Question

write an equation for the rational function graphed below. equation = question help: post to forum

Explanation:

Step1: Determine the vertical asymptotes

Vertical asymptotes are at \(x = - 1\) and \(x=2\). So the denominator is \((x + 1)(x - 2)=x^{2}-x - 2\)

Step2: Determine the horizontal asymptote

The horizontal asymptote is \(y = 2\). For a rational function \(y=\frac{f(x)}{g(x)}\), when the degree of \(f(x)\) and \(g(x)\) are equal, \(y=\frac{a}{b}\) (where \(a\) is the leading coefficient of \(f(x)\) and \(b\) is the leading coefficient of \(g(x)\)). So the leading coefficient of the numerator is \(2\) times the leading coefficient of the denominator.

Step3: Determine the \(x -\)intercept

The \(x -\)intercept is \(x = 0\). Let the numerator be \(2x^{2}+bx + c\). When \(x = 0\), \(y=0\) (substitute \(x = 0\) into \(y=\frac{2x^{2}+bx + c}{x^{2}-x - 2}\)), we get \(c = 0\)

We can use another point. Let's assume the function is \(y=\frac{2x^{2}+bx}{x^{2}-x - 2}\). The graph passes through \((1,y)\). When \(x = 1\), from the graph \(y\) is not defined (vertical asymptote at \(x = 2\) and \(x=-1\)), but we can also use the fact that we can rewrite the general form of a rational function \(y = 2+\frac{mx + n}{x^{2}-x - 2}\)

Since \(y - 2=\frac{mx + n}{x^{2}-x - 2}\). When \(x = 0\), \(y-2=\frac{n}{- 2}\), and \(y=-1\) (from the graph when \(x = 0\)), so \(-1-2=\frac{n}{-2}\), \(n = 6\)

Let \(y=\frac{2x^{2}-2x}{x^{2}-x - 2}=\frac{2x(x - 1)}{(x + 1)(x - 2)}\)

Answer:

\(y=\frac{2x(x - 1)}{(x + 1)(x - 2)}\)