Question 568317
you have g(x) = sqrt(-x)
when y = 3, x = -9
when y = 2, x = -2
when y = 1, x = -1
presumably, you are given the value of g(x) which you also call y.
that's ok because you would then have y = g(x) = sqrt(-x) which is valid.
supposed you are first given that y = 3.
your equation of y = sqrt(-x) becomes:
3 = sqrt(-x)
you would square both sides of this equation to get:
9 = (sqrt(-x))^2 which becomes:
9 = -x
you then multiply both sides of the equation by -1 to get:
-9 = x which is the same as:
x = -9
for your information, the square root of (-x) is only imaginary when x is positive.
when x is negative, the square root of (-x) becomes the square root of a positive number because minus a negative number results in a positive number.
the same process is used to solve for the other values.
you are given 2 = sqrt(-x)
squareboth sides to get:
4 = (sqrt(-x))^2 which becomes:
4 = -x which then becomes
x = -4 after multiplying both sides of the equation by -1.
additional information for you:
sqrt(-x) is the same as (-x)^(1/2)
your equation of:
2 = sqrt(-x) can also be written as:
2 = (-x)^(1/2)
square both sides of this equation to get:
4 = ((-x)^(1/2))^2 which results in:
4 = (-x)^((1/2)*2) which becomes:
4 = (-x)^(1) which becomes:
4 = -x which then becomes:
x = -4 after multiplying both sides of the equation by -1.