f(x)=ax²+bx+5,且f(x+1)=f(x)+8x+3,求a,b的值,
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f(x)=ax²+bx+5,且f(x+1)=f(x)+8x+3,求a,b的值,
f(x)=ax²+bx+5,且f(x+1)=f(x)+8x+3,求a,b的值,
f(x)=ax²+bx+5,且f(x+1)=f(x)+8x+3,求a,b的值,
因为f(x)=ax²+bx+5,
所以f(x+1)=a(x+1)²+b(x+1)+5
=ax²+2ax+a+bx+b+5
=ax²+(2a+b)x+a+b+5
f(x)+8x+3=ax²+bx+5+8x+3
=ax²+(b+8)x+8
可得:ax²+(2a+b)x+a+b+5 =ax²+(b+8)x+8
可构造方程组:
2a+b=b+8
a+b+5=8
解得:
a=4
b=-1
f(x+1)=f(x)+8x+3=ax²+bx+5+8x+3=ax^2+(b+8)x+8
f(x+1)=a(x+1)^2+b(x+1)+5=ax^2+(2a+b)x+(a^2+b+5)
ax^2+(b+8)x+8=ax^2+(2a+b)x+(a^2+b+5)
b+8=2a+b
8=a^2+b+5
a=4
b=-13
f(x+1)=a(x+1)^2+b(x+1)+5=ax^2+bx+5+2ax+a^2+b=f(x)+2ax+a^2+b
所以2a=8
a^2+b=3
a=4,b=-13
f(x+1)=a(x+1)²+b(x+1)+5
=ax^2+2ax+a+bx+b+5
=ax^2+bx+5+(2ax+a+b)
=f(x)+(2ax+a+b)
所以 2a=8
a+b=3
所以:a=4,b=-1
=>a(x+1)²+b(x+1)+5=ax²+bx+5 +8x+3
=>ax²+(2a+b)x+a+b+5=ax²+(b+8)x+8
=>2a+b=b+8 ; a+b+5=8
=> a=4 ;b=-1