CPM Homework Help : APCALC Problem 1-139

What value of $a$ will make $h$ continuous?

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

Hint:

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 $x = 2$ .

Steps:

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

Show Step 1:

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

Show Step 2:

$2-1=9a$