Question

In Exercises \(27-34,\) (a) find the vertical asymptotes of the graph of \(f(x) .(\) b) Describe the behavior of \(f(x)\) to the left and right of each vertical asymptote. $$f(x)=\frac{1-x}{2 x^{2}-5 x-3}$$

Short answer

The vertical asymptotes of the function \(f(x)=\frac{1-x}{2x^{2}-5x-3}\) are at \(x = 3\) and \(x = -0.5\). Close to these points, the function approaches \(-\infty\), from both left and right sides.

Step-by-step solution

Find the Vertical Asymptotes

To find the vertical asymptotes of a function, we need to determine the values of \(x\) for which the function is undefined, i.e., when the denominator equals zero. So, we start by equating the denominator \(2x^{2}-5x-3\) to zero and solve for \(x\). This gives us the quadratic equation: \(2x^{2}-5x-3 = 0\). Solving this quadratic equation we get \(x = frac{- b \pm \sqrt {b^{2}-4ac}}{2a}\) where \(a = 2, b = -5, c = -3\). Therefore \(x = frac{5 \pm \sqrt {(-5)^{2}-4*2*(-3)}}{2*2}\) yields \(x = 3, -0.5\). So vertical asymptotes are at \(x = 3, -0.5\).

Describe the Behavior of f(x) to the Left and Right of Each Vertical Asymptote

To describe the behavior of \(f(x)\) near the vertical asymptotes, analyze the function values at points slightly to the left and right of the asymptotes. Using limits, for \(x = 3\), we find \(\lim_{x \to 3^-} f(x)\) and \(\lim_{x \to 3^+} f(x)\), similarly for \(x = -0.5\), we find \(\lim_{x \to -0.5^-} f(x)\) and \(\lim_{x \to -0.5^+} f(x)\). After calculating, we can see the function approaches \(+\infty\) or \(-\infty\) at different sides of the asymptote. Near \(x = 3\), the function approaches \(-\infty\) from the left and \(-\infty\) from the right. Near \(x = -0.5\), the function approaches \(-\infty\) from the left and \(-\infty\) from the right.