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Work Shown:
(g°h)(x) is the same as g(h(x))
So, (g°h)(0) = g(h(0))
Effectively h(x) is the input to g(x). Let's first find h(0)
h(x) = x^2+3
h(0) = 0^2+3
h(0) = 3
So g(h(x)) becomes g(h(0)) after we replace x with 0, then it updates to g(3) when we replace h(0) with 3.
Now let's find g(3)
g(x) = -3x
g(3) = -3*3
g(3) = -9
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alternatively, you can plug h(x) algebraically into the g(x) function
g(x) = -3x
g( h(x) ) = -3*( h(x) ) ... replace all x terms with h(x)
g( h(x) ) = -3*(x^2 + 3) ... replace h(x) on right side with x^2+3
g( h(x) ) = -3x^2 - 9
Next we can plug in x = 0
g( h(0) ) = -3(0)^2 - 9
g( h(0) ) = -9
we get the same result.