The gradient of the tangent drawn at any point in a curve us given by m= dy/dx = 1-4x
integrate the derivative dy/dx = 1 - 4x.
Integration gives us,
y = x - 2x^2 + const
We are told thsat it passes through the point (x,y) = (-2, 4)
Therefore x = -2 and y = 4 satisfy the equation of the curve above.
Substituting fior x = -2 and y = 4 into the equation y = x - 2x^2 + const,
4 = -2 - 2(-2)^2 + const
4 = -2 -2*(4) + const
6 = -8 + const
const = 14
The equation of the curve then is: y = x -2x^2 + 14