Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 38

Determine the horizontal and vertical asymptotes of the curve $\displaystyle F(x) = \frac{x - 9}{\sqrt{4x^2 + 3x + 2}}$

Solving for the vertical asymptotes

We set the denominator equal to zero

$4x^2 + 3x + 2 = 0$

Using the discriminant of $ax^2 + bx + c$ which is $\Delta = b^2 - 4ac$.

(If $\Delta < 0$, then the equation $ax^2 + bx + c = 0$ has no real solution.)


$
\begin{equation}
\begin{aligned}

\Delta =& (3)^2 - 4 (4)(2)
\\
\\
\Delta =& 9 - 32
\\
\\
\Delta =& -23

\end{aligned}
\end{equation}
$


But $-23 < 0$

So $4x^2 + 3x + 2 = 0$ has no real solution.

Therefore,

$F(x)$ has no vertical asymptotes

Solving for the horizontal asymptotes

In the function $\displaystyle F(x) = \frac{x - 9}{\sqrt{4x^2 + 3x + 2}}$ we remove everything except the biggest exponents of $x$ found in the numerator an denominator.

So we have


$
\begin{equation}
\begin{aligned}

F(x) =& \frac{x}{\pm \sqrt{4x^2}} \qquad \text{where $F(x) = y$}
\\
\\
y =& \frac{\cancel{x}}{\pm 2 \cancel{x}}
\\
\\
y =& \pm \frac{1}{2}

\end{aligned}
\end{equation}
$



Thus, the horizontal asymptotes are $\displaystyle y = \frac{1}{2}$ and $\displaystyle y = \frac{-1}{2}$

Therefore,

The function $F(x)$ has no vertical asymptotes and have horizontal asymptote which is $\displaystyle y = \frac{1}{2}$ and $\displaystyle y = \frac{-1}{2}$

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