To analyze the function \( f(x) = \frac{4x^2 - 4x - 8}{2x + 2} \), we will first simplify it.
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Factor the numerator: \[ 4x^2 - 4x - 8 = 4(x^2 - x - 2) \] The quadratic \( x^2 - x - 2 \) can be factored further: \[ x^2 - x - 2 = (x - 2)(x + 1) \] Therefore, the numerator becomes: \[ 4(x - 2)(x + 1) \]
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Factor the denominator: The denominator can be factored as: \[ 2x + 2 = 2(x + 1) \]
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Rewrite the function: Now we can write \( f(x) \) as: \[ f(x) = \frac{4(x - 2)(x + 1)}{2(x + 1)} \] For \( x \neq -1 \), we can simplify this to: \[ f(x) = \frac{4(x - 2)}{2} = 2(x - 2) = 2x - 4 \]
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Identify the discontinuity: The function \( f(x) = 2x - 4 \) is defined everywhere except at \( x = -1 \). At \( x = -1 \), the original function has a removable discontinuity since the term \( (x + 1) \) cancels out.
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Find the value of \( f(-1) \): To find the value of the function at the discontinuity: \[ f(-1) = 2(-1) - 4 = -2 - 4 = -6 \]
Therefore, the graph of \( f(x) = 2x - 4 \) has a discontinuity at \( x = -1 \), with the point of discontinuity being \( (-1, -6) \).
Thus, the correct answer is the graph of \( 2x - 4 \), with discontinuity at \( (-1, -6) \).