3.3 Finding Equation of a Curve from its Gradient Function

3.3 Finding Equation of a Curve from its Gradient Function


Example 1:
Find the equation of the curve that has the gradient function   dy dx =2x+8 and passes through the point (2, 3).

Solution:
y= ( 2x+8 ) y= 2 x 2 2 +8x+C
y = x2 + 8x + C
3 = 22 +8(2) + C  (2, 3)
C = –17
Hence, the equation of the curve is
 y = x2 + 8x – 17


Example 2:
The gradient function of a curve is given by 2x – 4 and the curve has a minimum value of 3. Find the equation of the curve.

Solution:
At the point where a curve has a minimum value,  dy dx =0

dy dx =0
2x – 4 = 0
x = 2
Therefore minimum point = (2, 3).

dy dx =2x4 y= ( 2x4 )dx y= 2 x 2 2 4x+C y= x 2 4x+C

When x = 2, y = 3.
3 = 22 – 4(2) + c
c = 7

Hence, the equation of the curve is
y = x2 – 4x + 7