1)
[tex] \bf \begin{array}{c|ccccccc}
x&2&4&\boxed{6}&8&\boxed{10}&12&14\\\\
y&1&6&\boxed{11}&16&\boxed{21}&26&31
\end{array}
\\\\\\
(\stackrel{x_1}{6}~,~\stackrel{y_1}{11})\qquad
(\stackrel{x_2}{10}~,~\stackrel{y_2}{21})\\\\
------------------------------- [/tex]
[tex] \bf slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{21-11}{10-6}\implies \cfrac{10}{4}\implies \cfrac{5}{2}
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-11=\cfrac{5}{2}(x-6)
\\\\\\
y-11=\cfrac{5}{2}x-15\implies y=\cfrac{5}{2}x-4 [/tex]
2)
well, for radical with an even root, like the square root with a root of 2, it can't have a negative radicand.
so 3√(x-2) the x-2 part must never become negative, if it ever does, then you'd get an imaginary root or so-called complex root, which is another way of saying no real root for such radicand.
and for x-2 that occurs if x ever becomes less than 2. if x is 2, then we get 2-2 =0, and 0 is ok, but something like 1-2 = -1, and √-1 is just "i", and if we go down further, say x = -5, then we get -5-2 = -7 and √-7 is just i√7, yet another imaginary root.
so the restriction is that x can never become less than 2.
therfore, the domain is in Set Builder form {x | x ∈ ℝ, x ⩾ 2 }.
