The graph below shows the function f(x)=5x+10/x^2+7x+10 Where is the removable discontinuity of f(x) located?

For this case we have the following function:
 [tex] f (x) = (5x + 10) / (x ^ 2 + 7x + 10) [/tex] 
 Let's rewrite the function:
 [tex] f (x) = (5x + 10) / ((x + 2) (x + 5)) f (x) = (5 (x + 2)) / ((x + 2) (x + 5))[/tex]
 We set the denominator to zero to see the values of x for which it is not defined:
 [tex] (x + 2) (x + 5) = 0 [/tex] 
 From here, we get:
 [tex]x = -2 x = -5 [/tex]
 There is a removable discontinuity at x = -2, since by rewriting the function we have:
 [tex] f (x) = 5 / (x + 5) [/tex] 
 Answer:
 the removable discontinuity of f (x) is located at:
 x = -2