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Fine the horizontal asymptote of the graph of y = -4x^6+6x +3/8x^2=9x+3

User Montdidier
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Answer:

There are different methods to find the horizontal asymptote of a function, but one common approach is to take the limit of the function as x approaches positive or negative infinity. If the limit exists and is a finite number, then that is the horizontal asymptote.

To apply this method to the given function:

y = (-4x^6+6x +3)/(8x^2+9x+3)

As x becomes very large (either positive or negative), the highest power of x in the numerator and denominator dominates the other terms. This is because exponential functions grow or shrink much faster than polynomials as x goes to infinity or negative infinity.

Therefore, we can simplify the function by dividing both numerator and denominator by x^6:

y = (-4 + 6/x^5 + 3/x^6)/(8/x^4 + 9/x^5 + 3/x^6)

As x approaches infinity, all the terms with negative powers of x go to zero, leaving:

y ≈ (-4 + 0 + 0)/(0 + 0 + 0)

This is an indeterminate form of 0/0, which suggests that we need to do more algebraic simplification. One way to do this is to factor out the highest power of x in both numerator and denominator:

y ≈ (-4/x^6 + 6/x^11)/(8/x^4 + 9/x^5 + 3/x^6)

y ≈ (-4/x^6)(1 - 6x^5)/(8/x^4)(1 + 9x/x^4 + 3/x^2)

Now, as x approaches infinity, the fraction becomes dominated by the leading terms:

y ≈ (-4/x^6)(-6x^5)/(8/x^4)(9x/x^4)

y ≈ (24/72)(1/x)

As x goes to infinity, 1/x goes to zero, so:

y ≈ 1/3x

Therefore, the horizontal asymptote of the graph of y = (-4x^6+6x +3)/(8x^2+9x+3) is y = 1/3x.

User Shreedhar Bhat
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