Problem 47 - College Algebra with Modeling and Visualization

Question

Sketch a graph of the parabola. $$ (y-1)^{2}=-(x+1) $$

Step-by-Step Solution

Answer

The parabola opens to the left, with vertex (-1, 1) and focus at (-1.25, 1).

Step 1: Rewrite the Equation

Start by rewriting the given equation \((y-1)^{2}=-(x+1)\) into the standard form for a parabola. This equation is already close to the vertex form. Notice it's in the form \((y-k)^2 = 4p(x-h)\) where \(h\) is -1, \(k\) is 1, and \(4p = -1\).

Step 2: Identify the Vertex

The vertex form of a parabola \((y-k)^2 = 4p(x-h)\) readily reveals the vertex at \((h, k)\). For our equation, the vertex is at \((-1, 1)\).

Step 3: Determine the Parabola's Orientation

Since the equation involves \((y-k)^2\), the parabola's axis is horizontal. The negative sign of \(4p\) indicates that the parabola opens to the left.

Step 4: Find the Value of p

Solve for \(p\): Given \(4p = -1\), we divide both sides by 4 to find \(p = -\frac{1}{4}\). This tells us how far the focus and directrix are from the vertex, horizontally.

Step 5: Plot the Focus and Directrix

Since \(p = -\frac{1}{4}\), the focus is \(\frac{1}{4}\) unit to the left of the vertex at \(-\frac{5}{4}, 1\). The directrix is a vertical line \(\frac{1}{4}\) unit to the right of the vertex, at \(x = -\frac{3}{4}\).

Step 6: Sketch the Parabola

Draw the vertex at \((-1, 1)\), the focus at \(-\frac{5}{4}, 1\), and the directrix at \(x = -\frac{3}{4}\). Sketch the parabola opening to the left, ensuring it is equidistant from the focus to the directrix.

Key Concepts

Vertex Form of a Parabola

When dealing with parabolas, one of the most helpful expressions to use is the vertex form of a parabola. This form is given as \[(y-k)^2 = 4p(x-h)\]The vertex form is particularly useful because it instantly provides the vertex of the parabola, located at \((h, k)\).In the context of parabolas that open sideways, like the one in our original exercise, the equation involves \((y-k)^2\),indicating a horizontal axis.

• In the equation \((y-1)^2=-(x+1)\),we recognize \(h = -1\)and \(k = 1\).
• The vertex is therefore located at these coordinates, \((-1, 1)\).

Understanding the vertex form allows for easy plotting of critical attributes of the parabola, such as its vertex and orientation. This aids in accurately sketching and characterizing the parabola's shape on a graph.

Focus and Directrix

The focus and directrix are two fundamental features of a parabola. They provide insight into the set of points that make up the curve. The distance from any point on the parabola to the focus is equal to the distance from that point to the directrix.

• The focus lies at \((-\frac{5}{4}, 1)\), which is \(\frac{1}{4}\) unit to the left of the vertex in the exercise problem.
• The directrix is a vertical line at \(x = -\frac{3}{4}\), \(\frac{1}{4}\) unit to the right of the vertex.

A negative value of \(p\)means the parabola opens in the opposite direction along the x-axis (to the left in this case). Calculating the position of the focus and directrix helps in understanding and visualizing how the parabola is oriented as well as ensuring the graph is plotted correctly.

Graphing Conic Sections

Graphing conic sections such as parabolas involves recognizing their forms and properties. In the exercise, given the equation as \((y-1)^2 = -(x+1)\), the parabola opens to the left, indicative of its \((y-k)^2\) structure and negative \(4p\) value.

• To graph, plot the vertex at \((-1, 1)\), clearly showing the orientation.
• Draw and label the focus and directrix appropriately to serve as guides.
• Sketch the curve equidistant from these two components, ensuring the parabola mirrors around the vertex.

When drafting conic sections, maintaining accuracy in plotting its defining elements ensures that you capture its true geometric shape. Utilizing the characteristics derived from vertex form, focus, and directrix is key in creating accurate and informative graphs.