Question

1. write a function that has exactly one vertical asymptote at x = - 1 and a horizontal asymptote of y = 1. it crosses the y - axis at y = 3 and has exactly one x intercept at x = - 3.

Explanation:

Step1: Determine the form for vertical asymptote

A rational - function has a vertical asymptote at \(x = a\) when the denominator is zero at \(x = a\). So, if there is a vertical asymptote at \(x=-1\), the denominator of the rational function can be \(x + 1\). Let the function be \(y=\frac{f(x)}{x + 1}\).

Step2: Determine the form for horizontal asymptote

For a rational function \(\frac{f(x)}{g(x)}\) where the degrees of \(f(x)\) and \(g(x)\) are the same, the horizontal asymptote \(y = k\) is given by the ratio of the leading - coefficients. Since the horizontal asymptote is \(y = 1\), the degrees of the numerator and denominator must be the same and the ratio of their leading coefficients must be 1. Let the numerator be \(ax + b\). So the function is \(y=\frac{ax + b}{x + 1}\).

Step3: Use the \(x\) - intercept

The \(x\) - intercept occurs when \(y = 0\). If the \(x\) - intercept is at \(x=-3\), then when \(y = 0\), \(ax + b=0\). Substituting \(x=-3\) into \(ax + b = 0\), we get \(-3a + b = 0\), so \(b = 3a\). Then the function is \(y=\frac{ax+3a}{x + 1}=\frac{a(x + 3)}{x + 1}\).

Step4: Use the \(y\) - intercept

The \(y\) - intercept occurs when \(x = 0\). Substituting \(x = 0\) into \(y=\frac{a(x + 3)}{x + 1}\), we have \(y=\frac{3a}{1}\). Since the \(y\) - intercept is at \(y = 3\), then \(\frac{3a}{1}=3\), so \(a = 1\).

Answer:

\(y=\frac{x + 3}{x + 1}\)