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Sagot :
EXPLANATION
Since we have the function:
[tex]f(x)=\frac{x^2-8x+15}{x^2}[/tex]Vertical asymptotes:
[tex]For\:rational\:functions,\:the\:vertical\:asymptotes\:are\:the\:undefined\:points,\:also\:known\:as\:the\:zeros\:of\:the\:denominator,\:of\:the\:simplified\:function.[/tex]Taking the denominator and comparing to zero:
[tex]x+5=0[/tex]The following points are undefined:
[tex]x=-5[/tex]Therefore, the vertical asymptote is at x=-5
Horizontal asymptotes:
[tex]\mathrm{If\:denominator's\:degree\:>\:numerator's\:degree,\:the\:horizontal\:asymptote\:is\:the\:x-axis:}\:y=0.[/tex][tex]If\:numerator's\:degree\:=\:1\:+\:denominator's\:degree,\:the\:asymptote\:is\:a\:slant\:asymptote\:of\:the\:form:\:y=mx+b.[/tex][tex]If\:the\:degrees\:are\:equal,\:the\:asymptote\:is:\:y=\frac{numerator's\:leading\:coefficient}{denominator's\:leading\:coefficient}[/tex][tex]\mathrm{If\:numerator's\:degree\:>\:1\:+\:denominator's\:degree,\:there\:is\:no\:horizontal\:asymptote.}[/tex][tex]\mathrm{The\:degree\:of\:the\:numerator}=1.\:\mathrm{The\:degree\:of\:the\:denominator}=1[/tex][tex]\mathrm{The\:degrees\:are\:equal,\:the\:asymptote\:is:}\:y=\frac{\mathrm{numerator's\:leading\:coefficient}}{\mathrm{denominator's\:leading\:coefficient}}[/tex][tex]\mathrm{Numerator's\:leading\:coefficient}=1,\:\mathrm{Denominator's\:leading\:coefficient}=1[/tex][tex]y=\frac{1}{1}[/tex][tex]\mathrm{The\:horizontal\:asymptote\:is:}[/tex][tex]y=1[/tex]In conclusion:
[tex]\mathrm{Vertical}\text{ asymptotes}:\:x=-5,\:\mathrm{Horizontal}\text{ asymptotes}:\:y=1[/tex]
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