## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$p(4)=\text{does not exist} ,\\\\ p(10)=0 ,\\\\ p(12)=2 ,\\\\ p(0)=\text{does not exist}$
$\bf{\text{Solution Outline:}}$ Substitute the given function value in $p(z)=\sqrt{2z-20} .$ $\bf{\text{Solution Details:}}$ If $z=4 ,$ then \begin{array}{l}\require{cancel} p(z)=\sqrt{2z-20} \\\\ p(4)=\sqrt{2(4)-20} \\\\ p(4)=\sqrt{-12} \text{ (not a real number)} .\end{array} If $z=10 ,$ then \begin{array}{l}\require{cancel} p(z)=\sqrt{2z-20} \\\\ p(10)=\sqrt{2(10)-20} \\\\ p(10)=\sqrt{0} \\\\ p(10)=0 .\end{array} If $z=12 ,$ then \begin{array}{l}\require{cancel} p(z)=\sqrt{2z-20} \\\\ p(12)=\sqrt{2(12)-20} \\\\ p(12)=\sqrt{4} \\\\ p(12)=2 .\end{array} If $z=0 ,$ then \begin{array}{l}\require{cancel} p(z)=\sqrt{2z-20} \\\\ p(0)=\sqrt{2(0)-20} \\\\ p(0)=\sqrt{-20} \text{ (not a real number)} .\end{array} Hence, \begin{array}{l}\require{cancel} p(4)=\text{does not exist} ,\\\\ p(10)=0 ,\\\\ p(12)=2 ,\\\\ p(0)=\text{does not exist} .\end{array}