Can someone me with this? i need to find the points of discontinuity/limits for each of these. i think one point is 4, but i'm not sure about the others.

The answers are shown in the attached image

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Explanation:

Set the denominator x^4-8x^3+16x^2 equal to zero and solve for x

x^4-8x^3+16x^2 = 0
x^2(x^2-8x+16) = 0
x^2(x-4)^2 = 0
x^2 = 0 or (x-4)^2 = 0
x = 0 or x-4 = 0
x = 0 or x = 4

The x values 0 and 4 make the denominator zero

These x values lead to asymptote discontinuities because the numerator 8x-24 = 8(x-3) has no common factors which cancel with the denominator factors.

There are two vertical asymptotes

Let's see what happens when we plug in a value to the left of x = 0, say x = -1, we'd get
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(-1) = (8(-1)-24)/((-1)^4-8(-1)^3+16(-1)^2)
f(-1) = -1.28
So as x gets closer and closer to x = 0 from the left side, the f(x) is heading to negative infinity

Now plug in some value to the right of x = 0. I'm going to pick x = 1
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(1) = (8(1)-24)/((1)^4-8(1)^3+16(1)^2)
f(1) = -1.78 (approximate)
So as x gets closer and closer to x = 0 from the right side, the f(x) is heading to negative infinity

Overall, as x approaches 0 from either the left or right side of x = 0, the y value is heading off to negative infinity

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Repeat for values to the left and right of x = 4
We can't use x = 1 as it turns out that x = 3 is a root
But we can use something like x = 3.5 to find that...
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(3.5) = (8(3.5)-24)/((3.5)^4-8(3.5)^3+16(3.5)^2)
f(3.5) = 1.31 approx
So as x gets closer to x = 4 from the left, y is getting closer to positive infinity

Plug in x = 5 to find that
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(5) = (8(5)-24)/((5)^4-8(5)^3+16(5)^2)
f(5) = 0.64
which has the same behavior as the left side

So overall, as we approach x = 4, the y value is heading off to positive infinity

Again everything is summarized in the image attachment

Note: you could make a table of more values but they would effectively say what has already been said. It would be redundant busy work. However, its always good practice for function evaluation. 

Ithink your answer is c mark me brainliest