若直线y=2x是抛物线y=x^2+ax+b上点(2,4)处的切线,求a、b各等于多少?(用导数的方法)写出详细步骤
\u5df2\u77e5\u76f4\u7ebfy=2x\u662f\u629b\u7269\u7ebfy=x2 ax b\u4e0a\u70b9(2.3)\u5904\u7684\u5207\u7ebf,\u6c42a.ba=0\uff0cb=1
\u89e3\uff1a
\u629b\u7269\u7ebfY\uff1dAX²\u8fc7\u70b9a\uff082,4\uff09\u65f6
a\u00d72²\uff1d4
\u5219a\uff1d1
\u5219\u629b\u7269\u7ebf\u7684\u89e3\u6790\u5f0f\u4e3aY\uff1dX²
\u628aY\uff1dX+2\u4ee3\u5165\u629b\u7269\u7ebfY\uff1dX²\u4e2d\uff0c\u5f97
X²\uff1dX+2
X²-X-2\uff1d0
\uff08X-2\uff09\uff08X+1\uff09\uff1d0
X1\uff1d2,X2\uff1d-1
\u628aX2\uff1d-1\u4ee3\u5165Y\uff1dX²\u4e2d\uff0c\u5f97
Y\uff1d1
\u5219\u70b9B\uff08-1,1\uff09
\u76f4\u7ebfY\uff1d2X+b\u8fc7\u70b9B\uff08-1,1\uff09\u65f6
-2+b\uff1d1
b\uff1d3
\u5219\u76f4\u7ebf\u7684\u89e3\u6790\u5f0f\u4e3aY\uff1d2X+3
\u628aY\uff1d2X+3\u4ee3\u5165\u629b\u7269\u7ebfY\uff1dX²\u4e2d\uff0c\u5f97
X²\uff1d2X+3
X²-2X-3\uff1d0
\uff08X+1\uff09\uff08X-3\uff09\uff1d0
\u5219X3\uff1d-1,X4\uff1d3
\u628aX4\uff1d3\u4ee3\u5165Y\uff1d2X+3\u4e2d\uff0c\u5f97
Y\uff1d2\u00d73+9\uff1d9
\u5219\u70b9C\u7684\u5750\u6807\uff083,9\uff09
将点(2,4)带入抛物线4=2*2+(-2)*2+b,解得:b=4.
解
抛物线 y=x²+ax+b
求导, y'=2x+a
由题设可知,
点(2,4)在该抛物线上,且当x=2时,导数为2.
∴由此可得
4+2a+b=4
4+a=2
解得: a=-2, b=4
直线y=2x是抛物线y=x^2+ax+b上点(2,4)处的切线
所以
4=4+2a+b
y'=2x+a=斜率=2 此时x=2,即
4+a=2
所以
a=-2
b=4
解:因为y=x^2+ax+b,所以y‘=2x+a,
抛物线y=x^2+ax+b上点(2,4)处的切线为4=2*2+a
所以a=0,y'=2x
因为点(2,4)在抛物线y=x^2+ax+b上,所以4=2^2+b,所以b=0
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