MHB How to find domain of function in implicit form

[find_the_fun]

What is the domain of $$y^2-2y=x^2-x-1$$? I don't know how to find it for implicit functions.

 

[MarkFL]

Express as a quadratic in $y$ in standard form, then write the inequality requiring the discriminant to be non-negative...what do you find?

 

[find_the_fun]

Express as a quadratic in $y$ in standard form, then write the inequality requiring the discriminant to be non-negative...what do you find?

I think I get the idea but am a little stuck on how to express it as a quadratic. So $$y=1 \pm \frac{\sqrt{1-4(1)(-1)}}{2(1)}+2y$$

EDIT: what exactly do you mean express as a quadratic?

 

[Prove It]

What is the domain of $$y^2-2y=x^2-x-1$$? I don't know how to find it for implicit functions.

Complete the square on each of the x and y terms, it becomes a form you should recognise...

$\displaystyle \begin{align*} y^2 - 2y &= x^2 - x - 1 \\ y^2 - 2y + \left( -1 \right) ^2 - \left( -1 \right) ^2 &= x^2 - x + \left( -\frac{1}{2} \right) ^2 - \left( -\frac{1}{2} \right) ^2 - 1 \\ \left( y - 1 \right) ^2 - 1 &= \left( x - \frac{1}{2} \right) ^2 - \frac{1}{4} - 1 \\ \left( x - \frac{1}{2} \right) ^2 - \left( y - 1 \right) ^2 &= \frac{1}{4} \\ \frac{ \left( x - \frac{1}{2} \right) ^2 }{ \frac{1}{4}} - \frac{ \left( y - 1 \right) ^2 }{ \frac{1}{4}} &= 1 \\ \frac{ \left( x - \frac{1}{2} \right) ^2 }{ \left( \frac{1}{2} \right) ^2 } - \frac{ \left( y - 1 \right) ^2}{ \left( \frac{1}{2} \right) ^2 } &= 1 \end{align*}$

This looks like the standard form of a hyperbola. Can you get the domain from it?

 

[MarkFL]

I think I get the idea but am a little stuck on how to express it as a quadratic. So $$y=1 \pm \frac{\sqrt{1-4(1)(-1)}}{2(1)}+2y$$

EDIT: what exactly do you mean express as a quadratic?

What I mean is write the equation as:

$$y^2-2y-\left(x^2-x-1\right)=0$$

Now, require the discriminant to be non-negative:

$$(-2)^2-4(1)\left(-\left(x^2-x-1\right)\right)\ge0$$

$$4+4\left(x^2-x-1\right)\ge0$$

$$1+x^2-x-1\ge0$$

$$x^2-x\ge0$$

$$x(x-1)\ge0$$

Hence, the domain is:

$$(-\infty,0]\,\cup\,[1,\infty)$$

 

[find_the_fun]

Thanks I get what you're doing. But how do you get from

$$4+4\left(x^2-x-1\right)\ge0$$

to

$$1+x^2-x-1\ge0$$
?

What I mean is write the equation as:

$$y^2-2y-\left(x^2-x-1\right)=0$$

Now, require the discriminant to be non-negative:

$$(-2)^2-4(1)\left(-\left(x^2-x-1\right)\right)\ge0$$

$$4+4\left(x^2-x-1\right)\ge0$$

$$1+x^2-x-1\ge0$$

$$x^2-x\ge0$$

$$x(x-1)\ge0$$

Hence, the domain is:

$$(-\infty,0]\,\cup\,[1,\infty)$$

 

[MarkFL]

Thanks I get what you're doing. But how do you get from

$$4+4\left(x^2-x-1\right)\ge0$$

to

$$1+x^2-x-1\ge0$$
?

I divided through by 4. :D