Differentiation

Question

The tangent to the curve y=ax^2+bx+2 at (1,0.5) is parallel to the normal to the curve y=x^2+6x+4 at (-2,-4).Find the value of a and b.

Answer 1

y=x^2+6x+4

y'=2x+6

y'(-2)=2(-2)+6 = -4+6 = 2

 

The normal to the curve at this point would have a slope of -1/2 then. So the tangent to the first curve has a slope of -1/2 at (1,0.5).

 

y=ax^2+bx+2

y'=2ax+b

y'(1)=2a+b=-0.5

          a+b=-1.5

 

Subtract the two equations:

 

a=1

b=-2.5