How do you find the range of $y = z^{2} - 3 z$ $y = z^2 - 3z$ ; given Domain = {-1, 0, 1, 2}?

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Answer 1 · 2016-09-08T14:04:15

$The Range= {4, 0, - 2}$ $"The Range="{4,0,-2}$ .

$y = z^{2} - 3 z, and, let D = {- 1, 0, 1, 2}$ $y=z^2-3z", and, let "D={-1,0,1,2}$ .

By, Range of the given function, we mean, the Set of all values of

$y$ $y$ , to be determined by using the values of $z$ $z$ choosing from the

given Domain Set, here, $D$ $D$ .

Thus, in Notation, the Range $= {y = z^{2} - 3 z : z \in D}$ $={y=z^2-3z : z in D}$ .

Now, $z = - 1 \Rightarrow y = z^{2} - 3 z = {(- 1)}^{2} - 3 (- 1) = 1 + 3 = 4$ $z=-1 rArr y=z^2-3z=(-1)^2-3(-1)=1+3=4$

$z = 0 \Rightarrow y = 0$ $z=0 rArr y=0$

$z = 1 \Rightarrow y = - 2$ $z=1 rArr y=-2$

$z = 2 \Rightarrow y = - 2$ $z=2 rArr y=-2$

Therefore, $The Range= {4, 0, - 2}$ $"The Range="{4,0,-2}$ .

How do you find the range of y=z2−3zy = z^2 - 3z; given Domain = {-1, 0, 1, 2}?