Can someone help mean I know this is easy but it’s not to me lol

I. Not a function

II. Not a function

III. Yes, function

IIII. Yep!

(I messed up the numbering in the explanations, sorry 3 is 4 and 4 is 3)

Step-by-step explanation:

Hi!

For a function, every x value must have exactly 1 y value.

f(2) can't equal both 0 and 10. It can only equal 0 or 10. So, because of this, we have a test called the vertical line test. Basically, you drag a line through a graph, and if there is any x value that goes to two (or more) different y-values, then that graph is not the graph of the function.

Let me try to give a practical explanation.

I.

In this graph, everything looks fine until we reach x=6. (I'm assuming the graph goes by ones). Do you see the problem? Yep! x has two different y values.

y= 4 and y= 6.

So, this isn't a function. Why? Because there are two y values for each x value (aka, this doesn't pass the V.L.T)

II.

(1,2), (2,5) , (3,8) , (2,-5) (1,-2)

We have two groups of problem points.

(2,5) and (2,-5) and (1,2) and (1,-2)

Here, each x value goes to a different y value. Therefore, this isn't a function

III. y=x^2

This is your typical parabola. See the graph that I attached.

Notice that -2 and 2 both go to 4. Two different x-values going to the same y-value is fine, but one x can't go to to different y values.

IIII.

(-4,1) , (0,3) , (4,5) , (6,6)

Check 1: Does any x-value go to multiple y values? Nope! So, this is a function.

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yes

step-by-step explanation:

so of course you want to start by plugging the 5 in which goes to 7(5)+3(5)=50 because 7*5=35 and 3*5=15 and then add them together, then plug in the 5 to the other equation as well 9(5)+5=50 because 9*5= 45 and then add 5 and so yes they are equivalent.