Whakaoti i te =10+20x-2x^2+10x^2-40x+5

$$= 10 + 20 x - 2 x ^ { 2 } + 10 x ^ { 2 } - 40 x + 5$$

Aromātai

$$8x^{2}-20x+15$$

Ngā Upane Otinga

Pahekotia te -2x^{2} me 10x^{2}, ka 8x^{2}.

$$10+20x+8x^{2}-40x+5$$

Pahekotia te 20x me -40x, ka -20x.

$$10-20x+8x^{2}+5$$

Tāpirihia te 10 ki te 5, ka 15.

$$15-20x+8x^{2}$$

Kimi Pārōnaki e ai ki x

$$16x-20$$

Ngā Upane e Whakamahi ana i te Ture Pārōnaki mō te Tapeke

Pahekotia te -2x^{2} me 10x^{2}, ka 8x^{2}.

$$\frac{\mathrm{d}}{\mathrm{d}x}(10+20x+8x^{2}-40x+5)$$

Pahekotia te 20x me -40x, ka -20x.

$$\frac{\mathrm{d}}{\mathrm{d}x}(10-20x+8x^{2}+5)$$

Tāpirihia te 10 ki te 5, ka 15.

$$\frac{\mathrm{d}}{\mathrm{d}x}(15-20x+8x^{2})$$

Ko te pārōnaki o tētahi pūrau ko te tapeke o ngā pārōnaki o ōna kīanga tau. Ko te pārōnaki o tētahi kīanga tau pūmau ko 0. Ko te pārōnaki o te ax^{n} ko te nax^{n-1}.

$$-20x^{1-1}+2\times 8x^{2-1}$$

Tango 1 mai i 1.

$$-20x^{0}+2\times 8x^{2-1}$$

Whakareatia 2 ki te 8.

$$-20x^{0}+16x^{2-1}$$

Tango 1 mai i 2.

$$-20x^{0}+16x^{1}$$

Mō tētahi kupu t, t^{1}=t.

$$-20x^{0}+16x$$

Mō tētahi kupu t mahue te 0, t^{0}=1.

$$-20+16x$$