Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials. First, multiply the first two terms in each binomial group. Next, multiply the outer terms in each group, followed by the inner terms. Finally, multiply the last two terms in each group.

`f(x) = (x^2 + 6x + 9) - 4`

Remove the parentheses around the expression `x^2 + 6x + 9`

`f(x) = x^2 + 6x + 9 - 4`

Subtract `4` from `9` to get `5` .

`f(x) = x^2 + 6x + 5`

The maximum or minimum of a quadratic function occurs at `x = - b/(2a).` If `a` is negative, the maximum value of the function is `f(-b/(2a)).` If `a` is positive, the minimum value of the function is `f(-b/(2a)).`

`f_(min) (x) = ax^2 + bx + c` occurs at `x = - b/(2a)`

Remove the parentheses that are not needed from the expression.

`x = - 6/(2(1))`

Multiply `2` by `1` to get `2` .

`x = - 6/2`

Reduce the expression by canceling out all common factors from the numerator and denominator.

`x=-3 `

Multiply to simplify the expression `(-3)(-3)` .

`f(-3) = 9 + 6(-3) + 5)`

Multiply `6` by each term inside the parentheses.

`f(-3) = 9 - 18 + 5`

The minimum value of the function is at `x = - b/(2a).` This is a minimum value because `a` is greater than `0` .

`f(-3) = -4`

Find the maximum (or minimum) of the function `f(x)=(x+3)^2-4` without graphing:

This is a quadratic. The graph is a parabola. The function will have a minimum if the leading coefficient is positive and a maximum if the leading coefficient is negative.

The function is written in vertex form `y=a(x-h)^2+k` ; the vertex is at (h,k) and represents the maximum or minimum. If a>0 there is a minimum, and if a<0 there is a maximum.

Here the vertex is at (-3,-4) and a=1>0.

**Thus there is a minimum at (-3,-4).**

This is quadratic function.

`f(x)=(x+3)^2-4=x^2+6x+9-4=x^2+6x+5`

The maximum or minimum (as in this case) value of quadratic function is fund in vertex of parabola and coordinates of the vertex of function `q(x)=ax^2+bx+c` are

`V=(-b/(2a),(-b^2+4ac)/(4a))`

Hence in your case the minimum is fund at `x=-6/(2*1)=-3`

and minimal value is `min f(x)=(-6^2+4*1*5)/(4*1)=(-36+20)/4=-4`

**Minimal value is -4.**