Problem 23 - Precalculus: Mathematical for Calculus

Question

Use a graphing device to graph the parabola. $$4 x+y^{2}=0$$

Step-by-Step Solution

Answer

The parabola, \(x = -\frac{y^2}{4}\), opens leftward with vertex (0,0).

Step 1: Rearrange the Equation

Start by rearranging the given equation to make it easier to graph. The equation is provided as \(4x + y^2 = 0\). Rearrange it to express \(x\) in terms of \(y\) by moving all the terms except \(x\) to the other side: \(x = -\frac{y^2}{4}\).

Step 2: Identify the Vertex

The equation \(x = -\frac{y^2}{4}\) is in the form \(x = a(y-k)^2 + h\), which resembles the standard form of a parabola that opens horizontally. Here, \(a = -\frac{1}{4}\), \(h = 0\), and \(k = 0\), indicating the vertex of the parabola is at the point \((0, 0)\).

Step 3: Determine the Direction of Opening

Since \(a = -\frac{1}{4}\), the parabola will open to the left. In parabolas opening horizontally, a negative coefficient for the \(y^2\) term means the parabola opens to the left.

Step 4: Plot Key Points

Choose several values of \(y\) to find corresponding \(x\) values and plot these points. For example, if \(y = -2\), \(x = -1\) (since \(x = -\frac{(-2)^2}{4}=-1\)) and if \(y = 2\), \(x = -1\). Similarly, if \(y = -4\), \(x = -4\) and if \(y = 4\), \(x = -4\). Plot these points on the graph.

Step 5: Sketch the Parabola

Using the vertex and plotted points, sketch the parabola. It should pass through all of these key points, with the vertex at the origin and opening to the left, forming a U-shape turned horizontally.

Key Concepts

Parabola Vertex

The vertex of a parabola is a significant point, often thought of as the "turning point" where the curve changes direction. In the context of the equation given in the exercise, the formula is rearranged to \(x = -\frac{y^2}{4}\) which fits into the standard form for a horizontal parabola: \(x = a(y-k)^2 + h\). Here, the values of \(h\) and \(k\) give the coordinates of the vertex.

• For this problem, \(h = 0\) and \(k = 0\), thus the vertex is at the origin, \((0, 0)\).

Understanding where the vertex is located is essential, as it serves as the central reference point from which the parabola expands or contracts. Remember, the vertex can be either a minimum or a maximum point depending on the parabola's opening direction.

Direction of Opening

Determining the direction in which a parabola opens is crucial for accurately sketching its graph. This is heavily influenced by the coefficient \(a\) in the standard equation form. For the parabola represented by \(x = -\frac{y^2}{4}\), we focus on the sign of \(a\), which is given as \(-\frac{1}{4}\).

• A negative \(a\) value indicates that the parabola opens in the negative direction. Here, it means the parabola opens to the left.
• If \(a\) were positive, the parabola would open to the right.

These rules apply specifically for horizontally opening parabolas. Knowing how \(a\) affects the graph enables clear understanding and prediction of the parabola's orientation without needing a graphing tool.

Horizontal Parabola

Unlike the more commonly seen vertically opening parabolas, horizontal parabolas open sideways. In this exercise, we deal with a horizontal parabola structured as \(x = a(y-k)^2 + h\). This inherently means the parabola opens either to the left or right, depending on the coefficient of the \(y^2\) term.

• The focal width and direction depend on the value and sign of \(a\). The negative value here resulted in a leftward opening.
• The parabola's axis of symmetry is horizontal, corresponding to the line \(y = k\).

This structure affects plotting because, instead of calculating \(y\) for every \(x\), you find \(x\) for a series of \(y\) values. Therefore, visualizing horizontal parabolas requires slightly different thinking but follows similar principles of symmetry and vertex-centered graphing.