Question

Finding vertical asymptotes Find all vertical asymptotes \(x=\) a of the follow ing functions. For each value of \(a\), determine \(\lim _{x \rightarrow a^{+}} f(x), \lim _{x \rightarrow a^{-}} f(x),\) and \(\lim _{x \rightarrow a} f(x)\) $$f(x)=\frac{x^{2}-9 x+14}{x^{2}-5 x+6}$$

Short answer

Answer: The vertical asymptotes of the given function are at x = 2 and x = 3. For x = 2, the limits are \(\lim_{x \rightarrow 2^{+}} f(x) = -\infty\), \(\lim_{x \rightarrow 2^{-}} f(x) = \infty\), and \(\lim_{x \rightarrow 2} f(x)\) is undefined. For x = 3, the limits are \(\lim_{x \rightarrow 3^{+}} f(x) = \infty\), \(\lim_{x \rightarrow 3^{-}} f(x) = -\infty\), and \(\lim_{x \rightarrow 3} f(x)\) is undefined.

Step-by-step solution

Find the values of x for which the denominator is zero.

To find the values of x for which the denominator is zero, we need to solve the equation: $$x^{2}-5 x+6=0$$ This is a quadratic equation, so we can use the quadratic formula or factoring to solve for x. Factoring the equation, we get: $$(x-2)(x-3)=0$$ This gives us two possible values for x: $$x=2, x=3$$

Calculate the limits as x approaches 2 and 3.

We will now calculate the following limits for each of the values of x found in Step 1: 1. \(\lim_{x \rightarrow a^{+}} f(x)\) 2. \(\lim_{x \rightarrow a^{-}} f(x)\) 3. \(\lim_{x \rightarrow a} f(x)\) Starting with x=2: 1. \(\lim_{x \rightarrow 2^{+}} \frac{x^{2}-9 x+14}{x^{2}-5 x+6}\) 2. \(\lim_{x \rightarrow 2^{-}} \frac{x^{2}-9 x+14}{x^{2}-5 x+6}\) 3. \(\lim_{x \rightarrow 2} \frac{x^{2}-9 x+14}{x^{2}-5 x+6}\) Notice that the function can be simplified by factoring the numerator as well: $$f(x)=\frac{(x-2)(x-7)}{(x-2)(x-3)}$$ For all x ≠ 2, 3, the factors (x-2) in the numerator and denominator will cancel out. So, the simplified function is given by: $$f(x)=\frac{x-7}{x-3}$$ for x ≠ 2, 3 Now, we can calculate the limits using the simplified function: 1. \(\lim_{x \rightarrow 2^{+}} \frac{x-7}{x-3} = -\infty\) 2. \(\lim_{x \rightarrow 2^{-}} \frac{x-7}{x-3} = \infty\) 3. \(\lim_{x \rightarrow 2} \frac{x-7}{x-3}\) is undefined since the limits from the left and right are different. Therefore, x=2 is a vertical asymptote. Now, we'll move on to x=3: 1. \(\lim_{x \rightarrow 3^{+}} \frac{x-7}{x-3} = \infty\) 2. \(\lim_{x \rightarrow 3^{-}} \frac{x-7}{x-3} = -\infty\) 3. \(\lim_{x \rightarrow 3} \frac{x-7}{x-3}\) is undefined since the limits from the left and right are different. Therefore, x=3 is also a vertical asymptote.

State the vertical asymptotes and their associated limits.

Based on the above calculations, the vertical asymptotes for the given function are: 1. Vertical asymptote at x=2: - \(\lim_{x \rightarrow 2^{+}} f(x) = -\infty\) - \(\lim_{x \rightarrow 2^{-}} f(x) = \infty\) - \(\lim_{x \rightarrow 2} f(x)\) is undefined. 2. Vertical asymptote at x=3: - \(\lim_{x \rightarrow 3^{+}} f(x) = \infty\) - \(\lim_{x \rightarrow 3^{-}} f(x) = -\infty\) - \(\lim_{x \rightarrow 3} f(x)\) is undefined.

Key concepts

Limits of Functions

In calculus, limits of functions are fundamental to understanding the behavior of functions as they approach a specific point. This concept is critical in identifying points of discontinuity, such as vertical asymptotes, and determining the behavior of a function at those points. For example, to find the limit of a function as x approaches a certain value, we look at what value the function tends toward as x gets arbitrarily close to that number from both the left (x-approaches from the negative side) and the right (x-approaches from the positive side).

When both of these one-sided limits are different, or when the function grows without bound, the limit at that point is undefined. In our exercise, the limits as x approaches the points where the denominator is zero were necessary to find the vertical asymptotes of the function.

Factoring Quadratic Equations

Factoring quadratic equations is a method used to solve equations of the form ax² + bx + c = 0. To factor a quadratic equation means to break it down into simpler binomials that, when multiplied together, give back the original quadratic. When factoring, we look for two numbers that multiply to give the constant term c, and add to give the coefficient b.

In our example, the quadratic equation x² - 5x + 6 = 0 was successfully factored into (x - 2)(x - 3) = 0. Hence, by setting each factor equal to zero, we found the potential x-values where the function's denominator becomes zero, indicating possible locations for vertical asymptotes.

Undefined Limits

Undefined limits occur when a function does not approach a single, finite value as x approaches a specified point. In the context of vertical asymptotes, limits are typically undefined because the function's value either approaches infinity (∞) or negative infinity (-∞) as x approaches the asymptote.

In the exercise, we saw how the limits of f(x) as x approaches the values that made the denominator zero were undefined. This is a characteristic of vertical asymptotes, highlighting a point where the function's graph shoots up or down infinitely, never touching that x-value but still influencing the shape of the graph near it.

Solving Quadratic Equations

Solving quadratic equations is a crucial step in finding the zeros of the function, which can also point us to potential vertical asymptotes for rational functions. We often employ methods such as factoring, using the quadratic formula, completing the square, or graphing. Factoring, as seen in our example, is often the quickest method, but when factoring is not possible, the quadratic formula is a reliable alternative.

It's also worth mentioning that each solution to a quadratic equation can be associated with a specific point on the function's graph where the graph intersects the x-axis, provided the quadratic is part of the function's numerator. However, when the quadratic is in the denominator, as in our example, the zeros of this quadratic don't intersect the x-axis but instead, indicate where the function is undefined, and thus, where the vertical asymptotes occur.