How to find the values of a,b and c?

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Answer 1 · 2018-03-10T09:14:18

#a=3/2#, #b=-2# and #c=-5/2#

Before we solve it, two things.

Slope of tangent at a point #x=x_0# on a curve #y=f(x)# is equal to the value of #f'(x)=(df)/(dx)# at that point i.e. #f'(x_0)#.
Astationary point or critical point of a differentiable function of one variable is a point on the graph of the function where the function's derivative is zero.

As #f(x)=ax^2+bx+c#, #f'(x)=2ax+b#. Now as at #x=2#, tangent is parallel to #y=4x#, the slope of tangent at #x=2# is #4# i.e.

#4a+b=4# ............(1)

Further, stationary point is #(1,-3)#, hence #-3=a*1^2+b*1+c#

or #a+b+c=-3# ............(2)

and as at this point #x=1#, the derivative is #0#, we have #2ax+b=0# and hence

#2a+b=1# ............(3)

Subtracting (3) from (1), we get #2a=3# or #a=3/2#

and putting this in (3) we get

#3+b=1# or #b=1-3=-2#

Now putting these values in (2), we get

#3/2-2+c=-3# or #c=-5/2#

Hence, #a=3/2#, #b=-2# and #c=-5/2#

and polynomial is #3/2x^2-2x-5/2#