Problem 36 - Beginning and Intermediate Algebra

Question

Rewrite each equation in the form \(x=a(y-k)^{2}+h\) by completing the square and graph it. $$x=y^{2}+4 y-6$$

Step-by-Step Solution

Answer

The given equation \(x=y^{2}+4 y-6\) can be rewritten as \(x=(y+2)^2 - 10\) after completing the square. In this form, the vertex of the parabola is at \((-2, -10)\) and the graph opens horizontally. To graph this equation, plot the vertex and several points around it, such as \((-9, -1)\), \((-9, -3)\), \((-6, 0)\), and \((-6, -4)\), then connect them to form the shape of a horizontal parabola.

Step 1: Rewrite the equation ##

First, we rewrite the given equation \(x = y^2 + 4y - 6\) so it is in the form \(x = y^2 + 4y + (4/2)^2 - (4/2)^2 - 6\). This way, we are creating a perfect square trinomial for the part \(y^2 + 4y + (4/2)^2\).

Step 2: Complete the square for the y terms##

Next, we work on completing the square for the \(y\) terms. The equation in Step 1 becomes: \(x = y^2 + 4y + (2)^2 - (2)^2 - 6\), which simplifies to \(x = y^2 + 4y + 4 - 4 - 6\). Now, we have a perfect square trinomial for the \(y\) terms: \(y^2 + 4y + 4\). This trinomial can be factored as \((y + 2)^2\).

Step 3: Rewrite the equation in the desired form##

Rewrite the equation in the form \(x = a(y-k)^2 + h\), now that we have the completed square for the \(y\) terms: \(x = (y + 2)^2 - 4 - 6\) So now, the equation is in the desired form: \(x = (y - (-2))^2 + (-10)\). Here, we have \(a = 1\), \(k = -2\), and \(h = -10\).

Step 4: Graph the equation##

To graph the equation \(x = (y - (-2))^2 - 10\), we can first identify the key features of the graph. The vertex of the parabola is \((-2, -10)\), which is the point where the graph changes direction. The value of \(a = 1\) tells us that the graph opens horizontally, since \(a\) is positive. The graph is a horizontal parabola with the vertex at the point \((-2, -10)\). First, plot the vertex at the point \((-2, -10)\). Since the parabola opens horizontally, move horizontally from the vertex and calculate the \(x\) values for corresponding \(y\) values. For example, if we move one unit up or down from the vertex, we get: For \(y = -1\): \(x = (-1-(-2))^2 - 10 = 1 - 10 = -9\) For \(y = -3\): \(x = (-3-(-2))^2 - 10 = 1 - 10 = -9\) For \(y = 0\): \(x = (0-(-2))^2 - 10 = 4 - 10 = -6\) For \(y = -4\): \(x = (-4-(-2))^2 - 10 = 4 - 10 = -6\) Plot these points on the graph, along with the vertex, and connect the points to form the shape of a horizontal parabola. The final graph will show a parabola with the vertex at \((-2, -10)\) that opens horizontally. Keep in mind that the graph will be symmetric with respect to the axis \(y = k\), which in this case is \(y = -2\).

Key Concepts

Completing the Square

The method called "completing the square" is a powerful algebraic tool to transform a quadratic expression into a perfect square trinomial. This makes it easier to work with,particularly when converting quadratic equations into vertex form.

Here's how it works: if you start with a quadratic equation like\[ x = y^2 + 4y - 6 \],our goal is to manipulate the terms \( y^2 + 4y \) into a perfect square trinomial.

We add and subtract the square of half the linear coefficient of \( y \) to achieve this.The linear coefficient here is 4, so half of 4 is 2, and 2 squared is 4.

• Add and subtract 4: \( y^2 + 4y = y^2 + 4y + 4 - 4 \).
• This becomes: \( (y + 2)^2 - 4 \).

Now the equation simplifies to\[ x = (y + 2)^2 - 10 \],which makes it much easier to analyze and graph.

Vertex Form

The vertex form of a parabola's equation is notably helpful for understanding the graph's shape and orientation.For horizontal parabolas, the vertex form is given by:\[ x = a(y-k)^2 + h \],where \( (h, k) \) is the vertex of the parabola.

After completing the square on the given equation, you convert it as:\[ x = (y + 2)^2 - 10 \],identifying the vertex \( (h, k) \) as \( (-10, -2) \).Here:

• \( a = 1 \), indicating the parabola's direction and width (if \( a > 0 \), it opens to the right).
• \( k = -2 \) is the y-value where the vertex occurs.
• \( h = -10 \) is the x-value of the vertex.

Understanding the vertex form allows one to effortlessly plot the parabola and determine its key features.

Horizontal Parabolas

Unlike the more familiar vertical parabolas, horizontal parabolas open in the right or left direction. Their standard vertex form is \( x = a(y - k)^2 + h \), as opposed to the vertical form \( y = a(x - h)^2 + k \).

For the given equation:\[ x = (y + 2)^2 - 10 \],we see it's a horizontal parabola.

• The vertex is at \( (-2, -10) \), indicating that the parabola is centered there.
• The positive \( a \) implies it opens to the right.

When graphing, you start by plotting the vertex.Then calculate additional points by selecting various \( y \) values and solving for \( x \).For example:

• At \( y = -4 \), you get \( x = -6 \).
• At \( y = 0 \), you also get \( x = -6 \).

This symmetry around the center line, \( y = -2 \), is a key characteristic of horizontal parabolas, making them easy to graph and understand.