Problem 59 - College Algebra with Corequisite Support

Question

For the following exercises, express the equation for the hyperbola as two functions, with \(y\) as a function of \(x\). Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. \(-4 x^{2}-16 x+y^{2}-2 y-19=0\)

Step-by-Step Solution

Answer

Functions: \(y = 1 \pm 2\sqrt{(x+2)^2 + 1}\).

Step 1: Reorganize the Equation

Start by grouping the terms involving the same variable together: \[-4x^2 - 16x + y^2 - 2y = 19.\]

Step 2: Complete the Square for x-terms

Focus on completing the square for the \(x\)-terms. Factor out the negative and then complete the square:\[-4(x^2 + 4x) = -4((x+2)^2 - 4) = -4(x+2)^2 + 16.\]

Step 3: Complete the Square for y-terms

Now complete the square for the \(y\)-terms:\[y^2 - 2y = (y-1)^2 - 1.\]

Step 4: Substitute Squares in the Equation

Substitute these results back into the original equation:\[-4(x+2)^2 + 16 + (y-1)^2 - 1 = 19.\]

Step 5: Simplify the Constant Terms

Combine the constant terms on the right side to simplify:\[-4(x+2)^2 + (y-1)^2 = 4.\]

Step 6: Rewrite Equation in Standard Hyperbola Form

Divide the entire equation by 4 to convert into the standard form:\[\frac{(y-1)^2}{4} - \frac{(x+2)^2}{1} = 1.\]

Step 7: Solve for y as a Function of x

Solve the equation to express \(y\) as two functions of \(x\):\[(y-1)^2 = 4\frac{(x+2)^2}{1} + 4.\]\[y = 1 \pm \sqrt{4(x+2)^2 + 4}.\]

Step 8: Simplify Further If Possible

Simplify the expression further if possible and express:\[y = 1 \pm 2\sqrt{(x+2)^2 + 1}.\]

Step 9: Graphing Considerations

Use a graphing calculator to sketch both functions \(y_1 = 1 + 2\sqrt{(x+2)^2 + 1}\) and \(y_2 = 1 - 2\sqrt{(x+2)^2 + 1}\) on the same axes.

Key Concepts

Completing the Square

Completing the square is a key algebraic technique crucial for transforming quadratic expressions. When applied to hyperbolas, it allows us to rewrite the equation in a more manageable form. Here’s how the method works: first, take any quadratic expression, such as \(x^2 + 4x\). We want to add and subtract a particular constant that transforms it into a complete square.

• Identify the coefficient of the linear term. In this case, it is 4.
• Divide this coefficient by 2 and square the result. This gives \(\left(\frac{4}{2}\right)^2 = 4\).
• Add and subtract this square number within the expression: \(x^2 + 4x + 4 - 4\)
• Factor the perfect square trinomial: \(x^2 + 4x + 4 = (x+2)^2\).

In our problem, completing the square was applied to both \(x\) and \(y\) variables. This made it possible to restructure the hyperbola’s equation, facilitating the conversion to standard form.

Standard Form of Hyperbola

The standard form of a hyperbola is essential for recognizing the shape and direction of the curve. For hyperbolas centered at some point \( (h, k) \), the standard form of the equation is written as:
\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]\[ \text{or} \quad \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \]
The arrangement depends on whether the transverse axis is horizontal or vertical. In the presented equation, it initially had mixed terms. By completing the square, we transform it into:
\[ \frac{(y-1)^2}{4} - \frac{(x+2)^2}{1} = 1 \]
This equation depicts a hyperbola with its transverse axis oriented vertically. The values \(h = -2\) and \(k = 1\) mark the center. Parameters \(a = 2\) and \(b = 1\) represent half the transverse and conjugate axes, respectively, characterizing the shape and spread of the hyperbola on the graph.

Function of a Variable

Understanding how a hyperbola can be expressed as a function of a variable involves algebraic manipulation. From its standard form, solve for one variable in terms of the other. For this exercise, we needed \(y\) as a function of \(x\):
Starting with:\[ \frac{(y-1)^2}{4} - \frac{(x+2)^2}{1} = 1 \]
Rearrange to isolate \( (y - 1)^2 \):\[ (y-1)^2 = 4(x+2)^2 + 4 \]
To express \(y\) clearly:\[ y = 1 \pm 2\sqrt{(x+2)^2 + 1} \]
Now \(y\) is expressed as two functions, \(y = 1 + 2\sqrt{(x+2)^2 + 1} \) and \(y = 1 - 2\sqrt{(x+2)^2 + 1}\). The "plus-minus" symbol indicates that hyperbolas result in two separate curves, reflecting the function's nature to produce two values of \(y\) for a given \(x\).

Graphing with Calculator

Graphing the equations of a hyperbola with a calculator is not only insightful but also intuitive. Let's break down the process:

• Input the equations separately into the calculator. Begin with \(y_1 = 1 + 2\sqrt{(x+2)^2 + 1}\).
• Next, enter \(y_2 = 1 - 2\sqrt{(x+2)^2 + 1}\).
• Set an appropriate window range to ensure both branches of the hyperbola are visible.
• Examine the graph, noticing both upward and downward curves, courtesy of the \(\pm\) in the equations.

Graphing on a calculator highlights the hyperbola's reflective symmetry about its center. It also emphasizes the transformation of the hyperbola from algebraic form to a tangible curve. Remember, visualization bolsters comprehension, making abstract concepts appear more concrete.