Question

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

Answer 1

x = –2

Reason: f(x) = (5x+10)/(x^2+7x+10) = 5(x+2)/[(x+2)(x+5)] = 5/(x+5) for x ≠ -2, so x = -2 is a removable hole (x = -5 is a vertical asymptote).