Problem 44 - Beginning and Intermediate Algebra

Question

Graph each parabola. Give the vertex, axis of symmetry, domain, and range. $$ f(x)=-\frac{2}{3}(x+2)^{2}+1 $$

Step-by-Step Solution

Answer

Vertex: (-2,1); Axis of Symmetry: x = -2; Domain: (-∞,∞); Range: (-∞,1]

Step 1: Identify the Vertex Form

The given function is already in the vertex form, which is: \[ f(x) = a(x-h)^2 + k \] Here, we have: \[ a = -\frac{2}{3}, \, h = -2, \, k = 1 \]

Step 2: Find the Vertex

Using the vertex form formula, the vertex (h, k) is: \[ (-2, 1) \]

Step 3: Determine the Axis of Symmetry

The axis of symmetry is the vertical line that passes through the vertex and can be found using \(x = h\). Thus, the axis of symmetry is: \[ x = -2 \]

Step 4: Determine the Domain

The domain of any quadratic function is all real numbers, expressed as: \[ (-\infty, \infty) \]

Step 5: Determine the Range

Since the coefficient of the quadratic term \(a\) is negative, the parabola opens downwards. The maximum value of \(f(x)\) is the y-coordinate of the vertex, which is 1. Therefore, the range is: \[ (-\infty, 1] \]

Step 6: Graph the Parabola

To graph the parabola, plot the vertex at (-2,1), draw the axis of symmetry at \(x = -2\), and sketch the parabola opening downwards.

Key Concepts

Vertex Form

The vertex form of a quadratic function is one of the most useful ways to express a parabola. This form is written as: \[ f(x) = a(x-h)^2 + k \] Here, \(a\) determines the width and direction of the parabola, while \(h\) and \(k\) reveal the vertex of the parabola. The vertex form makes it easy to identify the vertex at \((h, k)\). For the given function, \[ f(x) = -\frac{2}{3}(x+2)^{2}+1 \], we can see that \(a = -\frac{2}{3}\), \(h = -2\), and \(k = 1\). This tells us that the vertex of this parabola is at \((-2, 1)\). The vertex is a critical point as it shows the highest or lowest point on the graph, depending on whether the parabola opens upwards or downwards.

Axis of Symmetry

The axis of symmetry of a parabola is a vertical line that runs through the vertex, effectively splitting the parabola into two mirrored halves. This line is represented by \(x = h\). For our function, \[ f(x) = -\frac{2}{3}(x+2)^{2}+1 \], we know that \(h = -2\), thus the axis of symmetry is \[ x = -2 \]. Plotting this line on the graph helps in accurately drawing the parabola because it gives us a reference line around which the parabola symmetrically wraps.

Domain and Range

When discussing quadratic functions and their graphs, understanding the domain and range is key. The domain refers to all the possible \(x\)-values that the function can take. For any parabola or quadratic function, the domain is always all real numbers: \[ (-\infty, \infty) \].The range, on the other hand, represents the possible \(y\)-values the function can output. For a downward-opening parabola like ours, \[ f(x) = -\frac{2}{3}(x+2)^{2}+1 \], where the vertex is at \((-2, 1)\), the maximum \(y\)-value is 1. Hence, the range is: \[ (-\infty, 1] \].This information is important for sketching the graph and understanding the extent of the function.

Parabola

A parabola is the graph of a quadratic function and has a characteristic U-shape, which can either open upwards or downwards. The direction in which it opens is determined by the coefficient \(a\) in the function's formula. If \(a\) is positive, the parabola opens upwards, and if \(a\) is negative, it opens downwards. In our case, with \[ f(x) = -\frac{2}{3}(x+2)^{2}+1 \], \( a = -\frac{2}{3} \), so the parabola opens downwards. This U-shape forms because the squared term, \((x-h)^2\), dominates the behavior of the function as \(x\) moves away from \(h\). Parabolas are symmetrical, making visualization easier once you know the vertex and axis of symmetry.