For each of these, we first write the function in blank parenthesis form \(h(~)=\dfrac{(~)^2+1}{2}\). Next, we place the given expression into the blank parentheses and simplify the resulting expression.
\(h(w+2)=\dfrac{(w+2)^2+1}{2}=\dfrac{w^2+4w+5}{2}\)
\(h(\sqrt{2x-1})=\dfrac{(\sqrt{2x-1})^2+1}{2}=\dfrac{2x-1+1}{2}=x\)
\(h(x+a)=\dfrac{(x+a)^2+1}{2}=\dfrac{x^2+2ax+a^2+1}{2}\)