The equation of Vertical asymptote is x+4=0
Step-by-step explanation:
Figure shows vertical asymptote at x=4=0
Where redline is f(x) and blueline is vertical asymptote
Given the function is f(x) = [tex]\frac{x^{2}+7x+10 }{x^{2}+9x+20 }[/tex]
Step 1 : Simplifying denominator and numerator
For denominator
[tex]x^{2}+9x+20=0[/tex]
[tex]x^{2}+5x+4x+20=0[/tex]
[tex]x(x+5)+4(x+5)=0[/tex]
[tex](x+4)(x+5)=0[/tex]
For numerator
[tex]x^{2}+7x+10=0[/tex]
[tex]x^{2}+5x+2x+10=0[/tex]
[tex](x+5)(x+2)=0[/tex]
Step2 : Finding Vertical asymptote
After simplification of f(x)
f(x) = [tex]\frac{(x+5)(x+2)}{(x+5)(x+4)}[/tex]
f(x) = [tex]\frac{(x+2)}{(x+4)}[/tex]
Here, Denominator will give Vertical asymptote.
Therefore, Vertical asymptote is x+4=0
