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)=(x^2-8x+15)/(x^2)`

Vertical asymptotes:

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=(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=(1)/(1)`
`\mathrm{The\:horizontal\:asymptote\:is:}`
`y=1`

In conclusion:

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