We can only have an oblique asymptote if the degree of the numerator is one more than the degree of the denominator.  This stipulates that \(\displaystyle n\) must equal \(\displaystyle m+1\).  

The slope of the asymptote is determined by the ratio of the leading terms, which means the ratio of \(\displaystyle a\) to \(\displaystyle b\) must be 3 to 1.  The actual numbers are not important.

Finally, since the value of \(\displaystyle n\) is at least three, we know there is no intercept to our oblique asymptote.

To find the y-intercept of \(\displaystyle y=\frac{x^2+4x+3}{(x+3)}\), simply substitute \(\displaystyle x=0\) and solve for \(\displaystyle y\).

\(\displaystyle y=\frac{0^2+4(0)+3}{(0+3)}=\frac{3}{3}=1\)

The y-intercept is 1.

The numerator, \(\displaystyle x^2+4x+3\), can be simplified by factoring it into two binomials.

\(\displaystyle x^2+4x+3=(x+3)(x+1)\)

\(\displaystyle y=\frac{x^2+4x+3}{(x+3)}=\frac{(x+3)(x+1)}{(x+3)} = x+1\)

There is a removable discontinuity at \(\displaystyle x=-3\), but there are no asymptotes at \(\displaystyle x=-3\) since the \(\displaystyle x+3\) terms can be canceled.

The correct answer is:  \(\displaystyle \textup{Y-intercept at: } y=1; \textup{ No asymptotes.}\)

The \(\displaystyle \small x\)-intercept(s) is/are the root(s) of the numerator of the rational functions.

In this case, the numerator is \(\displaystyle \small 2x^2-6x+4\).

Using the quadratic formula,

\(\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

\(\displaystyle x=\frac{6\pm\sqrt{(-6)^2-4(2)(4)}}{2(2)}\)

\(\displaystyle x=\frac{6\pm\sqrt{36-32}}{4}\)

\(\displaystyle x=\frac{6\pm2}{4}=1,2\)

the roots are \(\displaystyle \small x=1,2\).

Thus, \(\displaystyle \small x=1,2\) are the \(\displaystyle \small x\)-intercepts.

Finding the vertical asymptotes of the rational function \(\displaystyle \small \frac{8x^2}{4x^2+31x-8}\) amounts to finding the roots of the denominator, \(\displaystyle \small 4x^2+31x-8\).

It is easy to check, using the quadratic formula,

\(\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

\(\displaystyle x=\frac{-31\pm\sqrt {31^2-4(4)(-8)}}{2(4)}\)

\(\displaystyle x=\frac{-31\pm\sqrt {961+128}}{8}\)

\(\displaystyle x=\frac{-31\pm 33}{8}=\frac{1}{4}, -8\)

that the roots, and thus the asymptotes, are \(\displaystyle \small x=\frac{1}{4}, -8\).

Before we start to simplify the problem, it is crucial to immediately identify the domain of this function \(\displaystyle y=\frac{3x}{x}\).

The denominator cannot be zero, since it is undefined to divide numbers by this value.  After simplification, the equation is:

\(\displaystyle y=3\)

The domain is \(\displaystyle x\neq0\) and there is a hole at \(\displaystyle x=0\) since there is a removable discontinuity.  There are no asymptotes.  

Since it's not possible to substitute \(\displaystyle x=0\) into the original equation, the y-intercept also does not exist.

Therefore, the correct answer is:

\(\displaystyle \textup{There are no y-intercepts or asymptotes.}\)

To find the vertical asymptote of a function, we set the denominator equal to \(\displaystyle 0\). 

With our function, we complete this process. 

The denominator is \(\displaystyle x + 2\), so we begin: 

\(\displaystyle x + 2 = 0\)

\(\displaystyle x = -2\)

The y-intercept of a function is always found by substituting in \(\displaystyle x = 0\).

We can go through this process for our function. 

\(\displaystyle \frac{x^2 + 3x + 6}{x + 2}= \frac{0^2 + 3\cdot 0+ 6}{0 + 2} = \frac{ 6}{ 2} = 3\)

In order for the vertical asymptote to be \(\displaystyle x = 3\), we need the denominator to be \(\displaystyle x -3\). This gives us three choices of numerators:

\(\displaystyle x^2 - 11x + 24\)

\(\displaystyle x^2 - 11x + 7\)

\(\displaystyle x^2 + 11x + 9\)

If the slant asymptote is \(\displaystyle x - 8\), we will be able to divide our numerator by \(\displaystyle x - 3\) and get \(\displaystyle x - 8\) with a remainder.

Dividing the first one gives us \(\displaystyle x - 8\) with no remainder.

Dividing the last one gives us \(\displaystyle x + 14\) with a remainder.

The middle numerator would give us what we were after, \(\displaystyle x - 8\) with a remainder of -17.

The answer is

\(\displaystyle y = \frac{x^2 - 11x + 7 }{ x - 3 }\)

To find the information we're looking for, we should factor this equation:

\(\displaystyle y = \frac{(2 x - 1 )(x + 3 )}{(2 x - 1 )(x+ 2 )}\)

This means that it simplifies to \(\displaystyle y = \frac{x + 3 }{ x + 2 }\).

When the equation is in the form of a fraction, to find the zero of the function we need to set the numerator equal to zero and solve for the variable.

\(\displaystyle \\x+3=0\\x=-3\)

To find the asymptote of an equation with a fraction we need to set the denominator of the fraction equal to zero and solve for the variable.

\(\displaystyle \\x+2=0\\x=-2\)

Therefore our equation has a zero at -3 and an asymptote at -2.

To find the vertical asymptote, just set the denominator equal to 0:

\(\displaystyle x + 3 = 0\)

\(\displaystyle x = -3\)

To find the slant asymptote, divide the numerator by the denominator, but ignore any remainder. You can use long division or synthetic division.

               \(\displaystyle 2x-8\)

\(\displaystyle \\x+3\ |\overline{2 x^ 2 - 5 x + 1 }\)

          \(\displaystyle \\ \underline{-(2x^2+3x) }\)

                         \(\displaystyle -8x+1\)

                   \(\displaystyle \underline{-(-8x-24)}\)

                                       \(\displaystyle 25\)

The slant asymptote is

\(\displaystyle y = 2x - 8\).