亚洲精品不卡AV在线播放|国产性爱无码高清视频|国产成人 ,97人人色,国产免费一区二区三区,日本一区二区在线

To solve the problem of finding the quadratic function (f(x)) given the conditions (f(0)=1) and (f(x+1)-f(x)=2x), follow these steps:

Step 1: Assume the form of the quadratic function

Since (f(x)) is quadratic, let:
[f(x) = ax^2 + bx + c \quad (a \neq 0)]

Step 2: Use (f(0)=1) to find (c)

Substitute (x=0):
[f(0) = a(0)^2 + b(0) + c = c = 1]
Thus, (c=1).

Step 3: Compute (f(x+1)-f(x)) and equate coefficients

First, expand (f(x+1)):
[f(x+1) = a(x+1)^2 + b(x+1) + c = a(x^2 +2x+1) +b(x+1)+1]
[= ax^2 +2ax +a +bx +b +1]

Subtract (f(x)):
[f(x+1)-f(x) = [ax^2 +2ax +a +bx +b +1] - [ax^2 +bx +1]]
[=2ax + (a +b)]

Given this equals (2x), equate coefficients:

• Coefficient of (x): (2a =2 \implies a=1)
• Constant term: (a +b =0 \implies b=-1) (since (a=1))

Step 4: Write the final function

Substitute (a=1), (b=-1), (c=1):
[f(x) = x^2 -x +1]

Answer: (\boxed{x^2 -x +1})

（免責(zé)聲明：本文為本網(wǎng)站出于傳播商業(yè)信息之目的進(jìn)行轉(zhuǎn)載發(fā)布，不代表本網(wǎng)站的觀點(diǎn)及立場(chǎng)。本文所涉文、圖、音視頻等資料的一切權(quán)利和法律責(zé)任歸材料提供方所有和承擔(dān)。本網(wǎng)站對(duì)此資訊文字、圖片等所有信息的真實(shí)性不作任何保證或承諾，亦不構(gòu)成任何購(gòu)買(mǎi)、投資等建議，據(jù)此操作者風(fēng)險(xiǎn)自擔(dān)。） 本文為轉(zhuǎn)載內(nèi)容，授權(quán)事宜請(qǐng)聯(lián)系原著作權(quán)人，如有侵權(quán)，請(qǐng)聯(lián)系本網(wǎng)進(jìn)行刪除。