The equation of the tangent to the curve at the point P(2, -10) is:
y = 8x - 26
\(f(x)=x^{3}+3x^{2}-16x\)
\(f'(x)=3x^{2}+6x-16\)
\(f'(2)=12+12-16=8\)
We need to find the coordinates of point Q where the slope of the tangent to the curve f(x) must also be 8.
\(f'(x)=3x^{2}+6x-16=8\)
Now we have a quadratic:
\(3x^{2}+6x-16-8=0\)
which simplifies to:
\(x^{2}+2x-8=0\)
which factorizes to:(x + 4)(x - 2) = 0
Therefore x = -4, 2.
f(-4) = 48
Therefore the coordinates of Q are (-4, 48).
