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Bogdan [553]
3 years ago
6

MAX POINTS

Mathematics
1 answer:
mrs_skeptik [129]3 years ago
3 0

Let's do

\\ \rm\dashrightarrow y=\dfrac{3}{x-h}+k

Release k for some while

If

  • we take h=0

So

  • y=3/x

So vertical asymptote is at origin now

It mentioned that it's at x=-5 so we need to change x

  • put -5 in place of h

\\ \rm\dashrightarrow y=\dfrac{3}{x-(-5)}

\\ \rm\dashrightarrow y=\dfrac{3}{x+5}

  • Vertical asymptote at x=-5

Now

  • for k=0 horizontal asymptote at origin

But it's given

  • y is at 12

Same put y=12 in place of k

\\ \rm\dashrightarrow y=\dfrac{3}{x+5}+12

  • h=-5
  • k=12

Graph attached for verification

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