Problem 19 - Algebra for College Students

Question

Graph each quadratic function, and state its domain and range. $$f(x)=x^{2}+2$$

Step-by-Step Solution

Answer

The domain is \((-\infty, \infty)\) and the range is \([2, \infty)\).

Step 1 - Understand the Structure of the Quadratic Function

The given function is a quadratic function in the form of \(f(x) = ax^2 + bx + c\). Here, \(a = 1\), \(b = 0\), and \(c = 2\). This means the function will produce a parabolic graph opening upwards because \(a > 0\).

Step 2 - Identify the Vertex

For the quadratic function \(f(x) = x^2 + 2\), the vertex form is derived from \(f(x) = a(x - h)^2 + k\), where the vertex is \((h, k)\). Here, \(h = 0\) and \(k = 2\), so the vertex is at the point \((0, 2)\).

Step 3 - Plot the Vertex

Plot the vertex (0, 2) on the coordinate plane. This is a crucial point that defines the parabolic shape.

Step 4 - Determine Additional Points

Choose a few x-values to find additional points on the graph. For example, for \(x = 1\) and \(x = -1\), calculate the corresponding y-values. Thus, \(f(1) = 1^2 + 2 = 3\) and \(f(-1) = (-1)^2 + 2 = 3\). Plot these points: (1, 3) and (-1, 3).

Step 5 - Plot the Additional Points and Sketch the Parabola

Use the symmetry of the parabola around the vertex to plot additional points and draw a smooth curve passing through these points. The curve should open upward starting from the vertex (0, 2).

Step 6 - State the Domain

The domain of any quadratic function in standard form \(f(x) = ax^2 + bx + c\) is all real numbers. Therefore, the domain of \(f(x) = x^2 + 2\) is \((-\infty, \infty)\).

Step 7 - State the Range

The range of the function \(f(x) = x^2 + 2\) depends on the vertex and the direction in which the parabola opens. Since the vertex is the minimum point at (0, 2) and the function opens upward, the range is \([2, \infty)\).

Key Concepts

Domain and Range

Quadratic functions are graphically represented by parabolas. One key aspect of understanding these functions is determining their domains and ranges.
The **domain** of a quadratic function like \(f(x)=x^2+2\) includes all possible x-values. This is because you can input any real number into the function and get a corresponding y-value. Hence, the domain here is \((-\infty, \infty)\).
The **range** is all the possible y-values that the function can output. For our function, the vertex is the lowest point due to the parabola opening upwards. The y-value at the vertex for \(f(x)=x^2+2\) is 2, and y-values increase without bound from there. Therefore, the range is \[2, \infty\]].

Vertex of a Parabola

The vertex of a parabola is a critical point that stands either at the maximum or minimum value of the quadratic function. For quadratic functions in the form of \(f(x)=ax^2+bx+c\), the vertex helps us to locate this exact point.
In our example, the function is \(f(x)=x^2+2\). To find the vertex, we use the formula derived from the vertex form \(f(x)=a(x-h)^2+k\). Here, \(a=1\), \(b=0\), and \(c=2\). Therefore, our vertex is at \(h=0\) and \(k=2\), giving us the point (0, 2).
This vertex point represents the minimum value of our function because the parabola opens upwards (since \(a>0\)). Plotting this point gives a foundation for sketching the entire graph of the quadratic function.

Plotting Points

Plotting points other than the vertex is essential to accurately sketch a quadratic function's graph. We begin with plotting the vertex, which we already identified as (0, 2).
To build around this, we must select a few additional x-values and calculate their y-values using the function.

• For \(x=1\), \(f(1)=1^2+2=3\)
• For \(x=-1\), \(f(-1)=(-1)^2+2=3\)

We now have the points (1, 3) and (-1, 3), which we can plot. By leveraging the symmetry of the parabola, we can replicate points on either side of the vertex. After plotting these points, we draw a smooth curve passing through them, ensuring it opens upward from the vertex point (0, 2).
By connecting these points, we visualize a complete and accurate graph of the quadratic function \(f(x)=x^2+2\).