The graph below shows two polynomial functions, f(x) and g(x): Which of the following statements is true about the graph above? (4 points)
f(x) iS an even degree polynomial with a positive leading coefficient.
• g(x) is an even degree polynomial with a positive leading coefficient.
f(x) is an odd degree polynomial with a negative leading coefficient.
• g(x) is an odd degree polynomial with a negative leading coefficient.

f(x) is an even degree polynomial with a positive leading coefficient

Step-by-step explanation:

The answer is A because f(x) IS an even degree polynomial with both ends traveling in the same direction, and it also has a positive leading coefficient because both sides continue up.

1) f(x) is an even degree polynomial with a positive leading coefficient

TRUE - Both sides going to negative infinity and infinity are facing the same direction (up), which means that the polynomial has an even degree. Because both sides are facing up, the polynomial also has a positive leading coefficient.

2) g(x) is an even degree polynomial with a positive leading coefficient

FALSE - Both sides of the function are going opposite directions, meaning that it is not an even degree polynomial. It does have a positive leading coefficient, however, as evidenced by the graph going upwards on the right side.

3) f(x) is an odd degree polynomial with a negative leading coefficient

FALSE - Both sides of the function are going the same way, which tells us that it has an even degree. Both sides of the function are going upwards, which means that the leading coefficient is not negative.

4) g(x) is an odd degree polynomial with a negative leading coefficient

FALSE - The function is an odd degree polynomial because the sides are going in opposite directions, but it does not have a negative leading coefficient because the right side of the graph is going up.

Hope this helps!! :)

it should be a and d. lmk if i'm wrong xd

they disappear if your absent from to long. i found out that the hard way.

step-by-step explanation:

brainliest lol i just need one more brainliest.