15. [-/0.83 Points] DETAILS MY NOTES SCALCET9 2.6.049. Find the horizontal and vertical asymptotes of the curve. You may want to use a graphing calculator (or computer) to check your work by graphing the curve and estimating the asymptotes. (Enter your answers as comma-separated lists. If an answer does not exist, enter DNE.) $$ y=\frac{4 x^{2}+x-3}{x^{2}+x-2} $$ $$ x= $$ $$ y= $$

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Function by following steps:
- step0: Find the horizontal asymptotes:
  $$y=\frac{4x^{2}+x-3}{x^{2}+x-2}$$
- step1: Evaluate the limits $$\lim _{x\rightarrow +\infty}\left(y\right)$$ and $$\lim _{x\rightarrow -\infty}\left(y\right):$$
  $$\begin{align}&\lim _{x\rightarrow +\infty}\left(\frac{4x^{2}+x-3}{x^{2}+x-2}\right)\&\lim _{x\rightarrow -\infty}\left(\frac{4x^{2}+x-3}{x^{2}+x-2}\right)\end{align}$$
- step2: Calculate:
  $$\begin{align}&4\&4\end{align}$$
- step3: The finite values are horizontal asymptotes:
  $$\begin{align}&y=4\end{align}$$
Find the vertical asymptotes of $$ y=\frac{4 x^{2}+x-3}{x^{2}+x-2} $$.
Function by following steps:
- step0: Find the vertical asymptotes:
  $$y=\frac{4x^{2}+x-3}{x^{2}+x-2}$$
- step1: Evaluate the limit:
  $$\begin{align}&\lim _{x\rightarrow -2}\left(\frac{4x^{2}+x-3}{x^{2}+x-2}\right)\&\lim _{x\rightarrow 1}\left(\frac{4x^{2}+x-3}{x^{2}+x-2}\right)\end{align}$$
- step2: Calculate:
  $$\begin{align}&\textrm{The limit does not exist}\&\textrm{The limit does not exist}\end{align}$$
- step3: $$x=-2$$ is a vertical asymptote$$:$$
  $$\begin{align}&x=-2\textrm{ }\textrm{is a vertical asymptote}\&x=1\textrm{ }\textrm{is a vertical asymptote}\end{align}$$
- step4: List all vertical asymptotes of the function:
  $$\begin{align}&x=-2\&x=1\end{align}$$
To find the asymptotes of the curve given by the function

$$
y = \frac{4x^{2} + x - 3}{x^{2} + x - 2}
$$

we have determined the following:

### Horizontal Asymptote
The horizontal asymptote is found by examining the degrees of the polynomial in the numerator and the denominator. Since both the numerator and denominator are quadratic (degree 2), we can find the horizontal asymptote by taking the ratio of the leading coefficients.

The leading coefficient of the numerator $$4x^2$$ is $$4$$ and the leading coefficient of the denominator $$x^2$$ is $$1$$. Therefore, the horizontal asymptote is:

$$
y = 4
$$

### Vertical Asymptotes
Vertical asymptotes occur where the denominator is equal to zero (and the numerator is not zero at those points). We need to solve the equation:

$$
x^{2} + x - 2 = 0
$$

Factoring the quadratic gives:

$$
(x + 2)(x - 1) = 0
$$

Setting each factor to zero gives the vertical asymptotes:

$$
x = -2 \quad \text{and} \quad x = 1
$$

### Summary of Asymptotes
- **Horizontal Asymptote:** $$y = 4$$
- **Vertical Asymptotes:** $$x = -2, x = 1$$

Thus, the final answers are:
- $$x = -2, 1$$
- $$y = 4$$
Horizontal asymptote: $$ y = 4 $$ Vertical asymptotes: $$ x = -2, 1 $$

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