Problem 113 - A Graphical Approach to Precalculus with Limits

Question

Graph each parabola by hand, and check using a graphing calculator. Give the vertex, axis, domain, and range. $$2 x=y^{2}-2 y+9$$

Step-by-Step Solution

Answer

Vertex: (4, 1), Axis: y = 1, Domain: (-∞, ∞), Range: (-∞, ∞).

Step 1: Rewrite Equation in Vertex Form

The original equation is \( 2x = y^2 - 2y + 9 \). To convert it to a usable form for graphing, we should first complete the square for the \( y \) terms. We have the equation:\[ y^2 - 2y = (y^2 - 2y + 1) - 1 \]This can be rewritten as:\[ (y - 1)^2 - 1 \]Substitute this back into the equation:\[ 2x = (y - 1)^2 - 1 + 9 \]Which simplifies to:\[ 2x = (y - 1)^2 + 8 \]

Step 2: Solve for x to Identify the Vertex

Now, let's solve for \( x \):\[ x = \frac{1}{2}((y - 1)^2 + 8) \]This equation is useful as it tells us that the parabola opens horizontally (sideways), which means it is not a function in terms of \( y \) directly. This form tells us the vertex is at \( (4, 1) \).

Step 3: Determine the Axis of Symmetry

For this sideways parabola, the axis of symmetry is the horizontal line at \( y = 1 \), which is the \( y \)-coordinate of the vertex.

Step 4: Define the Domain

Since \( x = \frac{1}{2}((y - 1)^2 + 8) \), the \( x \)-values are unrestricted. The domain of the parabola is all real numbers, \( (-\infty, \infty) \).

Step 5: Define the Range

The range of this parabola is determined by \( y \)-values since the parabola opens horizontally. The range extends from \(-\infty\) to \(\infty\), as \((y - 1)^2\) can take any non-negative value.

Step 6: Graph the Parabola

To graph by hand, plot the vertex at \((4, 1)\). Since the parabola opens along the x-axis, draw a symmetric curve around \( y = 1 \), opening to the right and left. Verify with a graphing calculator to check that the graph aligns with expectations around these points.

Key Concepts

Vertex Form

The vertex form of a parabola is essential for graphing, as it directly shows the vertex of the parabola. It's generally expressed as:

• For a parabola opening upwards or downward: \[ y = a(x-h)^2 + k \] Here, \((h, k)\) is the vertex, and \(a\) indicates the direction and width of the parabola.
• For a sideways opening parabola like in our exercise: \[ x = a(y-k)^2 + h \]Here, \((h, k)\) is still the vertex, but the parabola opens horizontally.

In our example, once we complete the square, we converted the given equation to: \[ x = \frac{1}{2}((y - 1)^2 + 8) \]This reveals that the vertex is \((4, 1)\). The equation was modified to this form so we can easily identify the vertex and direction of the parabola.

Completing the Square

Completing the square is a technique used to rewrite a quadratic equation so it's easier to manipulate or graph. This method involves altering a quadratic expression into a perfect square trinomial plus or minus a constant.For our exercise:

• We started with \(y^2 - 2y\) and added and subtracted \(1\) to complete the square: \[ y^2 - 2y = (y^2 - 2y + 1) - 1 = (y - 1)^2 - 1 \]
• Substituting back into the original equation, we obtain: \[ 2x = (y - 1)^2 - 1 + 9 \]which simplifies to: \[ 2x = (y - 1)^2 + 8 \]

This method helped transform the equation into a form that makes finding the vertex straightforward, aiding both calculation and visualization in graphing.

Domain and Range

The domain and range of a parabola can tell us which values the variables are allowed to take.- **Domain** refers to all possible \(x\)-values the function (or relation) can handle.

• For our sideways parabola, there are no restrictions on \(x\), meaning the domain is all real numbers: \[ (-\infty, \infty) \]

- **Range** specifies all possible \(y\)-values.

• For our sideways opening parabola, the \(y\) values can also take any value, hence the range is: \[ (-\infty, \infty) \]

Understanding the domain and range is crucial for knowing the "extent" of your graph, especially if you wish to plot it by hand.

Axis of Symmetry

The axis of symmetry for a parabola is the imaginary line that perfectly divides the parabola into two mirror-image halves. Understanding where this line lies helps to accurately graph a parabola.For vertical parabolas, the axis of symmetry is vertical:

• \( x = h \) for the equation \( y = a(x-h)^2 + k \)

For horizontal parabolas, like in our example:

• The axis of symmetry is horizontal: \( y = k \) for the equation \( x = a(y-k)^2 + h \)
• In the exercise, the axis of symmetry is at \( y = 1 \), which is the \( y \)-coordinate of our vertex \((4, 1)\).

The axis of symmetry guides us in plotting the vertex and sketching the parabola accurately on either side of this line.