Acurve has the equation y = x^3 + 3x^2 − 16x + 2. a) find an equation of the tangent to the curve at the point p (2, −10). ive done that part and got it right. then it says the tangent to the curve at the point q is parallel to the tangent at the point p. b find the coordinates of the point q. im so confused as to how its (-4,50) for point q. could someone explain?

The equation of the tangent to the curve at the point P(2, -10) is:
y = 8x - 26



We need to find the coordinates of point Q where the slope of the tangent to the curve f(x) must also be 8.

Now we have a quadratic:

which simplifies to:

which factorizes to:(x + 4)(x - 2) = 0
Therefore x = -4, 2.
f(-4) = 48
Therefore the coordinates of Q are (-4, 48).

idk sorry but I hope everyone is having an amazing day