Determine the asymptotes of the function: y=x^3-5x^2+4x-25/x^2-4x+3 (horizontal, vertical or slant)

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  

Step-by-step 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

just take the area of the square and then the area of the circle then subtract the circle area from the squares area

step-by-step explanation: