What are the discontinuity and zero of the function f(x) = quantity x squared plus 5 x plus 4 end quantity over quantity x plus 4?a Discontinuity at (−4, −3), zero at (−1, 0)b Discontinuity at (−4, −3), zero at (1, 0)c Discontinuity at (4, 3), zero at (−1, 0)d Discontinuity at (4, 3), zero at (1, 0)

Answer:Option a -  Discontinuity at (−4, −3), zero at (−1, 0)Step-by-step explanation:Given : Function [tex]\dfrac{x^2+5x+4}{x+4}[/tex]To find : What are the discontinuity and zero of the function ?Solution : To find the discontinuity for our given function [tex]\dfrac{x^2+5x+4}{x+4}[/tex]We will equate denominator to 0, [tex]x+4=0[/tex] [tex]x=-4[/tex]Now we simplify the expression,[tex]f(x)=\dfrac{x^2+5x+4}{x+4}[/tex][tex]f(x)=\dfrac{(x+4)(x+1)}{(x+4)}[/tex][tex]f(x)=x+1[/tex]Since the denominator term is cancelled out, so our give function has a removable discontinuity.Now, we will find value of y by substituting x=-4 in function [tex]f(x)=x+1[/tex]. [tex]f(-4)=-4+1[/tex][tex]f(-4)=-3[/tex]The discontinuity of given function is at point (-4,-3) .For zeros of the function put function equate to zero. [tex]x+1=0[/tex][tex]x=-1[/tex]The zero is at (-1,0).Therefore, Option a is correct.