The curve with equation $y = x^3 - 3x^2 + 2x + 5$ is shown on the diagram - Scottish Highers Maths - Question 7 - 2018

To find the equation of the tangent at point P, we first need to calculate the derivative of the curve:

$y' = 3x^2 - 6x + 2$

Next, we evaluate this derivative at $x = 0$ to find the gradient at P:

$y'(0) = 3(0)^2 - 6(0) + 2 = 2$

Now, using the point-slope form, the equation of the tangent line at point P ($0, 5$) is given by:

$y - 5 = 2(x - 0)$

Simplifying this, we get:

$y = 2x + 5$

To find the coordinates of Q, we need to set the original curve equal to the tangent line:

$x^3 - 3x^2 + 2x + 5 = 2x + 5$

This simplifies to:

$x^3 - 3x^2 = 0$

Factoring gives:

$x^2(x - 3) = 0$

Thus, we have two solutions: $x = 0$ and $x = 3$. Since $x = 0$ corresponds to point P, we use $x = 3$ to find the coordinates of Q:

Substituting $x = 3$ back into the curve equation:

$y = 3^3 - 3(3)^2 + 2(3) + 5 = 27 - 27 + 6 + 5 = 11$

Therefore, the coordinates of Q are $(3, 11)$.