Solution
= Solution
At $x=3$, the point on the <parabola> is $(3,8)$ and the <tangent line> has slope $f'(3)=6$. The <normal line to a plane curve> therefore has slope $-1/6$. Its equation is
$$
y-8=-\frac16(x-3),\qquad y=-\frac x6+\frac{17}{2}.
$$
Thus the requested results are
$$
\boxed{\text{gradient }-\frac16,\qquad\text{vertical-axis intercept }(0,17/2).}
$$