How do you find the range of each function given the domain: $h (n) = 3 n^{2} - 2 n + 2; {- 1, 0, 1}$ $h(n)=3n^2-2n+2; {-1,0,1}$ ?

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Answer 1 · 2016-11-14T14:10:51

Just evaluate each of the $x$ $x$ -values within the function.

$h (- 1) = 3 {(- 1)}^{2} - 2 (- 1) + 2 = 3 + 2 + 2 = 7$ $h(-1) = 3(-1)^2 - 2(-1) + 2 = 3 + 2 + 2 = 7$

$h (0) = 3 {(0)}^{2} - 2 (0) + 2 = 2$ $h(0) = 3(0)^2 - 2(0) + 2 = 2$

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

Hence, the range is ${2, 3, 7}$ ${2, 3, 7}$ .

Hopefully this helps!

How do you find the range of each function given the domain: h(n)=3n2−2n+2;{−1,0,1}h(n)=3n^2-2n+2; {-1,0,1}?