Question

@Hero Functions f(x) and g(x) are shown below:

Answer 1

For the first one, the solution is already given

Answer 2

Since the equation is already in vertex form

\(y = a(x - h)^2 + k\) where (h,k) represents the vertex. The vertex being the highest or lowest point on the graph. In this case, it will be the highest point since a is negative.

Answer 3

f(x) is the answer

Answer 4

Whenever you have an quadratic equation in vertex form and a is negative, then k will be the max y-value.

Answer 5

Is f(x) the answer

Answer 6

I thought you wanted to go over this.

Answer 7

If you think you already have the answer, then I'm wasting my time.

Answer 8

I am not sure please help

Answer 9

Well, as I was saying, for the first one, \(y_{max} = k\) since a is negative

Answer 10

Which means for the first function the max value occurs when\(y = f(x) = 3\)

Answer 11

Now for the second one, you have to be familiar with properties of the cosine function

Answer 12

The max value for cosine occurs when cos(x) = 1

Answer 13

okay

Answer 14

In other words for

g(x) = 2 cos(2x - pi) + 4

assume that cos(2x - pi) = 1, then you'll have

g(x) = 2(1) + 4

Answer 15

Which means the max value of g(x) will occur when g(x) = 6

Answer 16

But, the truly easiest and fastest way to solve the problem is to graph both functions:

https://www.desmos.com/calculator/dvlsz0xp3g