Question

find the horizontal and vertical asymptotes of the curve. you may want to use a comma - separated lists. if an answer does not exist, enter dne.)
$y = \frac{2x^{2}+3}{3x^{2}+8x - 3}$
$x = - 3,\frac{1}{3}$
great work!
$y = \frac{2}{3}$
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Explanation:

Step1: Find vertical asymptotes

Vertical asymptotes occur where the denominator is zero.
Set \(3x^{2}+8x - 3=0\).
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 3\), \(b = 8\), \(c=-3\).
\(x=\frac{-8\pm\sqrt{8^{2}-4\times3\times(-3)}}{2\times3}=\frac{-8\pm\sqrt{64 + 36}}{6}=\frac{-8\pm\sqrt{100}}{6}=\frac{-8\pm10}{6}\).
\(x=\frac{-8 + 10}{6}=\frac{1}{3}\) and \(x=\frac{-8-10}{6}=-3\).

Step2: Find horizontal asymptotes

For rational functions \(\frac{f(x)}{g(x)}\) where \(f(x)=2x^{2}+3\) and \(g(x)=3x^{2}+8x - 3\), when the degree of \(f(x)\) and \(g(x)\) are equal (both degree 2).
The horizontal asymptote is \(y=\frac{\text{leading coefficient of }f(x)}{\text{leading coefficient of }g(x)}\).
Leading coefficient of \(f(x)\) is 2 and of \(g(x)\) is 3. So \(y=\frac{2}{3}\).

Answer:

\(x=-3,\frac{1}{3}\); \(y = \frac{2}{3}\)