Question

How do you find the average rate of change for f(x) from x=4 to x=7?

Answer 1

\[\sf \text{avg.rate of change} = \frac{f(b)-f(a)}{b-a}\]

Answer 2

You're missing your function though.

Answer 3

Do i use vertex or standard?

Answer 4

Just standard should be fine.

Answer 5

Would you mind posting your entire question up? :)

Answer 6

Oh, well then \[2x^2-4+21\] is my equation in standard form

Answer 7

I just don't know where to go from there. It's been awhile lol

Answer 8

Awesome :)

\[f(x)= 2x^2-4x+21\] right?

Answer 9

\[ax^2+bx+c = 0? \]

Answer 10

Identify Parts, sub into equation

Answer 11

Think you might be missing an x there.

Answer 12

Yes Jhan

Answer 13

Sorry if I barged in btw, counldn't resist *u*

Answer 14

Tis ok Cam

Answer 15

Alright.\[\frac{f(7)-f(4)}{7-4}\]What is \(f(7) =~?\)
What is \(f(4) =~?\)

Answer 16

/.\ Ugh I don't know

Answer 17

Your equation: \(f(x) = 2x^2-4x+21\)

Just plug in 7 and 4 into your \(x\) values.

Answer 18

OH... OH OH OH... So i plug them in one at a time and the answer would be the f(x)?

Answer 19

The answer will be a quantitative value :)

Answer 20

\(\color{blue}{\text{Originally Posted by}}\) @Jhannybean
Alright.\[\frac{f(7)-f(4)}{7-4}\]What is \(f(7) =~?\)
What is \(f(4) =~?\)
\(\color{blue}{\text{End of Quote}}\)

Answer 21

Answer those two questions and we'll go from there! :)

Answer 22

Hint:
To find f(7) and f(4) you plug them into the main equation
\[f(x)=2x^2−4x+21 \]\[f(7)=2(7)^2−4(7) +21 \]

Answer 23

.... would i multiply the 2*7 first or square 7

Answer 24

square 7

Answer 25

PEMDAS, remember?

Answer 26

Yes lol
So f(7)=91
and
f(4)=37
?

Answer 27

yep

Answer 28

Yep, Great Job! Now you plug those values into the fraction @Jhannybean provided c:
\[\frac{ f(7)-f(4) }{ 7-4 }\]

Answer 29

\[\frac{91-37}{7-4}\]Just evaluate that :)

Answer 30

Which would be 54/3

Answer 31

and what does that reduce to?

Answer 32

18?

Answer 33

Good job :)

Answer 34

Thats ur rate of change c:

Answer 35

xO Mind blown
Thanks for teaching me ladies :3

Answer 36

no problem :)

Answer 37

Sorry for cutting in cx