Functions with square roots: the domain and asymptotes

A square root has a value only where the expression under it is not negative, and a fraction only where its denominator is not 0. A function with roots is defined where all its parts are:

• $\displaystyle {f(x)}={\frac{\sqrt{x+2}}{x-1}}$: $\displaystyle x+2\geq0$ and $\displaystyle x\neq1$, that is $\displaystyle x\geq-2$, $\displaystyle x\neq1$;
• $\displaystyle {g(x)}={\sqrt{x}+\sqrt{4-x}}$: $\displaystyle x\geq0$ and $\displaystyle 4-x\geq0$, that is $\displaystyle 0\leq x\leq4$ – a closed interval with two ends;
• $\displaystyle {h(x)}={\sqrt{x^2-4}}$: $\displaystyle x^2-4\geq0$, that is $\displaystyle x\leq-2$ or $\displaystyle x\geq2$ – two pieces with a gap between them;
• $\displaystyle {k(x)}={\frac{1}{\sqrt{x-1}}}$: the root in the denominator must be positive, so $\displaystyle x>1$.

Unlike a rational function, which is defined everywhere but at a few points, a function with a root may be defined only on an interval, and its graph ends at the ends of that interval.

The square root is never negative, so the root of a square is the absolute value:

For example, $\displaystyle {\sqrt{(-3)^2}}={\sqrt{9}}={3}={|-3|}$, not $\displaystyle -3$. So for large $\displaystyle |x|$, $\displaystyle \sqrt{x^2+1}$ is close to $\displaystyle |x|$: to $\displaystyle x$ on the right, and to $\displaystyle -x$ on the left.

Because $\displaystyle {\sqrt{x^2}}={|x|}$, a function with a root may approach one line far to the right and another far to the left. The example below finds the asymptotes of $\displaystyle {f(x)}={\frac{x}{\sqrt{x^2+1}}}$:

• for large positive $\displaystyle x$, $\displaystyle \sqrt{x^2+1}$ is close to $\displaystyle x$, so $\displaystyle f(x)$ is close to $\displaystyle {\frac{x}{x}}={1}$: the asymptote $\displaystyle {y}={1}$ as $\displaystyle x\to\infty$;
• for large negative $\displaystyle x$, $\displaystyle \sqrt{x^2+1}$ is close to $\displaystyle -x$, so $\displaystyle f(x)$ is close to $\displaystyle {\frac{x}{-x}}={-1}$: the asymptote $\displaystyle {y}={-1}$ as $\displaystyle x\to-\infty$.

A function may also have a horizontal asymptote on one side and grow without end on the other. The example below shows it for $\displaystyle {f(x)}={\sqrt{x^2+4}-x}$:

• as $\displaystyle x\to\infty$ both terms grow, so the difference is not clear at once; multiplying and dividing by $\displaystyle \sqrt{x^2+4}+x$ gives $\displaystyle {f(x)}={\frac{(x^2+4)-x^2}{\sqrt{x^2+4}+x}}={\frac{4}{\sqrt{x^2+4}+x}}$, which approaches 0: the asymptote $\displaystyle {y}={0}$ on the right;
• as $\displaystyle x\to-\infty$, $\displaystyle \sqrt{x^2+4}$ grows and $\displaystyle -x$ grows too: $\displaystyle f(x)$ grows without end, and there is no asymptote on the left.

A root in a denominator is 0 at the end of the domain, so the function may grow without end there – but from one side only, the side of the domain. For $\displaystyle {f(x)}={\frac{1}{\sqrt{x-1}}}$:

• the domain is $\displaystyle x>1$;
• just right of 1 the denominator is a small positive number: $\displaystyle f(x)$ grows without end, and $\displaystyle {x}={1}$ is a vertical asymptote;
• as $\displaystyle x\to\infty$, $\displaystyle f(x)$ approaches 0: the asymptote $\displaystyle {y}={0}$.

Find the domain of each function.

$\displaystyle {f(x)}={\sqrt{x^2-x-6}}$

$\displaystyle {g(x)}={\sqrt{\frac{x+2}{x-3}}}$

The function is $\displaystyle {f(x)}={\frac{2x}{\sqrt{x^2+4}}}$.

Write on the board the equations of its two horizontal asymptotes.

Which asymptote does the graph approach far to the right, and which far to the left? Explain by $\displaystyle {\sqrt{x^2}}={|x|}$.

Each board shows the graph of one of these functions:

• $\displaystyle {f_1(x)}={\sqrt{4-x^2}}$
• $\displaystyle {f_2(x)}={x\sqrt{4-x^2}}$
• $\displaystyle {f_3(x)}={\sqrt{x^2-4}}$
• $\displaystyle {f_4(x)}={\frac{\sqrt{x^2-4}}{x^2}}$

Decide by the domain, the sign and the asymptotes, and write the function on the board.

The function is $\displaystyle {f(x)}={\sqrt{x^2+2x}-x}$.

Find the domain of $\displaystyle f$.

$\displaystyle {f(x)}={\frac{2x}{\sqrt{x^2+2x}+x}}$ for $\displaystyle x>0$.
Write on the board the equation of the asymptote that the graph approaches far to the right.

What does $\displaystyle f(x)$ do as $\displaystyle x\to-\infty$? Explain.