Substituting the values of a and b found in part (i),

We deduce that the equation of the curve is :

y = 2x² + 2(2)x + 4(2) -14
y = 2x² + 4x - 6

Gradient of tangent at some value of x

= dy/dx
= 2(2x) + 4
= 4x + 4

Gradient of the line y = -4x + 1 is : -4

Since the the tangent to the curve at point P is parallel to the line y = -4x + 1, their gradients are equal.

So, 4x + 4 = -4
4x = -8
x = -2

So the x-cordinate of P is -2.

y = 2(-2)² + 4(-2) - 6
y = 8 - 8 - 6
y = -6

Coordinates of P are (-2,-6)

Since we know the gradient of the tangent to the curve at point P, and we have found P's coordinates already,

substitute m = -4 , x = -2 and y = -6 into genersl equation of a line y = mx + c

-6 = -4(-2) + c
-6 = 8 + c
c = -14

So the y-intercept = -14

Therefore, equation of the tangent to the curve at point P is y = -4x - 14

Thank you so much.