设出直线,y=k(x-1),设切点(x,y)
切点处导数相等
3x^2=k
y=x^3
y=k(x-1)
解得
x=3/2,y=27/8,k=27/4
对y=ax^2+15/4x-9
由相切得
y'=2ax+15/4
2a*3/2+15/4=27/4
a=1