Question
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Analyze the graph of the function \( F(x)=\frac{x^{2}-12 x+32}{x^{2}-8 x+16} \).

Ask by Rodriquez Bush. in the United States
Feb 20,2025

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Answer

The function \( F(x)=\frac{x^{2}-12x+32}{x^{2}-8x+16} \) is undefined at \( x = 4 \). So, its domain is all real numbers except \( x = 4 \).

Solution

The function \( F(x)=\frac{x^{2}-12x+32}{x^{2}-8x+16} \) does not have any specific domain restrictions based on the provided information. To determine the domain of the function, we need to consider the values of \( x \) for which the denominator is not equal to zero. The denominator of the function is \( x^{2}-8x+16 \). To find the values of \( x \) for which the denominator is not equal to zero, we need to solve the equation \( x^{2}-8x+16 = 0 \). Let's solve the equation \( x^{2}-8x+16 = 0 \) to determine the domain of the function. Solve the quadratic equation by following steps: - step0: Solve by factoring: \(x^{2}-8x+16=0\) - step1: Factor the expression: \(\left(x-4\right)^{2}=0\) - step2: Simplify the expression: \(x-4=0\) - step3: Move the constant to the right side: \(x=0+4\) - step4: Remove 0: \(x=4\) The solutions to the equation \( x^{2}-8x+16 = 0 \) are \( x = 4 \). This means that the denominator of the function is zero when \( x = 4 \), and the function is undefined at this point. Therefore, the domain of the function \( F(x)=\frac{x^{2}-12x+32}{x^{2}-8x+16} \) is all real numbers except \( x = 4 \).

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Bonus Knowledge

To analyze the function \( F(x) = \frac{x^{2}-12x+32}{x^{2}-8x+16} \), let's start with some key insights. First, the denominator can be rewritten as \( (x-4)^2 \), indicating a potential vertical asymptote at \( x = 4 \). The numerator, \( x^2 - 12x + 32 \), simplifies to \( (x-6)(x-6) \) when factored, resulting in a double root at \( x = 6 \). This means the function touches and bounces off the x-axis at this point. Now, let’s also consider the behavior of \( F(x) \) as \( x \) approaches infinity. As \( x \) becomes very large or very small, the leading terms dominate, allowing us to evaluate the horizontal asymptote. Both the numerator and denominator are quadratic, meaning we can simplify the function to \( F(x) \approx 1 \) as \( x \to \pm \infty \). Therefore, the horizontal asymptote is \( y = 1 \). Overall, expect a function that approaches \( y = 1 \) with a significant behavior change around \( x = 4 \) and has a bounce off at \( x = 6 \).

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