Question

What value is a discontinuity of quantity x squared plus 3 x minus 4 divided by quantity x squared plus x minus 12

Answer 1

(x² + 3x + 4) / (x² + x - 12) 

Note that you can't divide by zero, and discontinuities are a result of that. We are looking for: 
x² + x - 12 = 0 
(x + 4)(x - 3) = 0 
x = 3, -4 

a. is one of those values, and is the answer.

Answer 2

Answer:

x=3 is the point of discontinuity.

Step-by-step explanation:

We have been given the expression:

$\frac{x^2+3x-4}{x^2+x-12}$

We have to find the discontinuity

To find the discontinuity we first have to factorize the given expression

$\frac{x^2+4x-x-4}{x^2+4x-3x-12}$

$\Rightarrow\frac{x(x+4)-1(x+4)}{x(x+4)-3(x+4)}$

$\Rightarrow\frac{(x-1)(x+4)}{(x-3)(x+4)}$

Cancel out the common factors from the numerator and denominator which is (x+4) so, we get:

$\Rightarrow\frac{(x-1)}{(x-3)}$

x=-4 is zero for both numerator and denominator.

Hence, put x=-4 in the above equation we get

$\Rightarrow\frac{(4-1)}{(4-3)}=3$

Hence, x=3 is the point of discontinuity.