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Find the vertical asymptote(s) of the graph of the function. \( f(x)=\frac{2-x}{(x-5)(x+2)} \)

Ask by Graham Parry. in the United States
Mar 20,2025

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The vertical asymptotes are at \( x=5 \) and \( x=-2 \).

Solución

1. The given function is \[ f(x)=\frac{2-x}{(x-5)(x+2)}. \] 2. Vertical asymptotes occur where the denominator is zero provided that the numerator is not zero at those points. 3. Set the denominator equal to zero: \[ (x-5)(x+2)=0. \] This gives: \[ x-5=0 \quad \text{or} \quad x+2=0, \] resulting in: \[ x=5 \quad \text{or} \quad x=-2. \] 4. Next, check that the numerator does not also equal zero at these \(x\)-values: - For \( x=5 \): \[ 2-5=-3 \neq 0. \] - For \( x=-2 \): \[ 2-(-2)=4 \neq 0. \] 5. Since the denominator is zero and the numerator is non-zero at \( x=5 \) and \( x=-2 \), the function has vertical asymptotes at these points. The vertical asymptotes are at \( x=5 \) and \( x=-2 \).

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Bonus Knowledge

To find the vertical asymptotes of the function \( f(x) = \frac{2-x}{(x-5)(x+2)} \), we need to determine where the denominator equals zero since that’s where the function can potentially be undefined. Setting the denominator equal to zero: \[ (x-5)(x+2) = 0 \] This gives us two equations to solve: 1. \( x - 5 = 0 \) → \( x = 5 \) 2. \( x + 2 = 0 \) → \( x = -2 \) So, the vertical asymptotes are at \( x = 5 \) and \( x = -2 \). Whenever the graph approaches these x-values, the function will head towards infinity (or negative infinity), indicating that \( x = 5 \) and \( x = -2 \) are indeed vertical asymptotes!

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