Which statement is true about the discontinuities of the function f(x)?

f (x) = StartFraction x + 1 Over 6 x squared minus 7 x minus 3 EndFraction
There are asymptotes at x = three-halves and x = negative one-third.
There are holes at x = three-halves and x = negative one-third.
There are asymptotes at x = negative three-halves and x = one-third.
There are holes at x = negative three-halves and x = one-third.

Using the discontinuity concept, it is found that:

There are asymptotes at and

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A function is discontinuous at a point outside it's domain, and at these points in the graph there will be asymptotes.

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The given function is:

It is a fraction, then the denominator cannot be zero, and the asymptotes will be at the values of x for which:

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Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

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This polynomial has roots such that , given by the following formulas:

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For this equation, . Thus:

Thus, there are asymptotes at and

The graph of the function is given at the end of the answer.

A similar problem is given at link

135 is your answer

step-by-step explanation:

try 180 times three then divide that number by 4 (sides) to get your answer. im glad to : )

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐