Whakaoti i te y=sqrt{-x}+1

$$y = \sqrt { - x } + 1$$

Whakaoti mō x

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

$$-\left(1-y\right)\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$\sqrt{-x}+1=y$$

Tangohia te 1 mai i ngā taha e rua.

$$\sqrt{-x}=y-1$$

Pūruatia ngā taha e rua o te whārite.

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

Whakawehea ngā taha e rua ki te -1.

$$\frac{-x}{-1}=\frac{\left(y-1\right)^{2}}{-1}$$

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

$$x=\frac{\left(y-1\right)^{2}}{-1}$$

Whakawehe \left(y-1\right)^{2} ki te -1.

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

Whakaoti mō x (complex solution)

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

$$y=1\text{ or }arg(1-y)\geq \pi$$

Whakaoti mō y

$$y=\sqrt{-x}+1$$

$$x\leq 0$$