homework assignment #6\ncategory: quiz\ncurrent learning objective: identifying vertical asymptotes of…

homework assignment #6\ncategory: quiz\ncurrent learning objective: identifying vertical asymptotes of rational functions\nquestion 23 practice similar questions\nscore: 0 of 1 point\nlet ( f(x)=\frac{3 x^{2}-11 x - 20}{3 x^{2}+8 x + 5} )\nfind the vertical asymptotes\na ( x=\frac{5}{3}, x = 1 )\nb ( x=-\frac{5}{3}, x = 1 )\nc ( x=-\frac{5}{3}, x=-1 )\nd ( x=\frac{5}{3}, x=-1 )\nsubmit answer attempts: 0/2\nfind this question difficult? do you know you can practice another version of this question?\nneed help?\nmangpt isnt available for this question, but you can still master it if your instructor enabled practice similar questions

homework assignment #6\ncategory: quiz\ncurrent learning objective: identifying vertical asymptotes of rational functions\nquestion 23 practice similar questions\nscore: 0 of 1 point\nlet ( f(x)=\frac{3 x^{2}-11 x - 20}{3 x^{2}+8 x + 5} )\nfind the vertical asymptotes\na ( x=\frac{5}{3}, x = 1 )\nb ( x=-\frac{5}{3}, x = 1 )\nc ( x=-\frac{5}{3}, x=-1 )\nd ( x=\frac{5}{3}, x=-1 )\nsubmit answer attempts: 0/2\nfind this question difficult? do you know you can practice another version of this question?\nneed help?\nmangpt isnt available for this question, but you can still master it if your instructor enabled practice similar questions

Answer

Explanation:

Step1: Factor numerator and denominator

For the numerator (3x^{2}-11x - 20), using the formula (ax^{2}+bx + c), we find two numbers (m) and (n) such that (m\times n=3\times(- 20)=-60) and (m + n=-11). We get (m=-15) and (n = 4). [3x^{2}-11x - 20=3x^{2}-15x+4x - 20=3x(x - 5)+4(x - 5)=(3x + 4)(x-5)] For the denominator (3x^{2}+8x + 5), we find two numbers (p) and (q) such that (p\times q=3\times5 = 15) and (p + q=8). We get (p = 3) and (q = 5). [3x^{2}+8x + 5=3x^{2}+3x+5x + 5=3x(x + 1)+5(x + 1)=(3x + 5)(x + 1)] So (f(x)=\frac{(3x + 4)(x - 5)}{(3x + 5)(x + 1)})

Step2: Find vertical asymptotes

Vertical asymptotes occur at the values of (x) that make the denominator equal to zero (while the numerator is not zero at those values). Set the denominator ((3x + 5)(x + 1)=0) (3x+5 = 0) gives (x=-\frac{5}{3}) (x + 1=0) gives (x=-1) We check that the numerator is not zero at (x =-\frac{5}{3}) and (x=-1). When (x=-\frac{5}{3}), ((3\times(-\frac{5}{3})+4)(-\frac{5}{3}-5)=(-5 + 4)(-\frac{5}{3}-5)\neq0) When (x=-1), ((3\times(-1)+4)(-1 - 5)=(1)(-6)\neq0)

Answer:

C. (x =-\frac{5}{3},x=-1)