Question

standard 9 homework
score: 6/27 answered: 6/20
question 7
find the vertical asymptote(s) of ( f(x)=\frac{8 x - 1}{x^{2}+9 x + 14} ).
the vertical asymptote(s) are ( x=)
if there is more than one asymptote, enter your answers separated by a comma.
question help: message instructor post to forum

Explanation:

Step1: Factor the denominator

We factor \(x^{2}+9x + 14\). Using the formula \(x^{2}+(a + b)x+ab=(x + a)(x + b)\), where \(a = 2\) and \(b=7\) (since \(2\times7 = 14\) and \(2 + 7=9\)). So, \(x^{2}+9x + 14=(x + 2)(x + 7)\).

Step2: Find the values that make the denominator zero

Set the denominator equal to zero: \((x + 2)(x + 7)=0\).
Using the zero - product property \(ab = 0\) implies \(a = 0\) or \(b = 0\).
If \(x+2=0\), then \(x=-2\). If \(x + 7=0\), then \(x=-7\).
The numerator \(8x-1\) is not zero when \(x=-2\) (\(8\times(-2)-1=-16 - 1=-17
eq0\)) and when \(x=-7\) (\(8\times(-7)-1=-56 - 1=-57
eq0\)).

Answer:

\(-2,-7\)