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Determine the asymptotes of the function: y=x^3-5x^2+4x-25/x^2-4x+3

(horizontal, vertical or slant)

2 Answers

5 votes

Answer: D

Explanation:

EDGE 2021

User Jonasb
by
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2 votes

Answer:

Vertical A @ x=3 and x=1

Horizontal A nowhere since degree on top is higher than degree on bottom

Slant A @ y=x-1

Explanation:

I'm going to look for vertical first:

I'm going to factor the bottom first: (x-3)(x-1)

So we have possible vertical asymptotes at x=3 and at x=1

To check I'm going to see if (x-3) is a factor of the top by plugging in 3 and seeing if I receive 0 (If I receive 0 then x=3 gives me a hole)

3^3-5(3)^2+4(3)-25=-31 so it isn't a factor of the top so you have a vertical asymptote at x=3

Let's check x=1

1^3-5(1)^2+4(1)-25=-25 so we have a vertical asymptote at x=1 also

There is no horizontal asymptote because degree of top is bigger than degree of bottom

There is a slant asympote because the degree of top is one more than degree of bottom (We can find this by doing long division)

x -1

--------------------------------------------------

x^2-4x+3 | x^3-5x^2+4x-25

- ( x^3-4x^2+3x)

--------------------------------

-x^2 +x -25

- (-x^2+4x-3)

---------------------

-3x-22

So the slant asymptote is to x-1

User Mhall
by
7.4k points

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