select the correct answer.\nrational function h is continuous, with a horizontal asymptote at y = 1. which…

select the correct answer.\nrational function h is continuous, with a horizontal asymptote at y = 1. which function could be function h?\na. \\( h(x) = \\frac{x^2 - 16}{x^2 + 16} \\)\nb. \\( h(x) = \\frac{x^2 + 16}{x^2 - 16} \\)\nc. \\( h(x) = \\frac{x^2 - 16}{x - 4} \\)\nd. \\( h(x) = \\frac{x + 4}{x^2 + 16} \\)

select the correct answer.\nrational function h is continuous, with a horizontal asymptote at y = 1. which function could be function h?\na. \\( h(x) = \\frac{x^2 - 16}{x^2 + 16} \\)\nb. \\( h(x) = \\frac{x^2 + 16}{x^2 - 16} \\)\nc. \\( h(x) = \\frac{x^2 - 16}{x - 4} \\)\nd. \\( h(x) = \\frac{x + 4}{x^2 + 16} \\)

Answer

Explanation:

Step1: Recall Horizontal Asymptote Rules

For a rational function ( h(x)=\frac{f(x)}{g(x)} ), if the degrees of ( f(x) ) and ( g(x) ) are equal, the horizontal asymptote is ( y = \frac{\text{leading coefficient of } f(x)}{\text{leading coefficient of } g(x)} ). If degree of ( f(x) < ) degree of ( g(x) ), horizontal asymptote is ( y = 0 ). If degree of ( f(x) > ) degree of ( g(x) ), no horizontal asymptote (or oblique asymptote). Also, check continuity (no vertical asymptotes, i.e., denominator never zero for all real ( x )).

Step2: Analyze Option A

Function: ( h(x)=\frac{x^2 - 16}{x^2 + 16} )

  • Degree of numerator (( f(x)=x^2 - 16 )): 2. Degree of denominator (( g(x)=x^2 + 16 )): 2 (equal).
  • Leading coefficients: numerator leading coefficient is 1, denominator leading coefficient is 1. So horizontal asymptote ( y=\frac{1}{1}=1 ).
  • Denominator: ( x^2 + 16 ). For all real ( x ), ( x^2 \geq 0 ), so ( x^2 + 16 \geq 16 > 0 ). No vertical asymptotes, so function is continuous (as domain is all real numbers).

Step3: Analyze Option B

Function: ( h(x)=\frac{x^2 + 16}{x^2 - 16} )

  • Denominator: ( x^2 - 16=(x - 4)(x + 4) ). At ( x = 4 ) or ( x = -4 ), denominator is zero. So vertical asymptotes at ( x = 4, -4 ), function is not continuous (discontinuous at these points). Eliminate B.

Step4: Analyze Option C

Function: ( h(x)=\frac{x^2 - 16}{x - 4} )

  • Simplify: ( \frac{(x - 4)(x + 4)}{x - 4}=x + 4 ) (for ( x \neq 4 )). But at ( x = 4 ), denominator is zero (hole at ( x = 4 ), but still discontinuous at ( x = 4 )). Also, degree of numerator (2) > degree of denominator (1), so no horizontal asymptote (oblique asymptote ( y = x + 4 ) after simplification). Eliminate C.

Step5: Analyze Option D

Function: ( h(x)=\frac{x + 4}{x^2 + 16} )

  • Degree of numerator (1) < degree of denominator (2). So horizontal asymptote ( y = 0 ), not ( y = 1 ). Eliminate D.

Answer:

A. ( h(x) = \frac{x^2 - 16}{x^2 + 16} )