$y=\frac{x^2-4}{x^2-5 x+6}$

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02-11-2026
Mathematics

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$y=\frac{x^2-4}{x^2-5 x+6}$

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-02T00:22:08-05:00

Factor the numerator as $(x-2)(x+2)$ and the denominator as $(x-2)(x-3)$ to simplify the expression to $\frac{x+2}{x-3}$ for $x eq 2, 3$.
Identify the domain as $\{x \in \mathbb{R} | x eq 2, 3\}$.
Locate the hole at $(2, -4)$ by evaluating the simplified function at $x=2$.
Determine the vertical asymptote at $x=3$ and the horizontal asymptote at $y=1$.
Final simplified function is $\boxed{f(x) = \frac{x + 2}{x - 3}, x eq 2, 3}$.

Explanation

Factoring and Determining the Domain

To analyze the function $y = \frac{x^2 - 4}{x^2 - 5x + 6}$, we begin by examining the numerator and the denominator. Factoring the numerator, we have $x^2 - 4 = (x - 2)(x + 2)$. Factoring the denominator, we have $x^2 - 5x + 6 = (x - 2)(x - 3)$. The function is defined for all real numbers except where the denominator is zero. Setting the denominator to zero, we find $x - 2 = 0$ or $x - 3 = 0$, which gives $x = 2$ or $x = 3$. Therefore, the domain of the function is all real numbers $x$ such that $x eq 2$ and $x eq 3$. In set notation, the domain is $\{x \in \mathbb{R} | x eq 2, x eq 3\}$.

Simplifying and Identifying the Hole

We can simplify the expression for the function by canceling the common factor $(x - 2)$ from the numerator and the denominator. For all $x eq 2$ and $x eq 3$, the function becomes:
\[f(x) = \frac{(x - 2)(x + 2)}{(x - 2)(x - 3)} = \frac{x + 2}{x - 3}\]
Because the factor $(x - 2)$ was removed, there is a point discontinuity, or "hole," at $x = 2$. To find the y-coordinate of this hole, we evaluate the simplified expression at $x = 2$:
\[f(2) = \frac{2 + 2}{2 - 3} = \frac{4}{-1} = -4\]
Thus, there is a hole at the point $(2, -4)$.

Identifying Asymptotes and Behavior

The vertical asymptote occurs where the denominator of the simplified expression is zero. Setting $x - 3 = 0$, we find a vertical asymptote at $x = 3$. As $x$ approaches 3, the function values grow without bound, indicating an infinite discontinuity. To find the horizontal asymptote, we evaluate the limit of the function as $x$ approaches infinity:
\[\lim_{x \to \infty} \frac{x + 2}{x - 3} = \lim_{x \to \infty} \frac{1 + \frac{2}{x}}{1 - \frac{3}{x}} = \frac{1 + 0}{1 - 0} = 1\]
This shows that as $x$ becomes very large or very small, $y$ approaches 1, confirming a horizontal asymptote at $y = 1$.

Conclusion

In summary, the function $y = \frac{x^2 - 4}{x^2 - 5x + 6}$ is equivalent to $f(x) = \frac{x + 2}{x - 3}$ for $x eq 2$ and $x eq 3$. The domain is $\{x \in \mathbb{R} | x eq 2, x eq 3\}$, there is a hole at $(2, -4)$, a vertical asymptote at $x = 3$, and a horizontal asymptote at $y = 1$. Final representation of the simplified function with restrictions is $\boxed{f(x) = \frac{x + 2}{x - 3}, x eq 2, 3}$.

Examples

This type of analysis is crucial in fields like physics and engineering, where we model physical systems using rational functions. For example, if you are designing a feedback controller for a mechanical system, the transfer function might be represented as a rational function. Identifying where the function is undefined (discontinuities) tells you where the system might become unstable or where sensors might fail. By simplifying the function, you can better understand the underlying dynamics, such as the asymptotic behavior (long-term stability) or specific points of failure.