anonymous
  • anonymous
Identify the vertical asymptotes of f(x) = 10 over quantity x squared minus 7x minus 30. x = 10 and x = 3 x = 10 and x = -3 x = -10 and x = 3 x = -10 and x = -3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[f(x) = \frac{ 10 }{ x^2 -7x -30 }\]
misty1212
  • misty1212
set \[x^2-7x-30=0\] solve for \(x\)
anonymous
  • anonymous
okay give me a sec

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anonymous
  • anonymous
fortunalely for you this factors as \[(x-10)(x+3)=0\] so the zeros are easy to find
misty1212
  • misty1212
yeah @satellite73 gave you a huge hint there
anonymous
  • anonymous
so its x = -10 and x = 3
misty1212
  • misty1212
no dear
misty1212
  • misty1212
\[x-10=0\\ x=10\]
misty1212
  • misty1212
also \[x+3=0\\ x=-3\]
anonymous
  • anonymous
oh so i have to bring it to the other side okay i see
misty1212
  • misty1212
riiiiiigght
anonymous
  • anonymous
thnx i have one more do you mind helping with?
misty1212
  • misty1212
sure
anonymous
  • anonymous
Identify the horizontal asymptote of f(x) = quantity 2 x minus 1 over quantity x squared minus 7 x plus 3. y = 0 y = 1 over 2 y = 2 No horizontal asymptote
anonymous
  • anonymous
\[f(x) = \frac{ 2x - 1 }{ x^2 - 7x +3 }\]
anonymous
  • anonymous
do i basically do the same thing as before?
misty1212
  • misty1212
no!!
misty1212
  • misty1212
horizontal not vertical
anonymous
  • anonymous
oh so what do i need to do?
misty1212
  • misty1212
the degree of the numerator is less than the degree of the denominator
misty1212
  • misty1212
the degree of the numerator is 1, the degree of the denominator i s2
misty1212
  • misty1212
that means the denominator grows faster than the numerator so the horizontal asymptote is \[y=0\]
anonymous
  • anonymous
thank you so much!!!
anonymous
  • anonymous
i gtg but i have lie 5 more questions to do later can you help with?
misty1212
  • misty1212
sure dear if i am on

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