How do I find the maximum value of the power function T(p)= -p^2+4p-3?

The -p² term indicates that the curve is an inverted parabola . . . so the vertex will be the curve's maximum.

Putting the equation in vertex form . . . T(p) = -(p-2)² + 1, tells us the vertex is at y = 1

This is a calc problem, right?

I'd do optimization...i think that's what it's called. I could be totally wrong lol.

But yeah, you're maximizing the value of p, you wannaa know what the value is on the very top of the curve. So you'll take the derivative and set that equal to zero to find that:

T(p)= -p^2+4p-3 .....so:

T'(p)= -2p + 4

-2p + 4 = 0

x= 2

And to make sure it's a maximum, plug it into a number line [you can do either T' or T''].

And it is the maximum, so plug 2 back into T(p) [original equation] and find the y value, and that's your answer.

Hope that helped!

t(p) = -p^2 +4*p-3

So t' =dt/dp = -2*p +4 ...(1)

For t(p) to be maximum or a minimum, dt/dp=0 that gives us p=2

If t(p) is to be a maximum at p=2, the t''(2) should be -ve.

Differentiating (1) once more, we find

t'' = -2 which is -ve for all values of p. So at p=2, t(p) is a maximum and has a value of -2^2 +4*2-3 = 1