Problem 12 - Precalculus with Calculus Previews

Question

Find the vertical and horizontal asymptotes for the graph of the given rational function. Find \(x\) - and \(y\) -intercepts of the graph. Sketch the graph of \(f\). $$ f(x)=\frac{1}{(x-1)^{2}} $$

Step-by-Step Solution

Answer

Vertical asymptote at \(x = 1\), horizontal asymptote at \(y = 0\); no x-intercepts, y-intercept at \((0, 1)\).

Step 1: Find Vertical Asymptotes

To find the vertical asymptotes, set the denominator equal to zero and solve for \(x\). The denominator of \(f(x)\) is \((x-1)^2\). Setting this equal to zero gives \((x-1)^2 = 0\), which simplifies to \(x = 1\). Therefore, the graph has a vertical asymptote at \(x = 1\).

Step 2: Determine Horizontal Asymptotes

Examine the degrees of the polynomial in the numerator and the denominator. Here, the numerator has degree 0 (constant 1) and the denominator has degree 2. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\).

Step 3: Find x-intercepts

The x-intercepts occur where \(f(x) = 0\). Since the only way for a fraction to be zero is if the numerator is zero, and our numerator is 1 (a non-zero constant), there are no x-intercepts for this function.

Step 4: Find y-intercepts

The y-intercept occurs where \(x = 0\). Plugging \(x = 0\) into \(f(x)\), we get \(f(0) = \frac{1}{(0-1)^2} = 1\). Thus, the y-intercept is at \((0, 1)\).

Step 5: Sketch the Graph

Use the asymptotes and intercepts found to sketch the graph. The graph approaches \(x = 1\) as a vertical asymptote and \(y = 0\) as a horizontal asymptote. It passes through the point \((0, 1)\). The graph will be positive on both sides of \(x = 1\) as the square in the denominator ensures this. As \(x\) moves further from \(1\), \(f(x)\) approaches 0.

Key Concepts

Vertical Asymptotes

Vertical asymptotes are an important aspect of understanding the behavior of rational functions. These occur at points where the function is undefined. For the function given, \(f(x)=\frac{1}{(x-1)^{2}}\), the key to finding vertical asymptotes is in the denominator. Set the denominator equal to zero to find where the function flips to infinity, making the function undefined.
In this case, solving \((x-1)^2 = 0\) gives us \(x = 1\). This is where the vertical asymptote is located. Remember, at \(x = 1\), the function does not exist. As \(x\) approaches \(1\), the function values will shoot up towards positive or negative infinity, depending on the direction from which \(x\) approaches.

• Vertical asymptote at \(x = 1\) is a result of the denominator becoming zero.
• Graphs typically never cross vertical asymptotes, but they can touch them in rare cases not applicable here.
• It's crucial to always verify potential asymptotes by solving the equation in its simplified form.

Horizontal Asymptotes

Horizontal asymptotes give insight into what happens to the function as \(x\) moves towards infinity or negative infinity. They depend on the degrees of the polynomials in the numerator and the denominator.
For the function \(f(x)=\frac{1}{(x-1)^{2}}\), the numerator has a degree of 0 since it is a constant (1), and the denominator has a degree of 2. When the degree of the numerator is less than that of the denominator, the horizontal asymptote is \(y = 0\). This indicates that as \(x\) grows large (in either the positive or negative direction), the function \(f(x)\) approaches 0.

• Degrees comparison: Numerator is degree 0; Denominator is degree 2.
• If numerator degree < denominator degree, horizontal asymptote is \(y = 0\).
• Horizontal asymptotes describe end-behavior of functions.