To find the vertical asymptotes of the function \( f(x) = \frac{4x + 8}{x^2 + 3x - 4} \), we need to determine where the denominator equals zero, as vertical asymptotes occur at these points (provided that the numerator does not also equal zero at the same points).
First, we'll factor the denominator \( x^2 + 3x - 4 \):
- We look for two numbers that multiply to \(-4\) (the constant term) and add up to \(3\) (the coefficient of \(x\)).
- Those two numbers are \(4\) and \(-1\) because \(4 \times -1 = -4\) and \(4 + (-1) = 3\).
Thus, we can factor the denominator: \[ x^2 + 3x - 4 = (x + 4)(x - 1) \]
Now, we set the denominator equal to zero to find the values of \(x\) that make the denominator zero: \[ (x + 4)(x - 1) = 0 \]
This gives us: \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
Next, we check if these points make the numerator non-zero:
The numerator is \(4x + 8\):
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At \(x = -4\): \[ 4(-4) + 8 = -16 + 8 = -8 \quad \text{(not zero)} \]
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At \(x = 1\): \[ 4(1) + 8 = 4 + 8 = 12 \quad \text{(not zero)} \]
Since the numerator does not equal zero at these points, both \(x = -4\) and \(x = 1\) are indeed vertical asymptotes.
Thus, the vertical asymptotes of the function \( f(x) \) occur at:
x = -4 and x = 1
This corresponds to the choice: x = 1 and x = −4.
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