Maxima and Minima of Quadratic function help?

The derivative of the function is 2ax + b
The second derivative is 2a, so as long as a is positive, the parabola will have a minimum, not a maximum.

The location of the minimum is where the first derivative is zero
2ax + b = 0, and we want that to be true at x = -1 so the location of the minimum is where
2a(-1) + b = 0
-2a + b = 0

The value of the function is y = ax^2 + bx + 5 and we want that to be 3 when x = -1 so
3 =a(-1)x^2 + b(-1) + 5
3 = a - b + 5
a - b = -2

We now have two equations and two unknowns
-2a + b = 0
a - b = -2

Multiply the second equation by 2
-2a + b = 0
2a - 2b = -4

Add and solve
-b = -4
b = 4

Solve for a using either of the equations
a - b = -2
a - 4 = -2
a = 2

Check:
y=ax^2 +bx+5
y = 2x^2 + 4x + 5

If x = -1
y = 2(-1)^2 + 4(-1) + 5
y = 2 - 4 + 5
y = 3
Check!

The coefficient of x² term is a.
The quadratic function that minimum y = 3 at x = -1 :
y = a(x + 1)² + 3 ...... [1]

The above be function can be expressed as :
y = ax² + bx + 5 ...... [2]

[2] = [1] :
ax² + bx + 5 = a(x + 1)² + 3
ax² + bx + 5 = a(x² + 2x + 1) + 3
bx + 5 = 2ax + (a + 3)
2ax - bx + (a + 3) - 5 = 0
(2a - b)x + (a - 2) = 0

a - 2 = 0
Hence, a = 2

(2a - b)x = 0
b = 2a
b = 2*2
b = 4

Hence, a = 2, b = 4

y = ax² + bx + 5

plug in (-1,3)
3 = a(-1)² + b(-1) + 5
a = b-2

dy/dx = 2ax + b = 0
2(b-2)x + b = 0
2bx - 4x + b = 0
-2b + 4 + b = 0
b = 4

a = b-2 = 2

y = ax^2 + bx + 5,

y ' = 2ax + b

For min

2ax + b = 0

x = - b/2a

At x = - 1

-1 = -b/2a

b = 2a .......................(1)

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

3 = a - b + 5

a - b = - 2..............(2)

From (1)

a - 2a = - 2

-a = - 2

a = 2

b = 4