Does anyone know how to solve the range for....?

Question

y = radical (4 - x^2)?

I do not know at all :( I know the correct answer is supposed to be
(- infinity, 2]

Answer 1

y = sqrt(4 - x^2)

Note that if we square both sides, we get

y^2 = 4 - x^2, and

x^2 + y^2 = 4

This is a circle with radius 2 and center at (0,0). The range of a circle is [-2, 2]

How this differs from our original equation is that the original equation is a *half* circle. The range shouldn't be (-infinity, 2]. The range should be [0, 2]. The domain is [-2, 2] though.

I do not know where you got (-infinity, 2] from.

Answer 2

The domain for y = √(4-x^2) should be [-2,2] . Otherwise, y cannot be a real number. Therefore, the range for y is [0,2].
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Reason:
4-x^2≧0,
Solve for x,
-2≧x≧2

y = √(4-x^2)≧0
y = √(4-x^2)≦2
Combine the two,
0≦y≦2

Answer 3

Maybe try factoring (4-x^2) into (2-x)(2+x)?

I don't know what a radical is, sorry.

Answer 4

sorry cant help