Question

HURRY PLEASE

Q.14

If
`y=3x^(2) -9`

A.
`y^(-1) = -\sqrt(x+9)/(3)` such that x is greater than or equal to negative nine

B.
`y^(-1) =-\sqrt(x+9)/(3)` such that x id less than or equal to negative nine

C.
`y^(-1) = -\sqrt (x)/(3) +9` such that x is less than or equal to zero

D.
`y^(-1) = -\sqrt (x)/(3) + 9` such that x is greater than or equal to zero

Answer 1

the answer is A

Explanation:

Adding on to Voidspace101's answer,

For inverse, we'll get,

`x=3y^2-9\\x+9=3y^2\\(x+9)/3=y^2\\y=√((x+9)/3) \\and\\y=\ -sqrt{(x+9)/3} \\`

now, to get a positive number inside the square root,

we have the condition,

x + 9 is greater than or equal to 0

so,

x+9 > 0

x>-9

so x has to be equal to or greater than -9

hence the correct answer is A

Answer 2

Answer: B

Explanation:

This is inverse.

Essentially, you switch y for x.

So;

`x=3y^2-9`

`x+9=3y^2`

`(x+9)/3=y^2`

`y=√((x+9)/3)`