f(x)=7x/(x² - 16)\nplot points between and beyond each x - intercept and vertical asymptote. find the value…
f(x)=7x/(x² - 16)\nplot points between and beyond each x - intercept and vertical asymptote. find the value of the function at the given value of x.\nx -7 -6 -1 1 6 7\nf(x)=7x/(x² - 16) (simplify your answers.)\nuse the information obtained in the previous steps to graph the function between and beyond the vertical asymptotes. choose the correct graph below\na. graph image\nb. graph image\nc. graph image\nd. graph image
Answer
Explanation:
Step1: Substitute x = - 7 into the function
$f(-7)=\frac{7\times(-7)}{(-7)^{2}-16}=\frac{-49}{49 - 16}=\frac{-49}{33}$
Step2: Substitute x = - 6 into the function
$f(-6)=\frac{7\times(-6)}{(-6)^{2}-16}=\frac{-42}{36 - 16}=\frac{-42}{20}=-\frac{21}{10}$
Step3: Substitute x = - 1 into the function
$f(-1)=\frac{7\times(-1)}{(-1)^{2}-16}=\frac{-7}{1 - 16}=\frac{-7}{-15}=\frac{7}{15}$
Step4: Substitute x = 1 into the function
$f(1)=\frac{7\times1}{1^{2}-16}=\frac{7}{1 - 16}=\frac{7}{-15}=-\frac{7}{15}$
Step5: Substitute x = 6 into the function
$f(6)=\frac{7\times6}{6^{2}-16}=\frac{42}{36 - 16}=\frac{42}{20}=\frac{21}{10}$
Step6: Substitute x = 7 into the function
$f(7)=\frac{7\times7}{7^{2}-16}=\frac{49}{49 - 16}=\frac{49}{33}$
Answer:
When $x=-7$, $f(-7)=\frac{-49}{33}$; when $x = - 6$, $f(-6)=-\frac{21}{10}$; when $x=-1$, $f(-1)=\frac{7}{15}$; when $x = 1$, $f(1)=-\frac{7}{15}$; when $x = 6$, $f(6)=\frac{21}{10}$; when $x = 7$, $f(7)=\frac{49}{33}$