Problem 72 - Intermediate Algebra

Question

Sketch the graph of each equation. If the graph is a parabola, find its vertex. If the graph is a circle, find its center and radius. $$x=y^{2}+6 y+2$$

Step-by-Step Solution

Answer

The vertex of the parabola is (-7, -3).

Step 1: Rewrite the Equation in Standard Form

The equation given is \( x = y^2 + 6y + 2 \). For easier handling of this equation, let's rewrite it in the format of a parabola with respect to \( y \). The standard form for a vertical parabola is \( x = a(y-h)^2 + k \) where \((h, k)\) is the vertex. To achieve this, we need to complete the square for the \( y \)-terms.

Step 2: Complete the Square

To complete the square for \( y^2 + 6y \):1. Take the coefficient of \( y \), which is 6, divide by 2, and square it: \( \left( \frac{6}{2} \right)^2 = 9 \).2. Rewrite the equation: \( x = (y^2 + 6y + 9) + 2 - 9 = (y + 3)^2 - 7 \).Now the equation is in the standard form \( x = (y+3)^2 - 7 \).

Step 3: Identify the Vertex

With the rewritten equation \( x = (y+3)^2 - 7 \), we can now identify the vertex \((h, k)\) of the parabola. In the equation \( x = (y - (-3))^2 + (-7) \), \( h = -3 \) and \( k = -7 \). Thus, the vertex of the parabola is \((-7, -3)\).

Step 4: Sketch the Graph

The graph of the equation \( x = (y+3)^2 - 7 \) is a horizontal parabola that opens to the right because the squared term \((y+3)^2\) is positive. The vertex is at \((-7, -3)\). This point is the lowest point on the graph in terms of the \( x \)-values, and the parabola extends rightwards from there.

Key Concepts

Completing the Square

When you're dealing with quadratic equations, completing the square is a handy technique to rewrite the equation into a form that makes certain properties, like the vertex of a parabola, more apparent. Let's delve into how this works in the context of our given problem.Take the quadratic expression in the equation: \( y^2 + 6y \). The goal is to transform this into a perfect square, a trinomial that can be expressed as \((y + b)^2\). Here's how:

• Identify the coefficient of \( y \), which is 6.
• Divide it by 2: \( \frac{6}{2} = 3 \).
• Square the result: \( 3^2 = 9 \).

Now, you adjust the equation by adding and subtracting this result inside it. This allows us to express it as a perfect square:\[ y^2 + 6y = (y^2 + 6y + 9) - 9 \]Which simplifies to:\[ (y + 3)^2 - 9 \]This expression naturally fits into the standard form needed for parabola analysis. Completing the square not only helps to identify the vertex quickly but also aids in graphing since it reveals the transformation of the basic parabola \( y^2 \) into its current form.

Vertex of a Parabola

Once you have the equation in its standard square-completed form \( x = (y + 3)^2 - 7 \), identifying the vertex becomes straightforward. We need to backtrack from the mathematical form to interpret it geometrically.The form \( x = a(y - h)^2 + k \) guides us to find the vertex at \((h, k)\). In our equation:

• \( h \) corresponds to the part inside the square, which is \(-3\) (remember, it comes from \( y - (-3) \)).
• \( k \) takes from the constant term outside, which is \(-7\).

Thus, the vertex of this parabola is located at \((-7, -3)\). This point is significant as it is the "tip" or turning point of the parabola where it starts to symmetrically extend. Knowing the vertex facilitates sketching the graph and can be crucial when solving optimization problems or understanding the geometry of the parabola.

Graphing Parabolas

Graphing a parabola involves understanding its orientation, shape, and how it fits on your coordinate system. With the equation in standard form, you can easily sketch it.For the problem at hand, the rewritten form \( x = (y + 3)^2 - 7 \) suggests a horizontal parabola because it is a function of \( x \) with respect to \( y \), not the usual \( y \) with respect to \( x \). Here's how to interpret and graph it:

• The squared term \((y + 3)^2\) indicates symmetry around the axis parallel to the y-axis through \( y = -3 \).
• This parabola opens to the right since the coefficient of the squared term is positive (which would be unseen or 1 here).
• The vertex, \((-7, -3)\), acts as the lowest point in this configuration, where any lateral changes in \( y \) relate to \( x \) increasing.

Thus, to plot this, start at \((-7, -3)\) on your graph and illustrate the symmetrical arms extending rightward from there. Horizontal parabolas like this are less common in basic algebra but appear frequently in advanced applications involving conic sections.