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lanedance

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just start with [tex] y = -2 \sqrt{ 3x-1 } -5[/tex]

and rearrange for x(y) this is your inverse function

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HallsofIvy

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HallsofIvy

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[tex]\frac{x+ 5}{-2}= \sqrt{3y- 1}[/tex]

Now you need to get rid of the square root. What is the opposite of taking the square root?

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HallsofIvy

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Well, you don't need that "-2 squared" in the denominator- it's just 4.

y= (x+5)^2/12+ 1.

y= (x+5)^2/12+ 1.

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Mentallic

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Be careful when manipulating equations in ways other than adding/subtracting. There is no jumping, leaping, or any other form of hopping in that sense :tongue2:

So you squared both sides:

[tex]3y-1=\frac{(y+5)^2}{(-2)^2}[/tex]

Now, when dividing, be sure to divide EVERY term on both sides of the equation.

e.g. [tex]ax+b=c[/tex]

solving for x: divide by a: [tex]x+\frac{b}{a}=\frac{c}{a}[/tex]

and now isolate x by 'jumping' that 'thing' to the other side: [tex]x=\frac{c}{a}-\frac{b}{a}=\frac{c-b}{a}[/tex]

It can also be done like this:

[tex]ax+b=c[/tex]

[tex]ax=c-b[/tex]

[tex]x=\frac{c-b}{a}[/tex]

ok. See if you can apply this to the question now

So you squared both sides:

[tex]3y-1=\frac{(y+5)^2}{(-2)^2}[/tex]

Now, when dividing, be sure to divide EVERY term on both sides of the equation.

e.g. [tex]ax+b=c[/tex]

solving for x: divide by a: [tex]x+\frac{b}{a}=\frac{c}{a}[/tex]

and now isolate x by 'jumping' that 'thing' to the other side: [tex]x=\frac{c}{a}-\frac{b}{a}=\frac{c-b}{a}[/tex]

It can also be done like this:

[tex]ax+b=c[/tex]

[tex]ax=c-b[/tex]

[tex]x=\frac{c-b}{a}[/tex]

ok. See if you can apply this to the question now

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