What are the discontinuity and zero of the function f(x) = quantity x squared plus 8 x plus 7 end quantity over quantity x plus 1?

Answer:The last one is the one you want.Step-by-step explanation:Discontinuities are found in the denominator, where a certain value of x causes the denominator to equal 0 (which is a very big NO NO in math!). These discontinuities are representative of the vertical asymptotes or "holes" of the function.  x cannot exist at these values so the function does not exist at these values.  Zeros are found when you set the numerator equal to 0 and factor, if needed, to solve for x.  These zeros are representative of the horizontal asymptotes of the function. Let's factor your function to see what we have for specific values of x:[tex]f(x)=\frac{(x+1)(x+7)}{(x+1)}[/tex]Now there's a bit of an explanation that needs to go along with this answer.  The discontinuities are found where the values of x cause the denominator to go to 0.  IF we can cancel out a term in the denominator with one exactly like it in the numerator, then what's left behind is a hole.  IF we cannot cancel out a term in the denominator with a term in the numerator then it is a vertical asymptote.  As you can see, our (x + 1) terms can cancel out in the numerator and the denominator, so this type of discontinuity is called a "removeable" discontinuity and looks like a hole in the function.  So the discontinuity is at an x value of -1.  We need the y value that goes along with that, so plug in a -1 for x and solve for y.  Keep in mind that we have canceled out the (x + 1)'s in both the numerator and the denominator:y = (-1 + 7) so y = 6The discontinuity is at (-1, 6).Now to find the zero.  When we cancel out the numerator with the denominator, what's left is (x + 7).  If we set that equal to 0 and solve for x we get that x = -7.  Subbing that in, we can solve for y:(-7 + 7) = 0.  The zero is at (-7, 0).Your choice for your answer is the last one listed.