which statement is true about the discontinuities of the function f(x)?\n$f(x) = \\frac{x - 5}{3x^2 - 17x…

which statement is true about the discontinuities of the function f(x)?\n$f(x) = \\frac{x - 5}{3x^2 - 17x - 28}$\n- there are holes at $x = 7$ and $x = -\\frac{4}{3}$.\n- there are asymptotes at $x = 7$ and $x = -\\frac{4}{3}$.\n- there are asymptotes at $x = -7$ and $x = \\frac{4}{3}$.\n- there are holes at $(-7, 0)$ and $x = \\frac{4}{3}$.

which statement is true about the discontinuities of the function f(x)?\n$f(x) = \\frac{x - 5}{3x^2 - 17x - 28}$\n- there are holes at $x = 7$ and $x = -\\frac{4}{3}$.\n- there are asymptotes at $x = 7$ and $x = -\\frac{4}{3}$.\n- there are asymptotes at $x = -7$ and $x = \\frac{4}{3}$.\n- there are holes at $(-7, 0)$ and $x = \\frac{4}{3}$.

Answer

Answer:

There are asymptotes at ( x = 7 ) and ( x = -\frac{4}{3} ).

Explanation:

Step1: Factor the denominator

Factor ( 3x^2 - 17x - 28 ). We need two numbers that multiply to ( 3\times(-28)= -84 ) and add to ( -17 ). The numbers are ( -21 ) and ( 4 ). So, [ \begin{align*} 3x^2 - 17x - 28&=3x^2 - 21x + 4x - 28\ &=3x(x - 7)+4(x - 7)\ &=(3x + 4)(x - 7) \end{align*} ]

Step2: Analyze the function's discontinuities

The function is ( f(x)=\frac{x - 5}{(3x + 4)(x - 7)} ). The numerator ( x - 5 ) and the denominator have no common factors (since ( x - 5 ) is not equal to ( 3x + 4 ) or ( x - 7 ) for all ( x )). So, the function has vertical asymptotes where the denominator is zero, i.e., when ( 3x + 4 = 0 ) (solving gives ( x = -\frac{4}{3} )) and when ( x - 7 = 0 ) (solving gives ( x = 7 )). There are no holes because there are no common factors between the numerator and the denominator.