Problem 22 - Beginning and Intermediate Algebra

Question

Graph each parabola. Give the vertex, axis of symmetry, domain, and range. $$ f(x)=-\frac{1}{3}(x+6)^{2}+3 $$

Step-by-Step Solution

Answer

Vertex: \((-6, 3)\). Axis of Symmetry: \(x = -6\). Domain: \(x \in \mathbb{R}\). Range: \(f(x) \leq 3\).

Step 1: Identify the Standard Form

The given function is \[f(x)=-\frac{1}{3}(x+6)^{2}+3\]. This is in the vertex form of a parabola, \[f(x) = a(x-h)^2 + k\], where \(a\) is the coefficient, and \( (h, k) \) is the vertex.

Step 2: Determine the Vertex

The vertex \((h, k)\) is given directly in the equation. Here,\(h = -6\), and \(k = 3\). Thus, the vertex is \((-6, 3)\).

Step 3: Determine the Axis of Symmetry

The axis of symmetry for a parabola in the form \[f(x) = a(x-h)^2 + k\] is the vertical line \(x = h\). For this function, the axis of symmetry is \(x = -6\).

Step 4: Establish the Domain

For any quadratic function, the domain is all real numbers, \( -\infty < x < \infty\) .

Step 5: Establish the Range

Since the coefficient \(a = -\frac{1}{3}\) is negative, the parabola opens downward. The maximum value of \(f(x)\) is the y-coordinate of the vertex, which is \(3\). So, the range is \( f(x) \leq 3\).

Step 6: Graph the Function

Using the vertex \((-6, 3)\) and the axis of symmetry \(x = -6\), plot the vertex and a few symmetric points around it. Since the parabola opens downward, draw the curve extending downward from the vertex.

Key Concepts

Vertex Form of a Parabola

When you see a quadratic function like \[\f(x) = a(x-h)^2 + k\], it's in the vertex form. The vertex form allows you to easily identify the vertex of the parabola, which is a key feature. In our example, \[\f(x) = -\frac{1}{3}(x+6)^2 + 3\], you can spot that \-h\ and \k\ are hidden in the equation. Here, \-h = -6\ so \h = -6\, and \k = 3\. Thus, the vertex is \(-6, 3\). The vertex represents the highest or lowest point on the graph, depending on whether the parabola opens upwards or downwards.

Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. For a parabola in vertex form \[\f(x) = a(x-h)^2 + k\], the axis of symmetry always passes through the vertex and is given by \x = h\. In our case, the axis of symmetry line is \x = -6\. This means that if you folded the parabola along \x = -6\, both sides would match perfectly. Understanding the axis of symmetry helps in graphing because any point on one side of the vertex will have a mirror point on the opposite side.

Domain and Range

The domain of a parabola, or any quadratic function, is always all real numbers. This is because you can plug any x-value into the equation and get a corresponding y-value. So, for our function \[ -∞ < x < ∞ \].
The range, however, depends on the direction the parabola opens. If \a\ is positive, the parabola opens upwards, and if \a\ is negative, it opens downwards. In this example, \a = -\frac{1}{3}\, indicating that the parabola opens downward.
The y-coordinate of the vertex gives the highest point of the parabola when it opens downward. For \f(x) = -\frac{1}{3}(x+6)^2 + 3\, the vertex is \y = 3\, making the maximum y-value 3. Therefore, the range is all y-values less than or equal to 3: \f(x) ≤ 3\.