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11111nata11111 [884]
3 years ago
8

Find the vertical and horizontal asympotote for (×-2) (×+3)/ (2×+2)

Mathematics
1 answer:
nekit [7.7K]3 years ago
6 0

\bf \cfrac{(x-2)(x+3)}{2x+2}\implies \cfrac{x^2+x-6}{2x+2}~~ \begin{array}{llll} \leftarrow \textit{2nd degree polynomial}\\ \leftarrow \textit{1st degree polynomial} \end{array} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{vertical asymptote}}{2x+2=0}\implies 2x=-2\implies x=-\cfrac{2}{2}\implies x=-1


when the degree of the numerator is greater than the denominator's, then it has no horizontal asymptotes.


quick note:

when the degree of the numerator is 1 higher than the degree of the denominator, then it has an slant-asymptote, so this one has a slant-asymptote.

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Answer:

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Step-by-step explanation:

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We know the width. Now find the length by using the area formula and inserting known values:

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The length of the rectangle is 36. (You can check: 4 times 9 is 36)

Now find the perimeter:

P=2l+2w

Insert values:

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