Problem 20 - Algebra and Trigonometry Real Mathematics, Real People

Question

Describe the graph of the function and identify the vertex. Use a graphing utility to verify your results. \(f(x)=-6-\frac{1}{4} x^{2}\)

Step-by-Step Solution

Answer

The graph of the function \(f(x)=-6-\frac{1}{4} x^{2}\) is a downward-opening parabola with the vertex at (0, -6).

Step 1: Identify the coefficient of \(x^2\) and the constant term

The coefficient of \(x^2\) is -1/4 and the constant term is -6. Since the coefficient of \(x^2\) is negative, the parabola will open downward.

Step 2: Determine the vertex

Since the quadratic function is in the form \(f(x) = a + bx^2\), the x–coordinate of the vertex is 0. Substitute x=0 into the equation to find the y-coordinate of the vertex, which is \(f(0) = -6\). Therefore, the vertex of the parabola is (0, -6)

Step 3: Describe the graph

The graph of the function is a parabola opening downward with a vertex at (0, -6). It is symmetric with respect to the y-axis and does not cross the x-axis, implying that the function has no real zeros.

Step 4: Verify with a graphing utility

Plot the function on a graphing utility to confirm the shape of the graph and the location of the vertex. The graph should match the description given in Step 3.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a crucial concept when graphing quadratic functions. It represents the highest or lowest point on the graph, depending on whether the parabola opens upward or downward. Here's a simple way to find the vertex: In a quadratic equation of the form \(f(x) = ax^2 + bx + c\), if \(a\) is positive, the parabola opens upward, and the vertex is at the lowest point. If \(a\) is negative, as it is in the exercise \(f(x)=-\frac{1}{4} x^{2} - 6\), the parabola opens downward, and the vertex is at the highest point.

In the given function, since there is no \(x\) term, the vertex's horizontal coordinate is 0. Substituting \(x = 0\) into the equation, we find the vertical coordinate, which is \(f(0) = -6\). Hence, the vertex of the parabola in our example is at the point (0, -6). Understanding where the vertex lies helps in sketching the parabola accurately.

Parabola Symmetry

Every parabola is symmetric with respect to a straight line called the axis of symmetry. This axis runs vertically through the vertex of the parabola. In other words, if you were to fold the parabola along this axis, both halves would match up perfectly. For the standard quadratic function \(f(x) = ax^2 + bx + c\), the axis of symmetry can be found using the formula \(x = -\frac{b}{2a}\).

Since the exercise's function \(f(x)=-\frac{1}{4} x^{2} - 6\) lacks a \(bx\) term, the axis of symmetry is simply the y-axis, or \(x = 0\). This means the parabola is mirrored perfectly across the y-axis. When graphing, you can check your work by ensuring that each point on one side of the parabola has a corresponding point at an equal distance on the opposite side.

Downward Opening Parabola

The direction in which a parabola opens is determined by the sign of its quadratic term's coefficient. If the coefficient is positive, the arms of the parabola extend upwards; if it's negative, the arms open downwards. In our exercise, the function is \(f(x)=-\frac{1}{4} x^{2} - 6\), where the coefficient of the \(x^2\) term (\(a\)) is negative, indicating a downward opening parabola.

This attribute of a parabola affects many of its features, like the maximum value it can reach, which is given by the y-coordinate of the vertex. Since the graph opens downwards, the vertex represents the parabola's maximum point. It's important to recognize that a downward opening parabola signifies that the values of \(f(x)\) will decrease as \(x\) moves away from the vertex, creating the characteristic 'frown' shape on the graph.