(a) Function Composition:

\( (f \circ g)(x) = f(g(x)) = f(x – 1) \)
\( = (x – 1)^2 + 4 \)
\( = \boxed{x^2 – 2x + 5} \)

(b) Vertex of Translated Function:

Step 1: Find vertex of \( (f \circ g) \)
\( y = x^2 – 2x + 5 \) completes to \( y = (x – 1)^2 + 4 \)
Vertex at \( (1, 4) \)

Step 2: Apply translation \( \begin{pmatrix}3\\-1\end{pmatrix} \)
New vertex: \( (1+3, 4-1) = \boxed{(4, 3)} \)

(c) Showing h(x):

Method 1: Using vertex form
Translated function: \( h(x) = (x – 4)^2 + 3 \)
Expanded form: \( x^2 – 8x + 16 + 3 = \boxed{x^2 – 8x + 19} \) ✓

Method 2: Direct substitution
\( h(x) = (f \circ g)(x – 3) – 1 \)
\( = (x – 3 – 1)^2 + 4 – 1 \)
\( = (x – 4)^2 + 3 \)
Expands to same result

(d) Tangent Point:

Method 1: Equal roots approach
Set \( h(x) = 2x – 6 \):
\( x^2 – 8x + 19 = 2x – 6 \)
\( x^2 – 10x + 25 = 0 \)
\( (x – 5)^2 = 0 \) ⇒ Double root at \( \boxed{5} \)

Method 2: Using derivatives
\( h'(x) = 2x – 8 \)
Tangent slope = 2 ⇒ \( 2x – 8 = 2 \)
\( 2x = 10 \) ⇒ \( x = \boxed{5} \)