### Home > CALC > Chapter 1 > Lesson 1.4.1 > Problem1-139

1-139.

Find values of a that that make $h(x)$ continuous.
$h ( x ) = \left\{ \begin{array} { l l } { \sqrt { x + 2 } - 1 } & { \text { for } x < 2 } \\ { a ( x + 1 ) ^ { 2 } } & { \text { for } x \geq 2 } \end{array} \right.$

If the function is continuous, both pieces of $h(x)$ will be connected at $x = 2$. That means that both pieces will have the same $y$-value at $x = 2$.

Equate both parts of the function for $x = 2$. Then solve for $a$.

$\sqrt{2+2}-1=a(2+1)^2$

$2-1=9a$