If #y=ax^2+bx# then
#y'=2ax+b#.
This gives us our slope of #y# at any given #x#. So at the point #(1,1)#, the slope must be
#y'=2a(1)+b=2a+b#
We know the slope must also be #3# at the point #(1,1)#, to match the linear equation given. Thus, these two slope values must be equal:
#2a+b=3# [1]
We also know that #(1,1)# is a point on the parabola, so it must satisfy the original parabola equation, thus:
#1=a(1)^2+b(1)#
#=>a+b=1# [2]
So if both [1] and [2] are true, we can combine them to create a system of linear equations, to solve for the unknowns #a# and #b#.
#" "2a+b=3#
#-(a+b=1)#
"==========="
#" "a" "=2#
And if #a=2#, then
#" "a+b=1#
#=>2+b=1#
#=>" "b="-"1#
So our parabola equation #y=ax^2+bx# becomes
#y=2x^2-x#.
Check:
#y=2x^2-x#
#=>y'=4x-1#
#=>y'|_(x=1)=4(1)-1=3# #" "#(Slope at #(1,1)# is 3.)
#y=m(x-x_1)+y_1" "#(Slope-point form of a line)
#=>y=3(x-1)+1#
#=>y=3x-3+1#
#=>y=3x-2" "#(Matches given line equation)
