請問這個怎麼算？ 將f(x)=2x²+6x+1配方?
回答 (2)
2016-08-28 1:30 pm
Sol
2x^2+6x+1
=2(x^2+3x)+1
=2[x^2+2*x*(3/2)]+1
=2[x^2+2*x*(3/2)+(3/2)^2]+1－2*(3/2)^2
=2(x+3/2)^2+1－9/2
=2(x+3/2)^2－7/2
2x^2+6x+1
=2(x^2+3x)+1
=2[x^2+2*x*(3/2)]+1
=2[x^2+2*x*(3/2)+(3/2)^2]+1－2*(3/2)^2
=2(x+3/2)^2+1－9/2
=2(x+3/2)^2－7/2
2016-11-12 4:31 am
f(x)=2x²+6x+1
f'(x)=4x+6
f'(x)=0
4x+6=0
x=-3/2
f(x)=2(-3/2)^2+6(-3/2)+1
f(x)=-7/2
f(x)=2(x+3/2)^2 -7/2
f'(x)=4x+6
f'(x)=0
4x+6=0
x=-3/2
f(x)=2(-3/2)^2+6(-3/2)+1
f(x)=-7/2
f(x)=2(x+3/2)^2 -7/2
收錄日期: 2021-04-18 15:28:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160828045507AA6A68W