Question

drag the key features to the correct location on the image. each key feature can be used more than once, but not all key features will be used. which key features are present in these three functions? domain: (-∞, 6) ∪ (6, ∞) domain: (-∞, -3) ∪ (-3, 6) ∪ (6, ∞) horizontal asymptote: y = -2 domain: (-∞, -3) ∪ (-3, 2) ∪ (2, ∞) horizontal asymptote: y = 1 domain: (-∞, -3) ∪ (-3, ∞) domain: (-∞, 2) ∪ (2, 6) ∪ (6, ∞) oblique asymptote f(x) = (-2x + 4)/(x - 6) g(x) = (x² - 2x)/(x² + x - 6) h(x) = (x² - 2x)/(x - 6)

Explanation:

Step1: Analyze \( f(x)=\frac{-2x + 4}{x - 6} \)

• Domain: Denominator \( x - 6=0\Rightarrow x = 6 \). So domain is \( (-\infty,6)\cup(6,\infty) \).
• Asymptotes: Degree of numerator (1) and denominator (1) are equal. Horizontal asymptote \( y=\frac{-2}{1}=-2 \). No oblique asymptote (since degrees are equal).

Step2: Analyze \( g(x)=\frac{x^{2}-2x}{x^{2}+x - 6} \)

• Factor denominator: \( x^{2}+x - 6=(x + 3)(x - 2) \). So \( x

eq - 3,2 \). Domain: \( (-\infty,-3)\cup(-3,2)\cup(2,\infty) \).

• Degrees of numerator and denominator are equal (2). Horizontal asymptote \( y=\frac{1}{1}=1 \) (ratio of leading coefficients).

Step3: Analyze \( h(x)=\frac{x^{2}-2x}{x - 6} \)

• Denominator \( x - 6 = 0\Rightarrow x = 6 \), numerator \( x^{2}-2x=x(x - 2) \), no common factors with denominator. Degree of numerator (2) > degree of denominator (1), so oblique asymptote. Divide \( x^{2}-2x \) by \( x - 6 \): \( x^{2}-2x=(x - 6)(x + 4)+24 \), so oblique asymptote \( y=x + 4 \). Domain: \( (-\infty,6)\cup(6,\infty) \)? Wait, no: Wait, denominator is \( x - 6 \), so \( x

eq6 \), but also numerator has no restrictions other than denominator? Wait, no, original function \( h(x)=\frac{x^{2}-2x}{x - 6} \), denominator is \( x - 6 \), so domain is \( (-\infty,6)\cup(6,\infty) \)? Wait, no, maybe I made a mistake. Wait, no, let's re - check:

Wait, for \( g(x) \), denominator is \( (x + 3)(x - 2) \), so domain \( (-\infty,-3)\cup(-3,2)\cup(2,\infty) \). For \( h(x) \), denominator is \( x - 6 \), so \( x
eq6 \), so domain \( (-\infty,6)\cup(6,\infty) \)? Wait, no, the given domain options: Wait, maybe I mis - analyzed \( h(x) \). Wait, no, the problem's domain options: Wait, the third function \( h(x) \): Wait, maybe the domain for \( h(x) \) is \( (-\infty,6)\cup(6,\infty) \), and it has an oblique asymptote. Also, when we look at the domain options, for \( h(x) \), the domain should be \( (-\infty,6)\cup(6,\infty) \)? Wait, no, the given domain options: Wait, the first domain option is \( (-\infty,6)\cup(6,\infty) \), which matches \( f(x) \) and \( h(x) \)? Wait, no, \( h(x) \)'s denominator is \( x - 6 \), so \( x
eq6 \), so domain \( (-\infty,6)\cup(6,\infty) \). And it has an oblique asymptote.

Now, mapping:

• \( f(x) \): domain \( (-\infty,6)\cup(6,\infty) \), horizontal asymptote \( y = - 2 \)
• \( g(x) \): domain \( (-\infty,-3)\cup(-3,2)\cup(2,\infty) \), horizontal asymptote \( y = 1 \)
• \( h(x) \): domain \( (-\infty,6)\cup(6,\infty) \), oblique asymptote

Answer:

• \( f(x)=\frac{-2x + 4}{x - 6} \): domain \( (-\infty,6)\cup(6,\infty) \), horizontal asymptote \( y=-2 \)
• \( g(x)=\frac{x^{2}-2x}{x^{2}+x - 6} \): domain \( (-\infty,-3)\cup(-3,2)\cup(2,\infty) \), horizontal asymptote \( y = 1 \)
• \( h(x)=\frac{x^{2}-2x}{x - 6} \): domain \( (-\infty,6)\cup(6,\infty) \), oblique asymptote