If $f(x)=x^{2}-7 x$ and $g(x)=x+3$, then the minimum value of $f(g(x))-3 x$ is

We have $f(x) = x^2 - 7x$ and $g(x) = x + 3$

To find $f(g(x))$, we substitute $g(x)$ into $f(x)$:

$f(g(x)) = f(x+3) = (x+3)^2 - 7(x+3)$

So our expression becomes:

$f(g(x)) - 3x = (x+3)^2 - 7(x+3) - 3x$

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$(x+3)^2 = x^2 + 6x + 9$

$f(g(x)) - 3x = x^2 + 6x + 9 - 7(x+3) - 3x$

$= x^2 + 6x + 9 - 7x - 21 - 3x$

Combining like terms:

$x^2$ terms: $x^2$

$x$ terms: $6x - 7x - 3x = -4x$

Constant terms: $9 - 21 = -12$

$f(g(x)) - 3x = x^2 - 4x - 12$

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We now need to find the minimum value of $h(x) = x^2 - 4x - 12$

Since this is a quadratic with a positive coefficient of $x^2$ (which is 1), the parabola opens upward, so it has a minimum point.

For any quadratic $ax^2 + bx + c$, the minimum occurs at $x = -\frac{b}{2a}$

In our case: $a = 1$, $b = -4$, $c = -12$

$x = -\frac{(-4)}{2(1)} = \frac{4}{2} = 2$

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We substitute $x = 2$ into $h(x) = x^2 - 4x - 12$:

$h(2) = (2)^2 - 4(2) - 12 = 4 - 8 - 12 = -16$

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Therefore, the minimum value of $f(g(x)) - 3x$ is $-16$.