Question

Consider two functions: g(x)=−x2−6x and the quadratic function ​f(x)​ shown in the table.

Which statements are true?

Select each correct answer.




The average rate of change of f(x) is greater than ​g(x)​ on the interval [0,3] .

​g(x)​ has a greater y-intercept than ​f(x)​ does.

​f(x)​ is greater than ​g(x)​ on the interval​ (0,3) ​.

​ g(3) ​​ is less than​ f(3) ​.

Answer 1

1). g(3) ​​ is less than​ f(3).

​2). The average rate of change of f(x) is greater than ​g(x)​ on the interval [0,3] .

Explanation:

g(x) = -x^{2} - 6x

so f(3) = 9

let's solve for g(3)

3^{2} = 9

g(3) = - 9 - 18 = -27

so -27 is less than 9

g(3) < f(3)

So our answer is correct: g(3) ​​ is less than​ f(3).

In second option:

put the value from 0 to 3 in function g(x) = - x^{2} - 6x

g(0) = 0

g(1) = -7

g(2) = -16

g(3) = -27

on the other hand values of f(x) are:

f(0) = 0

f(1) = 1

f(2) = 4

f(3) = 9

So average values of f(x) are greater than g(x) so our option is correct.

The average rate of change of f(x) is greater than ​g(x)​ on the interval [0,3] .

i hope you get the idea.

Answer 2

Hence average rate of change is greater for f(x) in (0,3).

Yes. f(x) is greater than g(x) in the interval (0,3)

Yes. g(3) <f(3)

Explanation:

To find average rate of change of f and g in (0,3)

f(3)-f(0)/3 = (9-0)/3 =3

g(3) = -9-18 = -27

g(0) = 0

Rate of change in(0,3) of g(x) = -27/3 =-9

Hence average rate of change is greater for f(x) in (0,3)

---

Both f and g have intercepts of y as 0

Hence both are equal.

3) Yes. f(x) is greater than g(x) in the interval (0,3)

Because g(x) <0 for all x in (0,3) while f(x) >0

4) g(3)= -9-18=-27 <f(3)