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

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#?

Luke Alvarez · Accepted Answer

# 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#

Emilia Aguilar · Answer

undefined To find the values of a and b, we need to use the given information that (2,6) lies on the curve f(x) = ax^2 + bx and y = x + 4 is a tangent to the curve at that point.

First, we substitute the coordinates of the point (2,6) into the equation f(x) = ax^2 + bx:
6 = a(2)^2 + b(2)

Simplifying this equation, we get:
6 = 4a + 2b

Next, we differentiate the equation f(x) = ax^2 + bx to find the derivative, which represents the slope of the curve at any given point:
f'(x) = 2ax + b

Since y = x + 4 is a tangent to the curve at (2,6), the slope of the tangent line should be equal to the slope of the curve at that point. Therefore, we set the derivative equal to the slope of the tangent line, which is 1:
2a(2) + b = 1

Simplifying this equation, we get:
4a + b = 1

Now, we have a system of two equations:
6 = 4a + 2b
4a + b = 1

Solving this system of equations, we find that a = 1 and b = -2.