For $p(x)=\frac{-3 x^{2}+5 x-1}{x^{2}+3}$ $\newline$ (a) Identify the horizontal asymptotes (if any). $\newline$ (b) If the graph of the function has a horizontal asymptote, determine the point (if any) where the graph crosses the horizontal asymptote(s). $\newline$ Separate multiple equations of asymptotes with commas as necessary. Select

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p(x)=\frac{-3 x^{2}+5 x-1}{x^{2}+3}

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(a) Identify the horizontal asymptotes (if any).

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(b) If the graph of the function has a horizontal asymptote, determine the point (if any) where the graph crosses the horizontal asymptote(s).

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Separate multiple equations of asymptotes with commas as necessary. Select

Question Prompt: The question_prompt for the new math problem is: ＂What are the horizontal asymptotes of the function $p(x) = \frac{-3x^2 + 5x - 1}{x^2 + 3}$ , and where does the graph of the function cross these asymptotes, if at all?＂
Identifying Asymptotes: To identify the horizontal asymptotes, we compare the degrees of the numerator and the denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$ . If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. $\newline$ Degree of numerator: $2$ (from $-3x^2$ ) $\newline$ Degree of denominator: $2$ (from $x^2$ ) $\newline$ Since the degrees are equal, the horizontal asymptote is $y = -\frac{3}{1} = -3$ .
Solving for Asymptotes: To determine if the graph crosses the horizontal asymptote, we need to solve the equation $p(x) = -3$ for $x$ . This means we set the function equal to the horizontal asymptote and solve for $x$ . $\frac{-3x^2 + 5x - 1}{x^2 + 3} = -3$ Multiplying both sides by $(x^2 + 3)$ to clear the denominator, we get: $-3x^2 + 5x - 1 = -3(x^2 + 3)$
Solving for x: Expanding the right side of the equation and simplifying, we get: $\newline$ $-3x^2 + 5x - 1 = -3x^2 - 9$ $\newline$ $5x - 1 = -9$ $\newline$ $5x = -8$ $\newline$ $x = -\frac{8}{5}$
Substitute x Value: Now we substitute $x = -\frac{8}{5}$ back into the original function to find the corresponding y-value. $p\left(-\frac{8}{5}\right) = \frac{-3\left(-\frac{8}{5}\right)^2 + 5\left(-\frac{8}{5}\right) - 1}{\left(-\frac{8}{5}\right)^2 + 3}$
Calculating $p\left(-\frac{8}{5}\right)$ : Calculating the value of $p\left(-\frac{8}{5}\right)$ :
$p\left(-\frac{8}{5}\right) = \left(-3\left(\frac{64}{25}\right) - \frac{40}{5} - 1\right) / \left(\frac{64}{25} + 3\right)$
$p\left(-\frac{8}{5}\right) = \left(-\frac{192}{25} - \frac{40}{5} - 1\right) / \left(\frac{64}{25} + \frac{75}{25}\right)$
$p\left(-\frac{8}{5}\right) = \left(-\frac{192}{25} - \frac{200}{25} - \frac{25}{25}\right) / \left(\frac{139}{25}\right)$
$p\left(-\frac{8}{5}\right) = \left(-\frac{417}{25}\right) / \left(\frac{139}{25}\right)$
$p\left(-\frac{8}{5}\right) = -\frac{417}{139}$
Checking Y-Value: We need to check if the y-value we found is equal to the horizontal asymptote $y = -3$ . $\newline$ $-\frac{417}{139}$ is not equal to $-3$ , so there must be a math error in the previous steps.