Which of the following function types exhibit the end behavior f(> 0 as x --> -infinity? power; y=x^n; n is even and greater than zero identity; y=x absolute value; y= absolute value of x reciprocal; y=1/x root; y=^n sort x; n is even and greater than zero exponential; y=b^x, b> 0 again, i know that two of these are correct but i'm not sure which ones. let me know!you!

Let's consider the functions one by one.

i) ,

x --> -infinity means that x is a very very small negative number. To model it in our mind let's think of . This number to an even power becomes ... etc.

Indeed, we can see that the smaller the x, the greater the value . In fact, as x --> -infinity,  f(x)-->+infinity.

ii)
y=x, 

this clearly means that the behaviors of x and y are identical.

As x --> -infinity, y --> -infinity as well.

iii) y=|x|,

We can think of x --> -infinity, again, as a very very small number, like . For this value of x, y is , that is .

Indeed, the smaller the x, the greater the y. As in the function in part (i), as x --> -infinity,  f(x)-->+infinity.

iv) 

y=1/x

consider the values of y for the following values of x:

for x=-10, -100, -1000, y is respectively -0.1, -0.01, -0.001.

We can see that the smaller the x, the closer y gets to 0.

Thus, f(x)-->0 as x --> -infinity

v) n'th root of x in not defined for negative x, when n is even.

vi) y=b^x, b>0

Here note that b cannot be equal to 1, otherwise the function is not exponential.

Let b=5, consider the values of y for the following values of x:

for x=-10, -100, -1000, y is respectively .

That is, the smaller the value of x, the closer y gets to 0.

Thus, f(x)-->0 as x --> -infinity