How do you find the domain and range for y = sqrt(x^2 -3x +2)y=√x2−3x+2?

1 Answer
Binayaka C.
Oct 22, 2017

Domain: x <= 1 and x >=2 or x| (-oo,1] uu [2,oo)x≤1andx≥2orx∣(−∞,1]∪[2,∞)
Range: y >=0 or y| [0,oo)y≥0ory∣[0,∞)

Explanation:

y= sqrt(x^2-3x+2) = sqrt((x-1)(x-2));y=√x2−3x+2=√(x−1)(x−2); Domain: under

root should be >=0 :. (x-1)(x-2)>=0

When 1 < x < 2 sign of y is (+)(-) = (-) :. < 0

Therefore for 1 < x < 2 ; y is undefined .

Domain: x <= 1 and x >=2 or x| (-oo,1] uu [2,oo)

Range: y >=0 or y| [0,oo) since square root of positive quantity

is also positive.

graph{(x^2-3x+2)^0.5 [-10, 10, -5, 5]}

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