If #(2,6)# lies on the curve #f(x) = ax^2+bx # and #y=x+4# is a tangent to the curve at that point. Find #a# and #b#?

Question

1240 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-31T21:22:37

# a=-1, b=5#

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

We have:

# f(x) = ax^2+bx #

We are given that the point #(2,6)# lies on the curve:

# => f(2)=6 #
# :. 4a+2b = 6 #
# :. 2a+b = 3 # ..... [A]

If we differentiate the parabola equation, then we have:

# f'(x) = 2ax+b #

The gradient of the tangent at #(2,6)# is given by:

# m_T = f'(2) = 4a+b #

We also know that the tangent equation is #y=x+4# and so comparing with the standard straight line equation #y=mx+c#:

# => m_T = 1 #
# :. 4a+b = 1 # ..... [B]

We now solve the equations [A] and [B] simultaneous:

# [B]-[A] => 2a = -2 => a=-1#

Substitute #a=-1# into [A]:

# b-2=3 => b=5#

Hence we have:

# a=-1, b=5#