Question

5 what are the vertical asymptotes for the graph of the rational function? ( y = \frac{x + 8}{x^2 - 9x + 14} ) (a) ( x = -8 ) (b) ( x = -2, x = -7 ) (c) ( x = 2, x = -7 ) (d) ( x = 2, x = 7 )

Explanation:

Step1: Recall Vertical Asymptote Rule

For a rational function \( y = \frac{f(x)}{g(x)} \), vertical asymptotes occur where \( g(x) = 0 \) (and \( f(x)
eq 0 \) at those points). Here, \( g(x)=x^2 - 9x + 14 \).

Step2: Factor the Denominator

Factor \( x^2 - 9x + 14 \). We need two numbers that multiply to \( 14 \) and add to \( -9 \). Those numbers are \( -2 \) and \( -7 \), so \( x^2 - 9x + 14=(x - 2)(x - 7) \)? Wait, no: \( (x - 2)(x - 7)=x^2 - 9x + 14 \)? Wait, \( 2\times7 = 14 \), \( 2 + 7 = 9 \), so actually, \( x^2 - 9x + 14=(x - 2)(x - 7) \)? Wait, no, \( (x - 2)(x - 7)=x^2 - 9x + 14 \)? Let's check: \( x\times x=x^2 \), \( x\times(-7)+(-2)\times x=-7x - 2x=-9x \), \( (-2)\times(-7)=14 \). Yes. Wait, but wait, the roots of \( (x - 2)(x - 7)=0 \) are \( x = 2 \) and \( x = 7 \). Wait, but let's check the numerator at these points. The numerator is \( x + 8 \). At \( x = 2 \), numerator is \( 10
eq0 \); at \( x = 7 \), numerator is \( 15
eq0 \). So vertical asymptotes are at \( x = 2 \) and \( x = 7 \). Wait, but let's re - factor. Wait, maybe I made a mistake. Wait, \( x^2-9x + 14 \): discriminant \( D=81 - 56 = 25 \), so roots are \( \frac{9\pm5}{2} \). \( \frac{9 + 5}{2}=7 \), \( \frac{9 - 5}{2}=2 \). So yes, roots are \( x = 2 \) and \( x = 7 \). So the vertical asymptotes are \( x = 2 \) and \( x = 7 \).

Answer:

D. \( x = 2 \), \( x = 7 \)