can input any value for \(x\).

Step 2: State the Domain

For any quadratic function of the form \(f(x) = ax^2 + bx + c\), the domain is all real numbers. Therefore, the domain of \(f(x) = x^2 + 6x + 4\) is \(\mathbb{R}\).

Step 3: Determine the Range

The range of a quadratic function depends on whether it opens upwards or downwards. Since \(a = 1\) is greater than 0, the parabola opens upwards. This means the range is from the minimum value of the function upwards.

Step 4: Find the Vertex

The vertex of a parabola in standard form \(ax^2 + bx + c\) can be found using \(x = -\frac{b}{2a}\). Substituting the values, we get \(x = -\frac{6}{2(1)} = -3\).

Step 5: Calculate the Minimum Value

Substitute \(x = -3\) back into the function to find the minimum value. Hence, \(f(-3) = (-3)^2 + 6(-3) + 4 = 9 - 18 + 4 = -5\).

Step 6: State the Range

The minimum value of the function is \(-5\), and since the parabola opens upwards, the range is \([-5, \infty)\).

Key Concepts

Domain and Range of Quadratic Functions

Every quadratic function is a parabola which can take any real number as an input. This characteristic gives the function an extensive domain, which means the domain is all real numbers, represented by \( \mathbb{R} \). This concept is fairly simple: whatever you choose for \( x \), you will get a corresponding value of \( f(x) \), without any restrictions.

However, the range of a quadratic function isn't always as broad as its domain. The range depends on the direction the parabola opens. If a parabola opens upwards, like our function with \( a = 1 \), the range starts from the lowest point (or minimum point) and goes infinitely upwards. Conversely, if the parabola opens downwards, the range starts from the highest point and stretches infinitely downwards.

For the function \( f(x) = x^2 + 6x + 4 \), since the parabola opens upwards, the range is dictated by the minimum value computed as \( -5 \). Hence, the range is \([-5, \infty)\).

Understanding Parabolas

A parabola is a symmetrical curve, and it is the graphical representation of a quadratic function. The shape and position of a parabola are determined by its quadratic equation \( ax^2 + bx + c \). Here are some key characteristics:

• The "a" value influences the direction and width of the parabola. If \( a > 0 \), it opens upwards. If \( a < 0 \), it opens downwards.
• Its symmetry is centered around the vertical line through the vertex of the parabola, known as the axis of symmetry.
• The vertex is either the minimum or maximum point, making it crucial for determining the range.

Our function \( f(x)=x^2+6x+4 \) with \( a = 1 \) indicates a parabola that opens upwards, creating a "U" shape on the graph.

Vertex of a Parabola: The Key to Its Properties

The vertex of a parabola is a significant point that represents the turning point or change in direction for the function's graph. For quadratic functions, this point can be calculated using the vertex formula \( x = -\frac{b}{2a} \).

Applying it to our function, \( f(x)=x^2+6x+4 \), we find the x-coordinate of the vertex as \( x = -\frac{6}{2(1)} = -3 \).

To find the y-coordinate or the function value at this x-value, substitute \( -3 \) back into the function. This calculation gives the minimum point of \( f(x) \), which is \( f(-3) = -5 \).

So, the vertex here is at \( (-3, -5) \), confirming this point is the minimal value from which the function \( f(x) \) starts to increase as we move either left or right on the graph.