How do you find $h (- 3)$ $h(-3)$ given $h (a) = a^{2} - 1 + a$ $h(a)=a^2-1+a$ ?

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Answer 1 · 2018-03-07T01:39:10

$h (- 3) = 5$ $h(-3)=5$

Substitute $- 3$ $-3$ everywhere you see $a$ $a$ : $h (a) \Leftrightarrow h (- 3)$ $h(a)<=>h(-3)$

$h (- 3) = {(- 3)}^{2} - 1 + (- 3)$ $h(-3)=(-3)^2-1+(-3)$

$h (- 3) = 9 - 1 - 3$ $h(-3)=9-1-3$

Use PEMDAS to solve...

$h (- 3) = 5$ $h(-3)=5$

How do you find h(−3)h(-3) given h(a)=a2−1+ah(a)=a^2-1+a?