Problem 22 - Contemporary Precalculus

Question

In Exercises \(17-28,\) determine the vertex, focus, and directrix of the parabola without graphing and state whether it opens upward, downward, left, or right. $$2 x-1=-6(y+1)^{2}$$

Step-by-Step Solution

Answer

Answer: The vertex of the parabola is \((-\frac{1}{2}, -1)\), the focus is \((-\frac{7}{12}, -1)\), the directrix is \(x = -\frac{5}{12}\), and the parabola opens to the left.

Step 1: Rewrite the equation in standard form

The given equation is: $$2x - 1 = -6(y + 1)^2$$ We want to rewrite this equation in the standard form of a parabola. To do this, we will isolate x: $$x = \frac{-6(y + 1)^2 + 1}{2}$$ Now, rewrite this equation as: $$x = -3(y + 1)^2 + \frac{1}{2}$$ From the given equation, we can observe that it represents a horizontal parabola because it is in the form of \(x = ay^2 + by + c\). Since the leading coefficient is negative, the parabola will open to the left.

Step 2: Find the vertex

The vertex of the parabola can be found from the standard form. In this case, the vertex is: Vertex: \((-\frac{1}{2}, -1)\)

Step 3: Find the focus

For a horizontal parabola in the form \(x = ay^2 + by + c\), the distance between the vertex and the focus is \(|d| = \frac{1}{4a}\) along the x-axis. In this case, \(a = -3\). So, \(d = \frac{1}{-12} = -\frac{1}{12}\). Since the parabola opens to the left, we need to subtract this distance from the x-coordinate of the vertex to find the focus. Focus: \((-\frac{1}{2} - \frac{1}{12}, -1) = (-\frac{7}{12}, -1)\)

Step 4: Find the directrix

For a horizontal parabola, its directrix is a vertical line at a distance of \(|d|\) from the vertex. Since the parabola opens to the left, the directrix will be to the right of the vertex. Thus, add the distance \(d\) to the x-coordinate of the vertex. Directrix: \(x = -\frac{1}{2} + \frac{1}{12} = -\frac{5}{12}\) In summary, the vertex of the parabola is \((-\frac{1}{2}, -1)\), the focus is \((-\frac{7}{12}, -1)\), the directrix is \(x = -\frac{5}{12}\), and the parabola opens to the left.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a fundamental concept in the study of quadratic functions, and it represents the highest or lowest point on the graph, depending on whether the parabola opens upwards or downwards. For horizontal parabolas, like the one in our exercise, which opens left or right, the vertex is the leftmost or rightmost point, respectively. It's essentially the turning point where the parabola changes direction.

When dealing with the standard form of a parabola, which is \(y - k = a(x - h)^2\) for vertical parabolas or \(x = ay^2 + by + c\) for horizontal ones, the vertex can be easily determined by looking at the values of \(h\) and \(k\), or by completing the square if necessary. In our particular problem, the vertex of the given parabola \(2x - 1 = -6(y + 1)^2\) is identified as the point \(\left(-\frac{1}{2}, -1\right)\). This vertex is a critical point that students should learn to find confidently, as it helps in graphing and understanding the parabola's direction of opening and its maximum or minimum value.

Focus of a Parabola

The focus of a parabola is a fixed point that, along with the directrix, defines the property of the parabola: each point on the parabola is equidistant from the focus and the directrix. The focus lies inside the parabola and is one of the key components of its definition based on the locus of points.

The position of the focus can be found using the formula \(d = \frac{1}{4a}\), where \(a\) is the coefficient of the squared term in the parabola's equation. For a parabola with the equation \(x = -3(y + 1)^2 + \frac{1}{2}\), the coefficient \(a = -3\), leading to a focus that lies in a distance of \(d = -\frac{1}{12}\) from the vertex along the x-axis. As our parabola opens to the left, the focus is found by moving to the left from the vertex by this distance, resulting in the coordinates \(\left(-\frac{7}{12}, -1\right)\). The focus is important for understanding the shape and orientation of the parabola.

Directrix of a Parabola

A directrix is a line that, together with the focus, defines the parabola. The directrix is perpendicular to the axis of symmetry of a parabola and is found on the opposite side of the vertex from the focus. For every point on the parabola, the distance to the focus is equal to the perpendicular distance to the directrix.

For our horizontal parabola, which opens to the left, the directrix is a vertical line. This means its equation will take the form of \(x = value\), and it will be placed at an equal distance from the vertex, but to the right since the parabola opens leftward. Using \(d = \frac{1}{4a}\), we find that the directrix for the equation \(2x - 1 = -6(y + 1)^2\) is given by the line \(x = -\frac{5}{12}\). This location serves as a reference to help graph the problem accurately and recognize the geometric property of the parabola.

Standard Form of a Parabola

The standard form of a parabola's equation is crucial for studying its properties and graphing it. For vertical parabolas, the standard form is \(y = ax^2 + bx + c\), and for horizontal parabolas, it's a similar equation with x and y variables switched, typically represented as \(x = ay^2 + by + c\). This format allows us to easily identify not only the vertex but also the direction in which the parabola opens, as well as other characteristics such as the axis of symmetry.

In the case of the parabola from our exercise, its standard form was found by rewriting \(2x - 1 = -6(y + 1)^2\) into \(x = -3(y + 1)^2 + \frac{1}{2}\), allowing us to identify that it opens to the left due to the negative coefficient in front of the squared term. The standard form is a powerful tool for students, enabling them to analyze and understand the graphical behavior of parabolic equations.