Question

what are the discontinuities of the function f(x)=x^2-36/4x+24?

Answer 1

I assume that is supposed to be\[f(x) = \frac{x^2-36}{4x+24}\], though what you have written is
\[f(x) = x^2-\frac{36}{4x}+24\]by the rules of operator precedence. If you won't use the equation editor, at least put in parentheses to make the meaning unambiguous, such as writing it as

f(x) = (x^2-36)/(4x+24)

The discontinuities of a fraction like this take place wherever the value of the denominator is equal to 0. We can't divide by 0, so we can't find the value of the function at such a point. So, to find the discontinuities of this function, we set the denominator equal to 0 and solve for the value(s) of \(x\) that make it happen:

\[4x+24 = 0\]\[x =\]

Easy, right?

Answer 2

@whpalmer4 x=4

Answer 3

0? @amoodarya

Answer 4

or 6

Answer 5

as "whpalmer4" says
find when denominator goes to zero
denominator is 4x+24
so
4x+24=0
x=-24/4=-6