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Question: $f(x)=\frac{-4(x+4)(x+3)}{(x+4)(x+1)(x+3)}$ $\newline$ Answer Attempt $2$ out of $3$ $\newline$ # of Horizontal Asymptotes: None $\newline$ # of Holes: None $\newline$ # of Vertical Asymptotes: None $\newline$ # of x-intercepts: None $\newline$ # of y-intercepts: None

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Function transformation rules

Full solution

Q. Question:

f(x)=\frac{-4(x+4)(x+3)}{(x+4)(x+1)(x+3)}

\newline

Answer Attempt

2

out of

3

\newline

# of Horizontal Asymptotes: None

\newline

# of Holes: None

\newline

# of Vertical Asymptotes: None

\newline

# of x-intercepts: None

\newline

# of y-intercepts: None

Horizontal Asymptotes: To find the horizontal asymptotes, we need to compare the degrees of the numerator and the denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$ .
Numerator and Denominator Degrees: The degree of the numerator is $2$ (since it's a quadratic expression after expanding $(x+4)(x+3)$ ), and the degree of the denominator is $3$ (since it's a cubic expression after expanding $(x+4)(x+1)(x+3)$ ). Since the degree of the numerator is less than the degree of the denominator, there is one horizontal asymptote at $y = 0$ .
Holes: To find holes, we need to factor both the numerator and the denominator and look for common factors. If a factor is common to both, it will create a hole at the $x$ -value that makes the factor equal to zero.
Common Factors: The factors $(x+4)$ and $(x+3)$ are common in both the numerator and the denominator. This means there will be holes at $x = -4$ and $x = -3$ .
Vertical Asymptotes: To find vertical asymptotes, we look for factors in the denominator that are not in the numerator. These factors, when set to zero, give the $x$ -values of the vertical asymptotes.
Remaining Factors: The factor $(x+1)$ is in the denominator but not in the numerator. Setting $x+1$ to zero gives $x = -1$ . Therefore, there is one vertical asymptote at $x = -1$ .
X-Intercepts: To find $x$ -intercepts, we set the numerator equal to zero and solve for $x$ . The $x$ -intercepts are the $x$ -values where the function crosses the $x$ -axis.
No X-Intercepts: Since the factors $(x+4)$ and $(x+3)$ are canceled out due to the holes, there are no remaining factors in the numerator to set to zero. Therefore, there are no $x$ -intercepts.
Y-Intercept: To find the y-intercept, we set $x = 0$ and solve for $f(0)$ . The y-intercept is the point where the function crosses the y-axis.
Evaluation at $x=0$ : Substituting $x = 0$ into $f(x)$ gives $f(0) = \frac{-4(0+4)(0+3)}{(0+4)(0+1)(0+3)}$ . However, since there are holes at $x = -4$ and $x = -3$ , we cannot evaluate the function at these points, and thus there is no $y$ -intercept.

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