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Does 6(x+2)>x-3 have a solution
1 answer:
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Given:
The function is
To find:
The asymptotes and zero of the function.
Solution:
We have,
For zeroes, f(x)=0.
Therefore, zero of the function is 0.
For vertical asymptote equate the denominator of the function equal to 0.
Taking square root on both sides, we get
So, vertical asymptotes are x=-4 and x=4.
Since degree of denominator is greater than degree of numerator, therefore, the horizontal asymptote is y=0.