To have a extreme point (Either maximum or minimum or inflection point), the function's first derivative must be zero at the given point. Also to have a maximum at that point, its second derivative must be negative at the given point. Now let's incorporate this in our calculations.

First we will calculate the first derivative.

`f(x) = ax^3+bx^2-5`

`f'(x) = 3ax^2+2bx`

f'(x) is zero at (-1.-4)

Therefore,

`0 = 3a(-1)^2+2b(-1)`

This gives,

`3a -2b = 0 ` -------- Equation 1

Calculating the second derivative,

`f''(x) = 6ax+2b`

now for a maximum f''(x) must be negative at (-1,4), then,

`6a(-1)+2blt0`

`3agtb`

**b<3a.**

Also we have been given that the value of f(x) at this local maximum is -4. (since the point is(-1,-4)

Therefore substituting in f(x),

`-4 = a(-1)^3+b(-1)^2-5`

`-a+b = 1` -------------Equation 2

Multiply equation 2 by 3 and add it to equation 1,

`3(-a+b)+3a-2b = 3+0`

`b = 3`

Therefore `-a+3 = 1`

`a = 2`

So we get an answer as a = 2 and b =3. Now this satisfies our requirement for maximum above. (b<3a).

Therefore for values a = 2 and b = 3, f(x) has a maximum at (-1,-4)

`f(x) = 2x^3+3x^2-5`