How do you solve for F(x) = 5 (x-2)^1/2 + 3

find transformation, domain and intercept

5 (x-2)^{ 1/2} + 3=0

5 (x-2)^{1/ 2} = -3

squaring both sides

25(x-2) = 9

(x-2) =9/25 = 0.36

x =2+0.36 =2.36 or 2 9/25

So x intercept = 2.36 or 2 9/25

Y intercept is not possible since x= 0 is not in the domain.

(y-3)^{2 } =25(x-2)

x= 2+(y-3)^{2} /25

Domain is { x | x>=2}

Range is { y | y is real number}

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