Problem 14 - BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Question

In Exercises 11–18, graph the function. State the domain and range. $$ y=\frac{1}{x+2} $$

Step-by-Step Solution

Answer

The graph of the function \( y = \frac{1}{x+2} \) is a hyperbola with a vertical asymptote at \(x = -2\). The function goes through the point (0, 1/2). The domain of the function is all real numbers except -2, and the range is all real numbers.

Step 1: Find the vertical asymptote

To find the vertical asymptote, we set the denominator equal to zero and solve for x. So, \(x + 2 = 0 \) which gives \(x = -2\). This means the function is undefined at \(x = -2\).

Step 2: Graph the function

Now we have to plot a few points and draw the hyperbola. From the equation, we know that as \(x\) approaches to -2 from either direction, \(y\) approaches infinity. Therefore, our function will have branches on both sides of the line \(x = -2\). Another point we can get from the function is when \(x = 0\), \(y = \frac{1}{2}\), so it will pass through the point (0,1/2).

Step 3: State the domain and range

The set of all possible x-values is called the domain. Since \(x\) cannot be -2 (our vertical asymptote), our domain is all real numbers except -2. So, the domain is \(x≠-2\). The function is defined for all y (y can take any real number value) and hence the range is all real numbers.

Key Concepts

Vertical Asymptote

A vertical asymptote is like an invisible barrier on a graph. It's where the function "blows up" or becomes undefined as the values of a given equation approach it. For the function \(y = \frac{1}{x + 2}\), we find the vertical asymptote by setting the denominator to zero:

• \(x + 2 = 0\) leads to \(x = -2\).

This tells us the function is undefined at \(x = -2\). As \(x\) gets closer to \(-2\) from either side, \(y\) shoots to positive or negative infinity. That's why, when graphing the function, you'll notice that the curve gets nearer and nearer to the line \(x = -2\), but never actually touches it. This line marks the vertical asymptote.

Domain and Range

The domain of a function includes all the possible input values (x-values) you can use without breaking the math. With the function \(y = \frac{1}{x + 2}\), our focus is on the denominator where the function becomes undefined.

• The only value that makes the denominator zero and thus undefined is \(x = -2\).

So the domain of the function is all real numbers except \(-2\). Visually, this means our graph will not cover \(x = -2\).

Now, let’s talk about the range, which consists of all the possible output values (y-values) the function can take. In this instance, since the graph approaches infinity positively or negatively but never truly hits a maximum or minimum value due to the asymptote, the range is all real numbers.

This means the line can reach as high or dip as low as it wants without restrictions, aside from infinity.

Undefined Values in Functions

When we talk about "undefined values" in the context of functions, we're almost always referring to points where the function

• stops making sense,
• typically due to division by zero.

For the function \(y = \frac{1}{x + 2}\), the value \(x = -2\) makes the denominator zero, leading to a division by zero that is undefined. This undefinable moment creates a gap or break in the graph at \(x = -2\).

Really, it's these undefined values that hint at the presence of vertical asymptotes where the function dips into theoretical math territory. We must always find and note these values because they determine the limits of the domain.