Problem 39 - Beginning and Intermediate Algebra

Question

Rewrite each equation in the form \(x=a(y-k)^{2}+h\) by completing the square and graph it. $$x=\frac{1}{3} y^{2}+\frac{8}{3} y-\frac{5}{3}$$

Step-by-Step Solution

Answer

The equation, rewritten in the form \(x = a(y-k)^2 + h\) by completing the square, is: $$x = \frac{1}{3} (y+4)^2 - \frac{53}{3}$$ The function is a horizontally-opening parabola with its vertex at \((-4, -\frac{53}{3})\), and a stretch factor of \(\frac{1}{3}\).

Step 1: Set Up the Completing the Square Form

We want to rewrite the expression $$\frac{1}{3} y^2+\frac{8}{3} y$$ in the form $$a(y-k)^2$$, where a is the coefficient of the quadratic term. First, factor out the coefficient of the quadratic term: $$\frac{1}{3} (y^2+8y)$$ To complete the square, we will find a constant term to add inside the parenthesis that forms a perfect square trinomial.

Step 2: Complete the Square

The trinomial in the parenthesis is $$y^2+8y+c$$ To find c, we use the formula c = (b/2a)^2, where b is the coefficient of the linear term (8 in this case) and a is 1 (since we factored out 1/3). c = \((\frac{8}{2(1)})^2\) = \(\frac{8}{2}\)^2 = 4^2 = 16 Now, rewrite the expression as: $$\frac{1}{3} (y^2+8y+16)$$ However, since we added 16 inside the parenthesis, we must also subtract 16 outside the parenthesis to maintain the original expression: $$\frac{1}{3} (y^2+8y+16) - \frac{1}{3} \cdot 16$$

Step 3: Write in Vertex Form

Now, rewrite the expression as a square of a binomial: $$\frac{1}{3} (y+4)^2 - \frac{1}{3} \cdot 16$$ Finally, add the constant term from the original equation to the expression: $$\frac{1}{3} (y+4)^2 - \frac{1}{3} \cdot 16 -\frac{5}{3}$$ Combine the constants: $$\frac{1}{3} (y+4)^2 - \frac{53}{3}$$ Now we have the equation in the desired form: \(x = a(y-k)^2 + h\) $$x = \frac{1}{3} (y+4)^2 - \frac{53}{3}$$ Where \(a = \frac{1}{3}\), \(k = -4\), and \(h = -\frac{53}{3}\).

Step 4: Graph the Function

To graph the function $$x = \frac{1}{3} (y+4)^2 - \frac{53}{3}$$, we notice that the graph is a parabola with the vertex at the point (-4, -53/3), opens horizontally to the right (since a > 0), and has a stretch factor of 1/3.

Key Concepts

Parabola

The term "parabola" represents a distinct shape formed when a plane intersects a right circular cone. This result is a U-shaped curve that is symmetric, meaning both sides reflect each other perfectly across a line of symmetry, known as the axis of symmetry.
In quadratic equations, a parabola provides a visual representation of these equations. The parabola can be oriented vertically (opening upwards or downwards) or horizontally. In the context of this exercise, our emphasis is on a horizontally opening parabola, which is less common in algebraic studies but essential for conic sections.
Key characteristics of a parabola in any orientation include:

• The vertex, which is the highest or lowest point depending on the parabola's direction of opening.
• The axis of symmetry, a line that vertically or horizontally divides the parabola into two mirror images.
• The direction of opening, which can be up, down, left, or right, determined by the sign and value of the leading coefficient in its equation.

Recognizing these features in an equation's graphical form helps to understand its behavior and the effects of different algebraic transformations.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, typically written in the form of: \[ ax^2 + bx + c = 0 \]However, when dealing with conic sections like parabolas, we often encounter quadratics in the format of another variable such as y, which helps in deriving their vertex form.
In this problem, we started with an equation involving y, \[ x = \frac{1}{3} y^2 + \frac{8}{3} y - \frac{5}{3} \] The quadratic term in a quadratic equation is crucial as it defines the parabola's width and direction.
Solving quadratic equations can involve various methods, including:

• Factoring
• The quadratic formula
• Completing the square

Completing the square is particularly beneficial for converting a standard quadratic equation into vertex form, making it easier to graph and understand particularly when working with conic sections.

Vertex Form

The vertex form of a quadratic equation highlights key components of the parabola, especially the vertex. Usually expressed as:
\[ x = a(y-k)^2 + h \] This form helps to recognize easily the parabola's vertex as (h, k).
In our example, by completing the square, we transformed the given equation into vertex form: \[ x = \frac{1}{3} (y+4)^2 - \frac{53}{3} \]
The constants \(a\), \(h\), and \(k\) play significant roles:

• \(a\): Indicates the parabola's stretch and direction. If \(a\) is positive, the parabola opens to the right; if negative, to the left.
• \(h\) and \(k\): Represent the vertex. In this case, the vertex is located at (-4, -53/3).

This form simplifies understanding and graphing parabolic equations because it directly provides the vertex point and the parabola's direction and stretch.