Question 980068
To find the discontinuity of the function {{{(-x^2+x+20)/(x+4)}}} we will look at the denominator and determine what value for x will result in the denominator equalling 0, since the denominator cannot equal 0.  To do this, we will set our denominator equal to 0:


x + 4 = 0


Subtract 4 from both sides, giving us:


x = -4


Therefore, our discontinuity is x = -4


To find our zeroes, we will set our function equal to 0:


{{{(-x^2+x+20)/(x+4)=0}}}


Next, multiply both sides of the equation by (x + 4) to rid ourselves of our fraction on the left side of the equal sign.  This will result in:


{{{-x^2+x+20=0}}}


Third, multiply the entire equation by -1 to make the equation easier to factor.  This will give us:


{{{x^2-x-20=0}}}


Fourth, factor the left side of the equation.  This will result in:


{{{(x-5)(x+4)=0}}}


Set each set of parentheses equal to zero and solve for x:


{{{(x-5)=0}}} -----> {{{x=5}}}


{{{(x+4)=0}}} -----> {{{x=-4}}}


Since we have determined that -4 is NOT a zero since it will result in a zero value in our denominator, our only zero is {{{x=5}}}


We can verify by substituting the x in our original equation with 5:


{{{(-(5^2)+(-5)+20)/(5+4)=0}}} -----> {{{(-25-5+20)/9=0}}} -----> {{{0/9=0}}} -----> {{{0=0}}}.


Since {{{0=0}}} is a true statement, 5 is, in fact, the zero of our function.