# Let f (x)= 2-x/3+x (a) State the domain. (b) State the vertical asymptote(s). (c) State the...

## Question:

Let {eq}f\left( x \right)=\frac{2-x}{3+x} {/eq}

(a) State the domain.

(b) State the vertical asymptote(s).

(c) State the horizontal asymptote.

(d) State the y-intercept.

(e) State the x-intercept(s).

## Rational Functions:

Rational functions as any type of function have domain, range and intercepts with the axes of the plane. To find each of these characteristics, the following must be done:

• Domain: Depends on the denominator.
• Vertical Asymptote: Depends on the denominator.
• Horizontal Asymptote: Depends on the degree of the polynomials.
• Range: Depends on the horizontal asymptote.

Function:

{eq}f\left(x\right)=\frac{2-x}{3+x} {/eq}

a. State the domain:

The domain of a rational function is given by the numbers that make the denominator zero. So:

{eq}\begin{align*} 3+x&=0 \\ x&=-3 \end{align*} {/eq}

Domain: {eq}\left(-\infty \:,\:-3\right)\cup \left(-3,\:\infty \:\right) {/eq}

b. State the vertical asymptote.

The vertical asymptote of a rational function is related to the domain. The values that are excluded by the domain are the vertical asymptotes of the function. The vertical asymptote of {eq}f\left(x\right)=\frac{2-x}{3+x} {/eq} is {eq}x=-3 {/eq}

c. State the horizontal asymptote.

As the degree of the numerator and the degree of the denominator are the same, the horizontal asymptote is given by the quotient between the coefficients of the term of the first degree, that is: {eq}y=\frac{-1}{1} \quad y=-1 {/eq}

d. State the y-intercept.

{eq}\begin{align*} f\left(x\right)&=\frac{2-x}{3+x} \\ y&=\frac{2-x}{3+x} \\ y&=\frac{2-0}{3+0} \\ y&=\frac{2}{3} \\ \end{align*} {/eq}

e. State the x-intercept.

{eq}\begin{align*} f\left(x\right)&=\frac{2-x}{3+x} \\ f\left(x\right)&=0 \\ \frac{2-x}{3+x}&=0 \\ 2-x&=0 \\ x&=2 \end{align*} {/eq}