I need Quick Math Help Please
Find the average rate of change. f(x) = x^2 - 4x + 1 from x = 3 and x = 6
Have you found "av erage rate of change" before? If so, what does the appropriate formula look like?
I have no idea , its my first time
Hints: What are the values of the given function at x=3 and at x=6?
Im not sure im sorry i just dont understand this :(
Please see http://www.mesacc.edu/~marfv02121/readings/average/ for an explanation of "average rate of change." Here, f(x) = x^2 - 4x + 1. Just supposing (for example) that x=0, f(0)=0^2 - 4(0) + 1. Therefore, the value of the function f(x) at x=0 is ... what?
If g(x) = x and x=0, then g(0) = 0. Is this familiar for you?
To find the average rate of change in the problem you've posted, you need to know how to evaluate a function f(x) at a given x value. I am leading you through a review of that topic now.
By observing the link you gave me we would setup f(3) - f(6) / 3 - 6
Is this right ?
Thank you for looking up that link. You're definitely on the right track!! I would write \[\frac{ f(6)-f(3) }{ 6-3 }=Average~rate~of~change\]
...but what you have written would produce the same result. Please type the function f(x) from the original problem statement, here.
The original one is F(x) = x^2 - 4x + 1
Yes. Please now evaluate that for x=6. Please share your work.
So would it be f(6) = 6^2 -4(6) + 1 ?
Yes, and that comes out to f(6) = 36 - 24 + 1 = ??
f(6) = 13
Yes. Please evaluate f(3) similarly.
f(3) = -2
Right. now find the average rate of change. be careful with the - sign on that -2. Average rate of change of f(x) from x=3 to x=6 is\[\frac{ 13-(?) }{ 6-3}=??\]
29/3
f(6) = 13, and f(3) = -2. Thus, f(6) - f(3) = 13 - (-2) = ? (not 29).
Oops 15
That 15 is your numerator and is correct. What is the average rate of change?
\[\frac{ f(6)-f(3) }{ 6-3 }=?\]
15 / 3 which would = to 5 if simplified
yes. You've got it. Any questions about "average rate of change" or about any of the details we've gone through?
Good idea to write down formulas such as this one, for later reference and review.
Nope seems like I got it now! Thank you so much for your help your a great teacher , and thank you for the link it helped me alot as well!
Yes certainly will be taking notes of what we just reviwed
Thanks a LOT; so happy that you were willing to check out the link and to act on my suggestions. Best wishes to you!
Join our real-time social learning platform and learn together with your friends!