Respuesta :
Answer:
D.
Step-by-step explanation:
To find the discontinuity given this is a fraction, we only have to worry about dividing by 0.
So we only have to worry about x-1 being 0.
x-1=0
Add 1 on both sides gives:
x=1
So there is a discontinuity when x=1.
Now there cannot also be a zero at this number since the function doesn't exist there.
However if you plug 1 into the fraction before simplification any you get 0/0 then you have a hole at x=1.
[tex]\frac{3(1)^2+(1)-4}{1-1}=\frac{3+1-4}{0}=\frac{4-4}{0}=\frac{0}{0}[/tex].
So we do indeed have a hole at x=1.
To find point that would make this function continuous at x=1, we will simplify our fraction and then plug in 1:
I already know x-1 is a factor of the top because x=1 made it zero on top:
[tex]\frac{(x-1)(3x+4)}{x-1}[/tex]
The (x-1)'s cancel:
[tex](3x+4)[/tex]
Now 3x+4 evaluated at x=1 gives 3(1)+4=3+4=7.
So the hole is at (1,7).
The choice is either C or D.
The zero is when 3x+4 is 0.
This is the numerator after simplification. The top of the fraction is what determines if the fraction itself will be 0 since 0/anything is 0 except when you are dividing by 0.
3x+4=0
Subtract 4 on both sides gives:
3x=-4
Divide both sides by 3:
x=-4/3
So the zero is at (-4/3,0).
