which statement is true about the discontinuities of the function f(x)? f(x) = (x - 5)/(3x² - 17x - 28)…

which statement is true about the discontinuities of the function f(x)? f(x) = (x - 5)/(3x² - 17x - 28) there are holes at x = 7 and x=-4/3. there are asymptotes at x = 7 and x=-4/3. there are asymptotes at x = -7 and x = 4/3. there are holes at (-7, 0) and x = 4/3.

which statement is true about the discontinuities of the function f(x)? f(x) = (x - 5)/(3x² - 17x - 28) there are holes at x = 7 and x=-4/3. there are asymptotes at x = 7 and x=-4/3. there are asymptotes at x = -7 and x = 4/3. there are holes at (-7, 0) and x = 4/3.

Answer

Explanation:

Step1: Factor the denominator

Factor $3x^{2}-17x - 28$. We need two numbers that multiply to $3\times(-28)=-84$ and add up to -17. The numbers are -21 and 4. So $3x^{2}-17x - 28=3x^{2}-21x + 4x-28=3x(x - 7)+4(x - 7)=(3x + 4)(x - 7)$. Then $f(x)=\frac{x - 5}{(3x + 4)(x - 7)}$.

Step2: Find the vertical - asymptotes

Vertical asymptotes occur where the denominator is zero and the numerator is non - zero. Set the denominator equal to zero: $(3x + 4)(x - 7)=0$. Solving $3x+4 = 0$ gives $x=-\frac{4}{3}$, and solving $x - 7=0$ gives $x = 7$. Since the numerator $x - 5$ is non - zero at $x=7$ and $x=-\frac{4}{3}$, the function has vertical asymptotes at $x = 7$ and $x=-\frac{4}{3}$.

Answer:

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