Problem 25 - 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. $$y=3 x^{2}+x-4$$

Step-by-Step Solution

Answer

Question: Determine the vertex, focus, and directrix of the parabola $$y=3x^2+x-4$$ without graphing, and state whether it opens upward, downward, left, or right. Answer: The vertex of the parabola is at $$\bigg(-\frac{1}{6}, \frac{35}{36}\bigg)$$, the focus is at $$\bigg(-\frac{1}{6}, \frac{3}{4}\bigg)$$, the directrix is at $$y=\frac{11}{12}$$, and the parabola opens upward.

Step 1: Write the equation in standard form

To write the given equation in the standard form of a parabola that opens upward, we'll complete the square. The standard form for a vertical parabola is: $$y=ax^2+bx+c$$ Given equation is $$y=3x^2+x-4$$, we can rearrange it as $$y-1=3(x^2+\frac{1}{3}x)$$ Now, we'll complete the square.

Step 2: Complete the square

To complete the square, we take the coefficient of x, divide it by 2, and then square it. So \(\big(\frac{\frac{1}{3}}{2}\big)^2=\frac{1}{36}\). Now, add and subtract this value to the equation. $$y-1+\frac{1}{36}=3\bigg(x^2+\frac{1}{3}x+\frac{1}{36}\bigg)$$ Now, the left side becomes a perfect square: $$y-\frac{35}{36}=3\bigg(x+\frac{1}{6}\bigg)^2$$ This is the standard form of the parabola that opens upward.

Step 3: Determine the vertex, focus, and directrix

From the standard form, we can see that the vertex of the parabola is: $$(h, k)=\bigg(-\frac{1}{6}, \frac{35}{36}\bigg)$$ Now, let's determine the focus and directrix. For a parabola that opens upward, the distance from the vertex to the focus and the vertex to the directrix is: $$\frac{1}{4a}$$ In our case, \(a=3\). So the distance is: $$\frac{1}{4(3)}=\frac{1}{12}$$ The focus of the parabola will be at a distance of \(\frac{1}{12}\) units above the vertex: Focus:$$\bigg(-\frac{1}{6}, \frac{35}{36}+\frac{1}{12}\bigg) = \bigg(-\frac{1}{6}, \frac{3}{4}\bigg)$$ The directrix of the parabola will be at a distance of \(\frac{1}{12}\) units below the vertex: Directrix:$$y=\frac{35}{36}-\frac{1}{12}=\frac{11}{12}$$

Step 4: State the direction of the parabola

Since the coefficient of \(x^2\) is positive, the parabola opens upward. In conclusion, the vertex of the parabola is at \(\bigg(-\frac{1}{6}, \frac{35}{36}\bigg)\), the focus is at \(\bigg(-\frac{1}{6}, \frac{3}{4}\bigg)\), the directrix is at \(y=\frac{11}{12}\), and the parabola opens upward.

Key Concepts

Vertex

The vertex of a parabola is a crucial point determining the shape and direction of the curve. It is the point where the parabola changes direction. Mathematically, the vertex is given by the coordinates \(h, k\) in the vertex form of a parabola, which is \(y = a(x-h)^2 + k\).
In the given equation \(y = 3(x+\frac{1}{6})^2 + \frac{35}{36}\), you can see that \(h = -\frac{1}{6}\) and \(k = \frac{35}{36}\). This means our vertex is located at:

• \([\frac{-1}{6}, \frac{35}{36}]\)

The vertex not only helps in locating the exact position of the parabola on the graph but also aids in understanding the symmetry of the graph. Since the coefficient of \(x^2\) is positive in our equation, the parabola opens upward with its vertex being the lowest point.

Focus

The focus of a parabola is a specific point that lies on the axis of symmetry. Light or signals that reflect off the parabola's surface converge at this point. The formula to find the focus when the parabola is in the form \(y - k = a(x-h)^2\) is:

• Focus: \([h, k + \frac{1}{4a}]\)

With our parameters, \(h = -\frac{1}{6}, k = \frac{35}{36},\) and \(a = 3\), we find the focus by calculating:

• \(k + \frac{1}{12} = \frac{3}{4}\)

This places the focus at:

• \([-\frac{1}{6}, \frac{3}{4}]\)

The focus determines where light or signals will concentrate after reflecting off the parabola, making it essential in applications like satellite dishes and headlights.

Directrix

The directrix of a parabola is a line that is equidistant from the focus point and reflects the geometric property of the parabola. It helps define the shape along with the focus, creating a perfect mirrored line at any point on the parabola.
The formula to find the directrix \(y = k - \frac{1}{4a}\) gives a line that lies parallel to the vertex.
From the given problem, since \(k = \frac{35}{36}\) and \(a = 3\), the directrix is calculated as:

• \(y = \frac{35}{36} - \frac{1}{12}\)
• \(y = \frac{11}{12}\)

This vertical line sits beneath the vertex, ensuring that the parabola remains constantly reflecting through its focus. The position of the directrix is balanced by the parabola's focus and showcases a fundamental property of parabolas connecting the path of light or projections.

Completing the Square

Completing the square is a method used to adjust quadratic equations into a more understandable form. It turns a standard quadratic equation into its vertex form, revealing the vertex directly. This is particularly useful for identifying the vertex and other properties of a parabola.
To complete the square, you take half of the coefficient of \(x\), square it, add and subtract it inside the equation to make a perfect square trinomial.
For the equation \(y=3x^2+x-4\):

• Rearrange into: \(y-1=3(x^2+\frac{1}{3}x)\)
• Find square: \(\left(\frac{1}{6}\right)^2 = \frac{1}{36}\)
• Add inside: \(y-1+\frac{1}{36}=3\left(x+\frac{1}{6}\right)^2\)
• Resulting form: \(y-\frac{35}{36}=3\left(x+\frac{1}{6}\right)^2\)

This process clarifies the core shape and direction of the parabola, providing insights into the characteristics like vertex, focus, opening direction, and more. By completing the square, one can transform an initially complex equation into a neatly summarized parabola.