cosine graph with max at 0,5 and min at pi/2, 1, what would the graph look like, and what would the equation be?

Answer 1

We'll build the desired function in three steps.

Step 1.

The plain vanilla cosine graph y = cos(x) hits its maximum at x=0 and it minimum at x=π.

You want a graph that hits its maximum at x=0 and its minimum at x=π/2.

In other words, you want the minimum to come twice as early as it usually would (at x=π/2 instead of at x=π) as we travel from left to right.

Therefore your graph would have to evolve (or vibrate, or swing down and up) twice as fast as it normally would.

So here's a first shot at your desired graph:

y = cos(2x)

It hits its minimum at x=π/2 instead of at x=π because the x value you plug in gets doubled.

Step 2.

The plain vanilla cosine graph y = cos(x) goes from a maximum of 1 (at x=0) down to a minimum of -1 (at x=π).

So it goes 2 units up and down (from 1 down to -1).

You want a cosine graph that goes from maximum of 5 (at x=0) down to a minimum of 1 (at x=π/2).

So the desired graph would go 4 units up and down (from 5 down to 1).

That's twice as far up and down as the plain vanilla graph.

So here's an improved version of your desired graph:

y = 2cos(2x). It goes twice as far up and down as the plain vanilla one does.

Step 3.

y = 2cos(2x) goes from a maximum of 2 (at x=0) down to a minimum of -2 (at x=π/2)

You want a graph that goes from a maximum of 5 (at x=0) down to a minimum of 1 (at x=π/2)

So the entire graph needs to be raised 3 units.

That's easy: we'll just add 3:

y = 2cos(2x) + 3

All done.

Answer 2

f(x) = 2cos2x + 3, because f(0) = 2cos0 + 3 = 5 and f(π/2) = 2cosπ + 3 = 1