we know that
the expression represent the function
[tex]f(x)=\frac{4x^2-4x-8}{2x+2}[/tex]
Factor [tex]4[/tex] in the numerator
Factor [tex]2[/tex] in the denominator
so
[tex]f(x)=\frac{4(x^2-x-2)}{2(x+1)}[/tex]
the domain of the function is all real numbers except for [tex]x=-1[/tex] because the denominator can not be zero
Simplify the function
[tex]f(x)=\frac{4(x^2-x-2)}{2(x+1)}=\frac{4(x+1)(x-2)}{2(x+1)}[/tex]
[tex]f(x)=2(x-2)=2x-4[/tex]
Remember that for [tex]x=-1[/tex] the function does not exist
so
find the value of f(x) for [tex]x=-1[/tex] in the simplified function
[tex]f(x)=2x-4[/tex]
[tex]f(-1)=2*(-1)-4=-6[/tex]
The function has a discontinuity at point [tex](-1,-6)[/tex]
therefore
the answer is the option
graph of 2 x minus 4, with discontinuity at negative 1, negative 6
