Problem 19 - Algebra 2

Question

Graph each equation. $$ x=5 y^{2}+25 y+60 $$

Step-by-Step Solution

Answer

It's a vertical parabola opening to the right with vertex \((\frac{115}{4}, -\frac{5}{2})\).

Step 1: Rewrite the Equation in a Standard Form

The given equation is \(x = 5y^2 + 25y + 60\). To make it easier to graph, we need to express it in the standard form of a parabola. Since this is a vertical parabola, the equation should look like \(x = a(y-k)^2 + h\). We start by completing the square for the terms involving \(y\).

Step 2: Completing the Square

Take the quadratic part of the equation, \(5y^2 + 25y\), and factor out 5 to get \(5(y^2 + 5y) + 60\). To complete the square, consider the expression \(y^2 + 5y\). Half of the coefficient of \(y\) is \(\frac{5}{2}\), and squaring this gives \(\left(\frac{5}{2}\right)^2 = \frac{25}{4}\). Add and subtract \(\frac{25}{4}\) inside the parentheses to maintain balance: \(5((y + \frac{5}{2})^2 - \frac{25}{4}) + 60\).

Step 3: Simplifying the Expression

Distribute 5 through, giving \(5(y + \frac{5}{2})^2 - \frac{125}{4} + 60\). To make it easier to understand, convert 60 into quarters: \(60 = \frac{240}{4}\). Thus, the expression simplifies to \(5(y + \frac{5}{2})^2 + \frac{115}{4}\). The equation is now \(x = 5(y + \frac{5}{2})^2 + \frac{115}{4}\).

Step 4: Identify Vertex and Direction

The equation \(x = 5(y + \frac{5}{2})^2 + \frac{115}{4}\) is in vertex form \(x = a(y - k)^2 + h\), where \(a = 5\), \(k = -\frac{5}{2}\), and \(h = \frac{115}{4}\). This tells us that the vertex of the parabola is at \((\frac{115}{4}, -\frac{5}{2})\), and since \(a > 0\), the parabola opens to the right.

Step 5: Plot Key Points and Graph

To graph the parabola, plot the vertex \((\frac{115}{4}, -\frac{5}{2})\). Also, find some additional points by selecting values for \(y\) and calculating corresponding \(x\) values using the simplified equation \(x = 5(y + \frac{5}{2})^2 + \frac{115}{4}\). Draw a smooth curve through these points starting from the vertex, showing the parabola opening to the right.

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic equation into a form that is easier to analyze or graph. In this process, we focus on the quadratic expression, which typically involves a term with a squared variable and a linear term. To complete the square:

• Identify the coefficient of the linear term.
• Take half of this coefficient and square it.
• Add and subtract this squared value within the expression.

For instance, consider the expression \(y^2 + 5y\). The coefficient of \(y\) is 5. Halving it gives \(\frac{5}{2}\), and squaring this value produces \(\left(\frac{5}{2}\right)^2 = \frac{25}{4}\). By adding and subtracting \(\frac{25}{4}\) within the parenthesis, we rewrite the quadratic part as a perfect square trinomial: \((y + \frac{5}{2})^2\). This helps in transforming the original quadratic expression into a more manageable form, making graphing much simpler.

Parabola Vertex Form

The vertex form of a parabola is a way of expressing the quadratic equation to highlight its vertex, which is its highest or lowest point, depending on direction. The general form is \(x = a(y - k)^2 + h\) or \(y = a(x - h)^2 + k\), where

• \((h, k)\) represents the vertex.
• \(a\) determines the direction and width of the parabola.

In the equation provided, \(x = 5(y + \frac{5}{2})^2 + \frac{115}{4}\), it fits the vertex format \(x = a(y - k)^2 + h\):

• The vertex \((h,k)\) is \(\left(\frac{115}{4}, -\frac{5}{2}\right)\).
• \(a = 5\) causes the parabola to open to the right, since it's greater than zero.

Having the equation in vertex form provides a straightforward way to quickly identify key features of the parabola, which is crucial for graphing.

Graphing Quadratic Equations

Graphing quadratic equations involves plotting a curve that represents the solutions to the equation. For a parabola, the graph is a smooth, symmetrical curve. Here’s how to graph using the vertex form:

• Start by locating the vertex on your graph, \((\frac{115}{4}, -\frac{5}{2})\), which acts as the "starting" or "turning" point.
• Since our parabola opens to the right, it will extend outward from the vertex.
• Select additional \(y\) values around the vertex to find corresponding \(x\) values using \(x = 5(y + \frac{5}{2})^2 + \frac{115}{4}\).

Plot these additional points and draw a smooth curve through them. Ensure the parabola is symmetrical around its axis, which is a vertical line passing through the vertex in this case. This method not only accurately reflects the shape of the quadratic function but also leverages the vertex form to simplify the drawing process.