Find p(q(x)) p(x)=x^2+9 q(x) = square root of x+8

$p (x) = x^{2} + 9$ $q (x) = \sqrt{x + 8}$

Set up the composite result function.

$p (q (x))$

Evaluate $p (\sqrt{x + 8})$ by substituting in the value of $q$ into $p$ .

$p (\sqrt{x + 8}) = {(\sqrt{x + 8})}^{2} + 9$

Rewrite ${\sqrt{x + 8}}^{2}$ as $x + 8$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 8}$ as ${(x + 8)}^{\frac{1}{2}}$ .

$p (\sqrt{x + 8}) = {({(x + 8)}^{\frac{1}{2}})}^{2} + 9$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$p (\sqrt{x + 8}) = {(x + 8)}^{\frac{1}{2} \cdot 2} + 9$

Combine $\frac{1}{2}$ and $2$ .

$p (\sqrt{x + 8}) = {(x + 8)}^{\frac{2}{2}} + 9$

Cancel the common factor of $2$ .

Cancel the common factor.

$p (\sqrt{x + 8}) = {(x + 8)}^{\frac{2}{2}} + 9$

Rewrite the expression.

$p (\sqrt{x + 8}) = (x + 8) + 9$

Simplify.

$p (\sqrt{x + 8}) = x + 8 + 9$

Add $8$ and $9$ .

$p (\sqrt{x + 8}) = x + 17$