# How do you graph and find the discontinuities of (x^(1/2) +1)/(x+1) ?

Feb 18, 2018

Horizontal asymptote at $y = 0$
Domain: $\left[0 , \infty\right)$

#### Explanation:

Given: $\frac{{x}^{\frac{1}{2}} + 1}{x + 1} = \frac{\sqrt{x} + 1}{x + 1}$

Discontinuities are caused by holes, jumps, vertical asymptotes, or specific function limitations.

This equation is a rational function in the form $R \left(x\right) = \frac{N \left(x\right)}{D \left(x\right)}$

Holes are found if any factors from the numerator can be canceled from the denominator. In this equation no factors can be canceled, so there are no holes.

There is a vertical asymptote at $x = - 1 , \left(D \left(x\right) = 0\right)$, but this is not a discontinuity for this function because the square root in the numerator limits the function's domain. The domain doesn't start until $x = 0$.

Domain is limited due to $\sqrt{x} : \text{ } x \ge 0$

To graph the function you would want to know that there is a horizontal asymptote.

Rational Functions $R \left(x\right) = \frac{N \left(x\right)}{D \left(x\right)} = \frac{{a}_{n} {x}^{n} + \ldots}{{b}_{m} {x}^{m} + \ldots}$

If $n = m$, the horizontal asymptote is $y = \frac{{a}_{n}}{{b}_{m}}$

If $n < m$, the horizontal asymptote is $y = 0$

If $n > m$, there is no horizontal asymptote.

If $n = m + 1$ there is a slant asymptote.

In the example given, $n = \frac{1}{2} < m = 1$; horizontal asymptote at $y = 0$

To graph the function find the $x$-intercept(s), $N \left(x\right) = 0$:

sqrt(x)+1 = 0; sqrt(x) = -1 " " No $x$-intercepts

Find the $y$-intercept, by letting $x = 0 : \text{ } y = 1$

Graph:
graph{(sqrt(x)+1)/(x+1) [-4.205, 15.795, -4.76, 5.24]}