Find the equation of the vertical asymptote and the equation of the slant asymptote of the rational function. $$ f(x)=\frac{12 x^{2}-38 x+33}{-2 x+4} $$ The equation of the vertical asymptote is $$ x=2 $$ The equation of the slant asymptote is $$ y=\square $$ Question Help: Video Message instructor

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Let’s start with the function

$$
f(x)=\frac{12x^2-38x+33}{-2x+4}.
$$

Since the denominator can be factored as

$$
-2x+4 = -2(x-2),
$$

we see that the denominator is zero when $$ x-2=0 $$, so

$$
x=2,
$$

which gives the vertical asymptote.

To find the slant asymptote, we perform polynomial long division of the numerator $$ 12x^2-38x+33 $$ by the denominator $$ -2x+4 $$.

1. **Divide the leading terms:**

The leading term of the numerator is $$ 12x^2 $$ and that of the denominator is $$ -2x $$. Dividing gives

$$
   \frac{12x^2}{-2x} = -6x.
   $$

2. **Multiply and subtract:**

Multiply the entire denominator by $$ -6x $$:

$$
   -6x(-2x+4) = 12x^2 - 24x.
   $$

Subtract this from the numerator:

$$
   (12x^2-38x+33) - (12x^2-24x) = -14x+33.
   $$

3. **Divide the new leading term:**

Now, divide $$ -14x $$ (the new leading term) by $$ -2x $$:

$$
   \frac{-14x}{-2x} = 7.
   $$

4. **Multiply and subtract again:**

Multiply the denominator by $$ 7 $$:

$$
   7(-2x+4) = -14x+28.
   $$

Subtract:

$$
   (-14x+33) - (-14x+28) = 5.
   $$

The division yields

$$
f(x) = -6x+7 + \frac{5}{-2x+4}.
$$

For large values of $$ x $$, the fraction $$ \frac{5}{-2x+4} $$ becomes negligible. Hence, the slant asymptote is given by the quotient:

$$
y=-6x+7.
$$

Thus, the vertical asymptote is

$$
x=2,
$$

and the slant asymptote is

$$
y=-6x+7.
$$
The vertical asymptote is $$ x=2 $$ and the slant asymptote is $$ y=-6x+7 $$.

Find the equation of the vertical asymptote and the equation of the slant asymptote of the rational function. \( f(x)=\frac{12 x^{2}-38 x+33}{-2 x+4} \) The equation of the vertical asymptote is \( x=2 \) The equation of the slant asymptote is \( y=\square \) Question Help: Video Message instructor

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