Question

Help with math please

Answer 1

I'm guessing this isn't calculus?

max value of a quadratic is where the vertex is.... \[\large f(x) = a(x-h)^2 +k\] has a vertex (h, k) where k is the maximum value

max value of a function like a*cos(x) + k will be a+k

Answer 2

No it's trig and I honestly have no clue what any of that means can you explain in like stupid terms? XD

Answer 3

if you wanted to find the max value of:
\[\Large f(x)=-3(x−6)^2+8\]
the max y-value would be 8. Literally nothing else matters.

Answer 4

So what's the max of the f(x) = -4(x-6)^2 + 3? only one number matters.

Answer 5

OHhhh okay, I understand :D and 3? o:

Answer 6

Yep! now, the max of a function like \[\Large g(x) = -3\cos(x- 7) + 5\] will be the 3 + 5 (ignore the negative in front of the 3, it has no impact on the max of a trig function)

Answer 7

ohhh so the max of g(X) would be 8 then? o:

Answer 8

yes, for that example. what about for yours?

Answer 9

Still the only numbers that matter are the one in front of the cos, and the last one.

Answer 10

for g(x) it would be 6 and for f(x) it would be 7 right?

Answer 11

No, two different methods for each function... go back up and see how to get the max for f.

Answer 12

g(x) will have a max of 6, f(x) won't be 7

Answer 13

OOHH okay f would be 3 and g would be 6

Answer 14

Yep!

Answer 15

Remember that the max (or min, if the a is positive) for \[\large f(x) = a(x-h)^2 +k\]is just the k.

For a sine or cosine, like \[\Large a \cos (bx-c) + d\] the max is the the sum of the number in front of sin or cos (ignore the negative), plus the number added on at the end... ie a+d

Answer 16

okay! :D thank you