Whakaoti i te y=-1/2sqrt{x+2}+4

$$y = \frac { - 1 } { 2 } \sqrt { x + 2 } + 4$$

Whakaoti mō x

$$x=4\left(4-y\right)^{2}-2$$

$$8-2y\geq 0$$

Ngā Upane mō te Whakaoti Whārite Pūtake

Ka taea te hautanga \frac{-1}{2} te tuhi anō ko -\frac{1}{2} mā te tango i te tohu tōraro.

$$y=-\frac{1}{2}\sqrt{x+2}+4$$

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

$$-\frac{1}{2}\sqrt{x+2}+4=y$$

Tangohia te 4 mai i ngā taha e rua.

$$-\frac{1}{2}\sqrt{x+2}=y-4$$

Me whakarea ngā taha e rua ki te -2.

$$\frac{-\frac{1}{2}\sqrt{x+2}}{-\frac{1}{2}}=\frac{y-4}{-\frac{1}{2}}$$

Mā te whakawehe ki te -\frac{1}{2} ka wetekia te whakareanga ki te -\frac{1}{2}.

$$\sqrt{x+2}=\frac{y-4}{-\frac{1}{2}}$$

Whakawehe y-4 ki te -\frac{1}{2} mā te whakarea y-4 ki te tau huripoki o -\frac{1}{2}.

$$\sqrt{x+2}=8-2y$$

Pūruatia ngā taha e rua o te whārite.

$$x+2=4\left(4-y\right)^{2}$$

Me tango 2 mai i ngā taha e rua o te whārite.

$$x+2-2=4\left(4-y\right)^{2}-2$$

Mā te tango i te 2 i a ia ake anō ka toe ko te 0.

$$x=4\left(4-y\right)^{2}-2$$

Whakaoti mō x (complex solution)

$$x=4\left(4-y\right)^{2}-2$$

$$y=4\text{ or }arg(8-2y)<\pi$$

Whakaoti mō y (complex solution)

$$y=-\frac{\sqrt{x+2}}{2}+4$$

Whakaoti mō y

$$y=-\frac{\sqrt{x+2}}{2}+4$$

$$x\geq -2$$