Which statement is true about the discontinuities of the function f(x) = x-5/3x^2-17x-28 a)there are holes at x = 7 and . b)there are asymptotes at x = 7 and . c)there are asymptotes at x = –7 and . d)there are holes at (–7, 0) and .

Discontinuity is the non-continuity of the function on some point in function's domain. The statements (true) for given function are

Suppose that we have the function f(x) such that it is continuous for all values < a or > a and have got the values of f(x) going  to infinity or -ve infinity (from either side of x = a) as x goes near a , and being not defined at x = a, then at that point, there can be constructed a vertical line x = a and it will be called as vertical asymptote for f(x) at x = a

Holes in the graph of function, where its undefined, or non-continuous, is called discontinuity.

The points in the domain of the function over which its not continuous, is called point of discontinuity for that function.

The given function is

f(x) = (x-5)/(3x^2-17x-28)

The only discontinuity that can arise is when denominator becomes 0, as for the rest of the case, rational numbers will come.

The denominator can be factored as:

3x^2 -17x - 28 = 3x^2 -21x + 4x - 28 = 3x(x-7) +  4(x-7) = (3x+4)(x-7)

Thus, if x = 7, or x = -4/3, then the denominator becomes 0 and the function becomes undefined as division by 0 is not defined.

Thus, there are two discontinuities, one at 7, and one at -4/3 (on domain axis, or say x axis).

Since f(x) is continuous near points of 7, but not on 7, and function grows unbounded near x=7, thus, there can be constructed a vertical asymptote as x = 7 (and on x = -4/3 too, by similar logic)

The holes and asymptotes of the given function are:

Learn more about discontinuities of a function here:

link

really really sorry if i'm

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