Question

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

vertical asymptote:
horizontal asymptote:
x-intercept:
y-intercept:
table
x | y

domaine:
range:

Explanation:

Step1: Analyze Vertical Asymptote

For a rational function \( f(x)=\frac{N(x)}{D(x)} \), vertical asymptote occurs where \( D(x) = 0 \) (and \( N(x)
eq0 \) at that point). Given \( f(x)=\frac{x + 4}{-x+4}=\frac{x + 4}{-(x - 4)} \), set denominator \( -x + 4=0\Rightarrow x = 4 \). So vertical asymptote is \( x = 4 \).

Step2: Analyze Horizontal Asymptote

For rational function \( f(x)=\frac{ax^n+...}{bx^m+...} \), if \( n=m \), horizontal asymptote is \( y=\frac{a}{b} \). Here \( n = 1,m = 1,a = 1,b=- 1 \), so \( y=\frac{1}{-1}=-1 \).

Step3: Analyze x - intercept

x - intercept is when \( y = 0 \), set \( f(x)=0\Rightarrow\frac{x + 4}{-x + 4}=0\Rightarrow x+4 = 0\Rightarrow x=-4 \). So x - intercept is \( (-4,0) \) (the original answer had a mistake here).

Step4: Analyze y - intercept

y - intercept is when \( x = 0 \), substitute \( x = 0 \) into \( f(x) \): \( f(0)=\frac{0 + 4}{-0+4}=\frac{4}{4}=1 \). So y - intercept is \( (0,1) \) (the original answer had a mistake here).

Step5: Analyze Domain

Domain is all real numbers except where denominator is zero. Denominator \( -x + 4=0\Rightarrow x = 4 \), so domain is \( \{x|x
eq4,x\in\mathbb{R}\} \).

Step6: Analyze Range

For \( y=\frac{x + 4}{-x + 4} \), solve for \( x \): \( y(-x + 4)=x + 4\Rightarrow -yx+4y=x + 4\Rightarrow -yx-x=4 - 4y\Rightarrow x(-y - 1)=4 - 4y\Rightarrow x=\frac{4y - 4}{y + 1} \). Denominator \( y + 1=0\Rightarrow y=-1 \), so range is \( \{y|y
eq - 1,y\in\mathbb{R}\} \).

Answer:

• Vertical Asymptote: \( x = 4 \)
• Horizontal Asymptote: \( y=-1 \)
• x - intercept: \( (-4,0) \)
• y - intercept: \( (0,1) \)
• Domain: \( \{x|x

eq4\} \)

• Range: \( \{y|y

eq - 1\} \)