Problem 59 - College Algebra

Question

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

The functions are \(y = 1 \pm \sqrt{4(x + 2)^2 + 4}\).

Step 1: Rearrange the Equation

Start by rearranging the given equation: \(-4x^2 - 16x + y^2 - 2y - 19 = 0\).Move all the terms involving \(y\) to one side:\(y^2 - 2y = 4x^2 + 16x + 19\).

Step 2: Complete the Square for y

Complete the square for the \(y\) terms to form a perfect square trinomial.Factor and manipulate it:\(y^2 - 2y = (y - 1)^2 - 1\).Thus, the equation is:\((y - 1)^2 = 4x^2 + 16x + 20\).

Step 3: Simplify the x Terms

Complete the square for the \(x\) terms to simplify the expression further.\(4x^2 + 16x + 20\).Factor out the 4 from \(4x^2 + 16x\):\(4(x^2 + 4x) + 20\).Complete the square inside the parentheses:\((x^2 + 4x)to (x + 2)^2 - 4\).Therefore, it becomes:\(4((x + 2)^2 - 4) + 20 = 4(x + 2)^2 + 4\).The new equation is:\((y - 1)^2 = 4((x + 2)^2 + 1)\).

Step 4: Solve for y in Terms of x

Convert the equation to express \(y\) as a function of \(x\):\((y - 1)^2 = 4(x + 2)^2 + 4\).Take the square root of both sides:\(y - 1 = \pm\sqrt{4(x + 2)^2 + 4}\).So, \(y = 1 \pm\sqrt{4(x + 2)^2 + 4}\), giving us two functions:\(y = 1 + \sqrt{4(x + 2)^2 + 4}\) and\(y = 1 - \sqrt{4(x + 2)^2 + 4}\).

Step 5: Sketch the Graphs Using a Graphing Calculator

Using a graphing calculator, plot the equations:\(y = 1 + \sqrt{4(x + 2)^2 + 4}\) and \(y = 1 - \sqrt{4(x + 2)^2 + 4}\).These functions represent the upper and lower halves of the hyperbola. Observe the symmetry along the horizontal axis.

Key Concepts

Completing the Square

Completing the square is a technique used to transform a quadratic expression into a perfect square trinomial. This is particularly useful in graphing conic sections, such as hyperbolas, by allowing us to rewrite equations in a form that reveals their structure. In the original exercise, we completed the square for both the x and y terms.

For the y term, the equation \( y^2 - 2y \) is transformed as follows:

• Initiate completing the square by taking half of the coefficient of y, which is \(-2\), square it to get \(1\), and then form \((y - 1)^2 - 1\).

For the x terms, in \( x^2 + 4x \):

• Take half of the coefficient of x, \(4\), square it, and add the result \(4\) inside the parentheses to form \((x + 2)^2 - 4\).

Completing the square allows us to easily convert the quadratic terms into a format suitable for solving as functions or graphing. Once we have \((y - 1)^2 = 4((x + 2)^2 + 1)\), it becomes much easier to manage and graph.

Functions of a Variable

In mathematics, expressing an equation with one variable as a function of another variable simplifies our understanding of relationships between variables. With hyperbolas, we often want to express y in terms of x. This is because graphing equations in terms of one variable can help in visualizing the curve they form.

In our exercise, we derived two expressions for y in terms of x after completing the square. These were:

• \( y = 1 + \sqrt{4(x + 2)^2 + 4} \)
• \( y = 1 - \sqrt{4(x + 2)^2 + 4} \)

The plus-minus symbol (±) indicates that for each x-value, there are two corresponding y-values. This signifies the symmetric nature of hyperbolas that have both upper and lower branches or halves. Such functional expressions give us an explicit means to plot points on the hyperbola's path.

Graphing Calculator Usage

A graphing calculator is an essential tool when working with complex functions and their graphs. It assists in visualizing mathematical equations and solving problems interactively. By inputting the functions derived from our hyperbola, the graphing calculator can efficiently draw their respective curves.

For this exercise, after obtaining the y in terms of x, we plugged the equations into the calculator:

• \(y = 1 + \sqrt{4(x + 2)^2 + 4}\)
• \(y = 1 - \sqrt{4(x + 2)^2 + 4}\)

The graph shows two curves that are symmetrical along the horizontal axis, illustrating the top and bottom halves of the hyperbola. Practicing how to input such functions into the calculator and interpreting the output enhances both our understanding of hyperbolas and graphing capabilities. It's a wonderful way to see mathematical concepts come to life.