九年級(jí)數(shù)學(xué)二次函數(shù)y=a(xh)2k的圖象

二次函數(shù)y=a(x–h)2的圖象和性質(zhì).

當(dāng)h>0時(shí),向右平移當(dāng)h<0時(shí),向左平移y=ax2y=a(x–h)2復(fù)習(xí)回顧1.如何同y=-x2的圖象得到y(tǒng)=-x2-3的圖象。并說(shuō)明后者圖象的頂點(diǎn)，對(duì)稱(chēng)軸，增減性。2.如何y=2x2的圖象得到y(tǒng)=2(x-3)2的圖象。并說(shuō)明后者圖象的頂點(diǎn)，對(duì)稱(chēng)軸，增減性。Oxy1234512345–5–4–3–2–1–5–4–3–2–1y頂點(diǎn)從(0,0)移到了(0,–2)，即x=0時(shí)，y取最大值–2頂點(diǎn)從(0,0)移到了(0,2)，即x=0時(shí)，y取最大值2Oxy1234512345–5–4–3–2–1–5–4–3–2–1y頂點(diǎn)從(0,0)移到了(2,0)，即x=2時(shí)，y取最大值0頂點(diǎn)從(0,0)移到了(–2,0)，即x=–2時(shí)，y取最大值0xy112233445567-1-1-2-3-4-50-2二次函數(shù)y=a(x–h)2+k的圖像和性質(zhì)Oxy1234512345–5–4–3–2–1–5–4–3–2–1向上y軸(0,0)向上直線x=1(1,0)向上直線x=1(1,1)y=2x2y=2(x–1)2y=2(x–1)2+1上Oxy1234512345–5–4–3–2–1–5–4–3–2–1畫(huà)出函數(shù)的圖像二次函數(shù)y=a(x–h)2+k的圖象和性質(zhì).

當(dāng)k>0時(shí),向上平移當(dāng)k<0時(shí),向下平移a＞0時(shí)，開(kāi)口_____,最____點(diǎn)是頂點(diǎn);a＜0時(shí)，開(kāi)口_____,最____點(diǎn)是頂點(diǎn);對(duì)稱(chēng)軸是_______，頂點(diǎn)坐標(biāo)是______。y=a(x–h)2y=a(x–h)2+k的圖象y=a(x–h)2+k

向上低向下高直線x=h(h，k)指出下列函數(shù)圖象的開(kāi)口方向,對(duì)稱(chēng)軸和頂點(diǎn)坐標(biāo).開(kāi)口對(duì)稱(chēng)軸頂點(diǎn)坐標(biāo)向上直線x=3(3,–5)向下直線x=–1(–1,0)向下直線x=0(0,–1)向上直線x=2(2,5)向上直線x=–4(–4,2)向下直線x=3(3,0)課堂練習(xí)1.拋物線y=0.5(x+2)2–3可以由拋物線

先向

平移2個(gè)單位，在向下平移

個(gè)單位得到。2.已知s=–(x+1)2–3，當(dāng)x為

時(shí)，s取最

值為

。3.頂點(diǎn)坐標(biāo)為(1,1)，且經(jīng)過(guò)原點(diǎn)的拋物線的函數(shù)解析式是()y=(x+1)2+1B.y=–(x+1)2+1C.y=(x–1)2+1D.y=–(x–1)2+1y=0.5x2左3–1大–3D4.已知一條拋物線的形狀與開(kāi)口方向都與拋物線y=–x2相同，它的頂點(diǎn)在直線y=2x+1上，且經(jīng)過(guò)這條直線與x軸的交點(diǎn)，求這條拋物線的解析式。5.如何來(lái)求與坐標(biāo)軸的交點(diǎn)？求y=x2+2x-8與坐標(biāo)軸的交點(diǎn)。根據(jù)圖象回答何時(shí)y<0?何時(shí)y>0?考點(diǎn)訓(xùn)練6.已知二次