Enter the letter of the graphed
function.
Plz help! And plz explain thx!!!

Option a

Step-by-step explanation:

One way to solve part of this is to look at the values of y between each x intercept.

Our x intercepts (or when y=0) occur when x=-2, x=1, and x=3. We know this by looking at the graph.

First, we have the space when x is between -∞ and -2. This is negative on the graph, and as a result, there should be an odd number of negatives as an odd number of negative numbers multiplied by each other is negative.

For a, (x+2) is negative when x is between -∞ and -2, (x-1) is negative, and (x-3) is negative as well. So far, this works

For b, we have to count (x-3) twice because it is squared. (x+2) is negative, (x-1) is negative, and (x-3) is negative twice. This is 4 negatives, so it doesn't work. This implies that between x=-∞ and x=-2, y is positive, which isn't the case

For c, (x+2) is negative, (x-3) is negative, and (x-1) is positive twice, making it 4 negatives and letting us rule it out

For d, (x-1) is negative, (x-3) is negative, and (x+2) is negative twice, making it 4 negatives and letting use rule it out

For e, (x-1) is negative, (x-3) is negative, and (x+2) is negative 3 times, making it 5 negatives and letting us keep it

After our first set of checks, we can see that only a and e are possibilities. Because the only thing separating the two possibilities is (x+2)², which is always positive past x=-2, and multiplying a number by a positive keeps its sign (e.g -5 * 5 = -25, which is still negative), all other rounds of checks would result in the same answer.

Another way to figure it out is to take an example point. We can see on the graph that when x=0, y=6. We can test that in each of our two remaining possibilities.

For a, we have (0+2)(0-1)(0-3) = (2)(-1)(-3) = 6

For e, we have (0-1)(0-3)(0+2)³ =  (-1)(-3)(2)³ = (3)(8) = 24

Therefore, option a is the only one that fits, and is your answer

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]