Question

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 .

Answer 1

1) function f(x)

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

2) factor the denominator:

3x^2 - 17x  - 28 = (3x + 4)(x - 7)

                        x - 5
=> f(x) = -----------------------
                (3x + 4) (x  - 7)

3) Find the limits when x → - 4/3 and when x → 7

Lim of f(x) when x → - 4/3 = +/- ∞

=> vertical assymptote x = - 4/3

Lim of f(x) when x → 7 = +/- ∞

=> vertical assymptote x = 7

Answer: there are assympotes at x = 7 and x = - 4/3

Answer 2

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

• A)There are holes at x = 7 and x = -4/3.
• B)There are asymptotes at x = 7 and x = -4/3 .

When do we get vertical asymptote for a function?

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

What are discontinuities?

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:

• A)There are holes at x = 7 and x = -4/3.
• B)There are asymptotes at x = 7 and x = -4/3 .

Learn more about discontinuities of a function here:

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