ƒ(−1) if ƒ(x) = 3(x −1)² + 5?
ƒ(−1) if ƒ(x) = 3(x −1)² + 5? we're just substituting "-1" into this for "x" and evaluating...so we have....
3 (-1-1)2 + 5 =
3(-2)2 + 5 =
3(4) + 5 =
12 + 5 =
17
