The tangent to the curve at #y=x^2+ax+b# where a and b are constants is #2x+y=6# at the point where x=1 and how do you find a and b?

1 Answer
Jul 19, 2016

#a=-4, b=7 -> y = x^2-4x+7#

Explanation:

#y=x^2+ax+b#

#y' =2x+a#

We are told that the tangent at #x=1# is #2x+y=6#
#-> y=-2x+6#
Therefore the slope of the tangent #=-2#

Since the slope of y at #x=1# is given by #y'(1)#
#2(1) + a = -2#
#a=-4#

Now notice that #y(1) = 1^2+(-4)*1 +b#
#y(1)= -3+b#

Since the tangent touches #y# at #x=1#
#y(1)=-2(1)+6 -> y(1)=4#

Therefore: #-3+b =4 -> b=7#

Hence: #a=-4, b=7#
#y = x^2-4x+7#