Question

19) Given that f(x)x² - 8x+ 15x² - 25find the horizontal and vertical asymptotes using the limits of the function.A) No Vertical or Horizontal asymptotesB) No Vertical asymptotesHorizontal asymptote aty - 1Vertical asymptote at x = 5Horizontal asymptote at y = 1D) Vertical asymptote at x = -5Horizontal asymptote at y = 1

Answer 1

EXPLANATION

Since we have the function:

$f(x)=\frac{x^2-8x+15}{x^2}$

Vertical asymptotes:

$For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.$

Taking the denominator and comparing to zero:

$x+5=0$

The following points are undefined:

$x=-5$

Therefore, the vertical asymptote is at x=-5

Horizontal asymptotes:

$\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.$$If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.$$If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}$$\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}$$\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1$$\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}$$\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1$$y=\frac{1}{1}$$\mathrm{The\:horizontal\:asymptote\:is:}$$y=1$

In conclusion:

$\mathrm{Vertical}\text{ asymptotes}:\:x=-5,\:\mathrm{Horizontal}\text{ asymptotes}:\:y=1$