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:

.

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

6 on both sides and twelve on the top and the bottom

step-by-step explanation:

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !