Only hole of function [tex]f(x) = \frac{x^{2}+7x+10 }{x^{2}+9x+20 }[/tex] is at x=(-4)
Step-by-step explanation:
Given the function is [tex]f(x) = \frac{x^{2}+7x+10 }{x^{2}+9x+20 }[/tex]
In order to find holes of any function, you should find when function is becoming undefined or say " infinity"
Given function is polynomial function.
It will become undefined become denominator become zero
[tex]x^{2}+9x+20=0[/tex]
Solving for x value when denominator become zero
[tex]x^{2}+9x+20=0\\x^{2}+5x+4x+20=0\\x(x+5)+4(x+5)=0\\(x+4)(x+5)=0[/tex]
we get possible holes at x=(-4) and x=(-5)
Check whether you can eliminate any holes
Now, Solving for x value when numerator become zero
[tex]x^{2}+7x+10=0\\x^{2}+5x+2x+10=0\\(x+5)(x+2)=0[/tex]
x=(-5) and x=(-2)
x=(-5) is common is both numerator and denominator.
So that, we can eliminate it.
[tex]f(x) = \frac{(x+5)(x+2)}{(x+5)(x+4)}[/tex]
[tex]f(x) = \frac{(x+2)}{(x+4)}[/tex]
Therefore, Only hole of function [tex]f(x) = \frac{x^{2}+7x+10 }{x^{2}+9x+20 }[/tex] is at x=(-4)