Answer
$\frac{1}{\sqrt{17}}(i+4j)$
Work Step by Step
The equation of the tangent line to $f$ at the point $(a,f(a))$ is:
$$y=f'(a)(x-a)+f(a)$$
so:
$$y=2a(x-a)+f(a)$$
$$y=2(2)(x-2)+f(2)$$
$$y=4(x-2)+4$$
$$y=4x-4$$
Find the directional vector of the tangent line by choosing a point that is on the tangent line except $(2,4)$.
for $x=1$ then $y=0$ so the point is $(1,0)$.
so the directional vector is:
$$\lt2-1,4-0\gt=\lt1,4\gt$$
the unit vector is:
$$\frac{1}{\sqrt{ 1^{2}+(4)^{2}}}\lt1,4\gt=\frac{1}{\sqrt{17}}\lt1,4\gt=\lt\frac{1}{\sqrt{17}},\frac{4}{\sqrt{17}}\gt=\frac{1}{\sqrt{17}}(i+4j)$$
