Problem 40 - College Algebra

Question

For the following exercises, graph the parabola, labeling the focus and the directrix. $$ 3 x^{2}+30 x-4 y+95=0 $$

Step-by-Step Solution

Answer

The vertex is (-5/2, 105/16), focus is (-5/2, 107/16), and the directrix is y = 103/16. Graph opens upwards.

Step 1: Rearrange Equation in Standard Form

The given equation of the parabola is \(3x^2 + 30x - 4y + 95 = 0\). Let's first rearrange the equation to isolate the \(y\) term: \(-4y = -3x^2 - 30x - 95\). Divide the entire equation by \(-4\): \(y = \frac{3}{4}x^2 + \frac{15}{2}x + \frac{95}{4}\). This resembles the quadratic form \(y = ax^2 + bx + c\).

Step 2: Complete the Square

To graph the parabola, we need to complete the square for the terms involving \(x\). Start with \(\frac{3}{4}x^2 + \frac{15}{2}x\). Factor out \(\frac{3}{4}\): \(\frac{3}{4}(x^2 + 5x)\). To complete the square inside the parenthesis, add and subtract \((\frac{5}{2})^2 = \frac{25}{4}\): \(\frac{3}{4}(x^2 + 5x + \frac{25}{4}) - \frac{3}{4} \cdot \frac{25}{4}\).

Step 3: Simplify the Equation

The equation becomes \(y = \frac{3}{4}(x + \frac{5}{2})^2 - \frac{75}{16} + \frac{95}{4}\). Combining \(-\frac{75}{16}\) and \(\frac{95}{4}\): \(y = \frac{3}{4}(x + \frac{5}{2})^2 + \frac{105}{16}\). Now, the equation is in the vertex form \(y = a(x - h)^2 + k\), with \(h = -\frac{5}{2}\) and \(k = \frac{105}{16}\).

Step 4: Identify the Vertex, Focus, and Directrix

The vertex of the parabola \(y = \frac{3}{4}(x + \frac{5}{2})^2 + \frac{105}{16}\) is \((-\frac{5}{2}, \frac{105}{16})\). For a parabola \(y = a(x - h)^2 + k\), the focus lies at \((h, k + \frac{1}{4a})\) and the directrix at \(y = k - \frac{1}{4a}\). Find \(a = \frac{3}{4}\), calculate \(\frac{1}{4a} = \frac{1}{3}\), hence the focus at \((-\frac{5}{2}, \frac{107}{16})\), and the directrix \(y = \frac{103}{16}\).

Step 5: Plot the Graph

Graph the parabola using the vertex form \(y = \frac{3}{4}(x + \frac{5}{2})^2 + \frac{105}{16}\). Draw the vertex at \((-\frac{5}{2}, \frac{105}{16})\), plot the focus \((-\frac{5}{2}, \frac{107}{16})\), and draw the directrix line at \(y = \frac{103}{16}\). The parabola opens upwards since \(\frac{3}{4} > 0\). Draw the U-shaped curve, ensuring the vertex is the point of symmetry.

Key Concepts

Completing the Square

Completing the square is a handy technique used to transform a quadratic expression into a perfect square trinomial. This is helpful when you're working to rewrite a quadratic equation in a form that's straightforward to analyze and understand, like the vertex form of a parabola.

Here's a step-by-step guide on how you can complete the square:

• Start by identifying the quadratic and linear coefficients in the expression. For instance, in the expression \(\frac{3}{4}x^2 + \frac{15}{2}x\), we recognize the coefficients are \(\frac{3}{4}\) for \(x^2\) and \(\frac{15}{2}\) for \(x\).
• Factor out the coefficient of the quadratic term from the terms involving \(x\). In doing so, your expression will look like \(\frac{3}{4}(x^2 + 5x)\).
• Look closely at the expression \(x^2 + 5x\), and identify the term that completes the square. This involves taking half of the linear coefficient (in this case 5, which gives \(\frac{5}{2}\)), squaring it to get \(\frac{25}{4}\), then adjusting the original expression accordingly.
• Insert this squared term inside the parenthesis, then balance out the constants outside the parenthesis. Addition and subtraction of the same value ensures equality. The expression becomes \(\frac{3}{4}(x^2 + 5x + \frac{25}{4}) - \frac{3}{4} \cdot \frac{25}{4}\).

By using this technique, the quadratic expression is rewritten into a form that's much easier to work with for graphing or solving.

Vertex Form of a Parabola

The vertex form of a parabola is an essential expression which displays a quadratic equation, making it convenient to identify important properties like the vertex. The equation is typically written as \(y = a(x - h)^2 + k\).

Understanding the elements:

• **\(a\):** This coefficient affects the direction and "width" of the parabola. If \(a > 0\), the parabola opens upwards; if \(a < 0\), it opens downwards. A larger absolute value of \(a\) makes the parabola narrower, while a smaller value means it is wider.
• **\(h\):** This denotes the x-coordinate of the vertex. It's essentially the point which, when used inside the expression \(x - h\), yields zero.
• **\(k\):** The y-coordinate of the vertex. This value directly shifts the parabola up or down along the y-axis.

By converting the equation of a parabola to its vertex form, you can easily identify where the vertex is placed relative to the coordinate axes. For the equation \(y = \frac{3}{4}(x + \frac{5}{2})^2 + \frac{105}{16}\), the vertex is at \((-\frac{5}{2}, \frac{105}{16})\).

Focus and Directrix

The concepts of focus and directrix are fundamental for understanding the geometric definition of a parabola. A parabola is defined as the set of all points that are equidistant from a fixed point, called the focus, and a fixed line, called the directrix.

Here's how you can identify these elements for a parabola:

• The **focus** of a parabola provides valuable information about its curvature. For a parabola given by the vertex form \(y = a(x - h)^2 + k\), the focus can be found at \((h, k + \frac{1}{4a})\).
• The **directrix** is a line that lies opposite to the opening of the parabola, positioned "below" the vertex in case of an upward-opening parabola. The equation for the directrix is \(y = k - \frac{1}{4a}\). This measures as far "below" the vertex as the focus does above.

For the parabola \(y = \frac{3}{4}(x + \frac{5}{2})^2 + \frac{105}{16}\), the focus is at \((-\frac{5}{2}, \frac{107}{16})\) and the directrix at \(y = \frac{103}{16}\). The linkage between the focus and directrix demonstrates the symmetrical attributes of the parabola and aids in its precise graphing.