# 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  $\frac{dy}{dx}=2x+8$ and passes through the point (2, 3).

Solution:
$\begin{array}{l}y=\int \left(2x+8\right)\\ y=\frac{2{x}^{2}}{2}+8x+C\end{array}$
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,  $\frac{dy}{dx}=0$
$\frac{dy}{dx}=0$
2x – 4 = 0
x = 2

Therefore minimum point = (2, 3).

$\begin{array}{l}\frac{dy}{dx}=2x-4\\ y=\int \left(2x-4\right)dx\\ y=\frac{2{x}^{2}}{2}-4x+C\\ y={x}^{2}-4x+C\end{array}$

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

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