The graph below shows the function f (x) = StartFraction 5 x + 10 Over x squared + 7 x + 10 EndFraction.

On a coordinate plane, a hyperbola is shown. Both curves approach x = negative 5. There is a hole at x = negative 2.

Where is the removable discontinuity of f(x) located?
x = –5
x = 0
x = –2
x = 5

A removable discontinuity happens where y=(0/0).

First factor the numerator and denominator.

Factor the GCF from the numerator, which would be 5:

Factoring 5 out of 5x gives us x, and factoring 5 out of 10 gives us 2, which now gives us 5(x+2)

Now to factor the denominator, we want factors of c, 10, that sum to b, 7: 5*2=10 and 5+2=7, so this will give us (x+5)(x+2).

Now we have:

5(x+2) / (x+5)(x+2)

Find the common factors in both the numerator and denominator:

(x+2) is the common factor. Set this equal to zero and solve for x:

x+2=0 x = -2

The answer would be x = -2