Whakaoti i te f(x)=-x^2-4x+12

$$f ( x ) = - x ^ { 2 } - 4 x + 12$$

Tauwehe

$$\left(2-x\right)\left(x+6\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei -x^{2}+ax+bx+12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-4 ab=-12=-12$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -12.

$$1,-12 2,-6 3,-4$$

Tātaihia te tapeke mō ia takirua.

$$1-12=-11 2-6=-4 3-4=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -4.

$$a=2 b=-6$$

Tuhia anō te -x^{2}-4x+12 hei \left(-x^{2}+2x\right)+\left(-6x+12\right).

$$\left(-x^{2}+2x\right)+\left(-6x+12\right)$$

Tauwehea te x i te tuatahi me te 6 i te rōpū tuarua.

$$x\left(-x+2\right)+6\left(-x+2\right)$$

Whakatauwehea atu te kīanga pātahi -x+2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(-x+2\right)\left(x+6\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.

$$-x^{2}-4x+12=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-\left(-4\right)±\sqrt{16+4\times 12}}{2\left(-1\right)}$$

Whakareatia 4 ki te 12.

$$x=\frac{-\left(-4\right)±\sqrt{16+48}}{2\left(-1\right)}$$

Tāpiri 16 ki te 48.

$$x=\frac{-\left(-4\right)±\sqrt{64}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{-\left(-4\right)±8}{2\left(-1\right)}$$

Ko te tauaro o -4 ko 4.

$$x=\frac{4±8}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{4±8}{-2}$$

Nā, me whakaoti te whārite x=\frac{4±8}{-2} ina he tāpiri te ±. Tāpiri 4 ki te 8.

$$x=\frac{12}{-2}$$

Whakawehe 12 ki te -2.

$$x=-6$$

Nā, me whakaoti te whārite x=\frac{4±8}{-2} ina he tango te ±. Tango 8 mai i 4.

$$x=-\frac{4}{-2}$$

Whakawehe -4 ki te -2.

$$x=2$$

Tauwehea te kīanga taketake mā te whakamahi i te ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Me whakakapi te -6 mō te x_{1} me te 2 mō te x_{2}.

$$-x^{2}-4x+12=-\left(x-\left(-6\right)\right)\left(x-2\right)$$

Whakamāmātia ngā kīanga katoa o te āhua p-\left(-q\right) ki te p+q.

$$-x^{2}-4x+12=-\left(x+6\right)\left(x-2\right)$$

Aromātai

$$\left(2-x\right)\left(x+6\right)$$