Find the vertical, horizontal, and slant asymptotes, if any, for f(x)=5x^3+29x^2-140x+21/x^2+6x-27

Vertical asymptotes: x=-9 and x=3

Slant asymptote: y=5x-1

Step-by-step explanation:

Given

f(x)=(5x^3+29x^2-140x+21)/(x^2+6x-27)

For vertical asymptote, the denominator is put equal to zero,so

x^2+6x-27=0

Factorizing

x^2+9x-3x-27=0

x(x+9)-3(x+9)=0

(x+9)(x-3)=0

So,

x=-9 ;x=3

As the degree of the numerator is greater than the denominator the function will not have horizontal asymptote but it will have a slant asymptote which will be calculated by long division.

After dividing 5x^3+29x^2-140x+21  by x^2+6x-27  we get

Quotient: 5x-1

Remainder: x-6

We only need the quotient for the slant asymptote,

So the slant asymptote is y = 5x -1  ..

ad = ae + ed = y + 7

similar triangles therefor

ac/ab = ad/ae

6/4 = (y + 7) / y

cross multiply

6y = 4(y + 7)

6y = 4y + 28

2y = 28

y = 14

step-by-step explanation:

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step-by-step explanation: