Question

Write the equation of a possible rational function with each set of characteristics. a) vertical asymptotes at \(x=\pm 5\) and \(x\) -intercepts of -10 and 4 b) a vertical asymptote at \(x=-4,\) a point of discontinuity at \(\left(-\frac{11}{2}, 9\right)\) and an \(x\) -intercept of 8 c) a point of discontinuity at \(\left(-2, \frac{1}{5}\right)\) a vertical asymptote at \(x=3,\) and an \(x\) -intercept of -1 d) vertical asymptotes at \(x=3\) and \(x=\frac{6}{7},\) and \(x\) -intercepts of \(-\frac{1}{4}\) and 0

Short answer

a) \(f(x) = \frac{(x+10)(x-4)}{(x-5)(x+5)}\) b) \(f(x) = \frac{(x-8)(x+\frac{11}{2})}{(x+4)(x+\frac{11}{2})}\) c) \(f(x) = \frac{(x+1)(x+2)}{(x-3)(x+2)}\) d) \(f(x) = \frac{x(4x+1)}{(x-3)(7x-6)}\)

Step-by-step solution

Identify Vertical Asymptotes and x-Intercepts for Part (a)

The rational function will have vertical asymptotes at the values of x that make the denominator zero. Similarly, the x-intercepts are the values of x that make the numerator zero. For part (a), vertical asymptotes are at \(x=\pm 5\), so the denominator includes the factors \((x-5)(x+5)\). The x-intercepts are at \(x=-10\) and \(x=4\), so the numerator includes the factors \((x+10)(x-4)\).

Construct the Rational Function for Part (a)

The rational function can be written as: \[f(x) = \frac{A(x+10)(x-4)}{(x-5)(x+5)}\] where A is any constant. For simplicity, let’s choose A = 1: \[f(x) = \frac{(x+10)(x-4)}{(x-5)(x+5)}\]

Identify Vertical Asymptote, Discontinuity, and x-Intercept for Part (b)

The vertical asymptote is at \(x=-4\), so we include the factor \((x+4)\) in the denominator. There is a discontinuity at \(x = -\frac{11}{2}\), meaning \((x+\frac{11}{2})\) cancels out in both the numerator and denominator. The x-intercept is 8, indicating the numerator includes the factor \((x-8)\).

Construct the Rational Function for Part (b)

The rational function considering discontinuity at \(x = -\frac{11}{2}\), vertical asymptote at \(x=-4\), and x-intercept at 8 can be written as: \[f(x) = \frac{A(x-8)(x+\frac{11}{2})}{(x+4)(x+\frac{11}{2})}\] For simplicity, let’s choose A = 1: \[f(x) = \frac{(x-8)(x+\frac{11}{2})}{(x+4)(x+\frac{11}{2})}\]

Identify Discontinuity, Vertical Asymptote, and x-Intercept for Part (c)

The point of discontinuity at \(x=-2\) means we have the factor \((x+2)\) in both numerator and denominator. The vertical asymptote at \(x=3\) involves a factor \((x-3)\) in the denominator. The x-intercept at \(x=-1\) includes the factor \((x+1)\) in the numerator.

Construct the Rational Function for Part (c)

The rational function considering the given properties is: \[f(x) = \frac{A(x+1)(x+2)}{(x-3)(x+2)}\] For simplicity, let’s choose A=1: \[f(x) = \frac{(x+1)(x+2)}{(x-3)(x+2)}\]

Identify Vertical Asymptotes and x-Intercepts for Part (d)

The vertical asymptotes are at \(x=3\) and \(x=\frac{6}{7}\), so the factors \((x-3)\) and \((7x-6)\) are in the denominator. The x-intercepts at \(x=-\frac{1}{4}\) and \(x=0\) imply the numerator includes the factors \((4x+1)\) and \(x\).

Construct the Rational Function for Part (d)

The rational function can be written as: \[f(x) = \frac{A x (4x+1)}{(x-3)(7x-6)}\] For simplicity, let’s choose A=1: \[f(x) = \frac{x(4x+1)}{(x-3)(7x-6)}\]

Key concepts

Vertical Asymptotes

A vertical asymptote occurs where a rational function's denominator is zero, making the function undefined at those points. To determine the vertical asymptotes for a given rational function, identify the values of x that make the denominator zero while ensuring these points do not also make the numerator zero. For example, if you have asymptotes at \(x = \pm 5\), then the factors \((x-5)\) and \((x+5)\) must be in the denominator, ensuring the function approaches infinity as x approaches these values.

x-Intercepts

The x-intercepts of a rational function are the points where the graph crosses the x-axis. These occur where the numerator is zero while the denominator is not zero. To find the x-intercepts, set the numerator equal to zero and solve for x. For instance, if the x-intercepts are at \(x = -10\) and \(x = 4\), the numerator must include the factors \((x + 10)\) and \((x - 4)\), causing the function's value to be zero at these points.

Points of Discontinuity

Points of discontinuity in a rational function occur when a factor is common to both the numerator and the denominator and cancels itself out, leaving a hole in the graph at that x-value. To identify these points, determine the common factors of the numerator and the denominator and solve for the x-value. For instance, given a discontinuity at \(x = -2\), both the numerator and the denominator should include the factor \((x + 2)\), resulting in a removable discontinuity where the function is undefined at this point, but the limit exists.

Constructing Rational Functions

To construct a rational function with specified characteristics like vertical asymptotes, x-intercepts, and points of discontinuity, start by identifying the necessary factors for the numerator and denominator based on the given properties. For example, if a function must have a vertical asymptote at \(x = 3\), a point of discontinuity at \(x = -2\), and an x-intercept at \(x = -1\), you can form the rational function by including the respective factors: \((x+1)\) for the numerator and \((x-3)\) and \((x+2)\) for the denominator. Be sure to arrange these factors appropriately based on standard algebraic rules and whether there are cancellations.