If the function f(x) = (x^2-9)/(x-3) were factored to f(x)= (x-3)(x+3)/(x-3) and then simplified to f(x)=(x+3)/1. How would you describe the point x=3?

Answer 1

if you graphed the original function f(x) = (x^2 -9)/(x-3) it's the same straight line as

the simplified f(x) = x+3,

BUT the original f(x) has a "hole" in it,

a discontinuity where the simplified line fills in the "hole" at the point (3,6)

(3,6) is not in the original function,

but it is in the "simplified" function.

x=3 makes the original function undefined. it makes the denominator = 0. You can't divide by zero. The resutl is an "error" message if you try it on a calculator. Divide 10/0 on a calculator and you get an "E" or similar redsult. It's undefined. It could be negarive infinity or positive infinity, but neither are real numbers.

Answer 2

The graph of y = f(x) is the same as the graph of y = x+3 except for a "hole" in the graph at the point (3,6). There is a removable discontinuity when x = 3.