The curve C has equation $y = \frac{1}{3}x^3 - 4x^2 + 8x + 3$ - Edexcel - A-Level Maths Pure - Question 10 - 2005 - Paper 1

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

Given $y = \frac{1}{3}x^3 - 4x^2 + 8x + 3$, we differentiate:
$\frac{dy}{dx} = x^2 - 8x + 8$
Now we evaluate the derivative at $x = 3$:
$\frac{dy}{dx} (3) = (3)^2 - 8(3) + 8 = 9 - 24 + 8 = -7$
Thus, the slope (m) of the tangent at point P is -7.

Next, we use the point-slope form of a line:
$y - y_1 = m(x - x_1)$
Substituting P (3, 0) into this equation:
$y - 0 = -7(x - 3)$
Simplifying gives:
$y = -7x + 21$
Therefore, the equation of the tangent at P is:
$y = -7x + 21$

For point Q, we need to find another point on the curve where the tangent is parallel to the tangent at P, which has a slope of -7.
To find where the derivative equals -7, we set up the equation:
$x^2 - 8x + 8 = -7$
Rearranging gives:
$x^2 - 8x + 15 = 0$
Factoring this, we find:
$(x - 3)(x - 5) = 0$
Thus, $x = 3$ or $x = 5$.
Since $x = 3$ corresponds to point P, we take $x = 5$.
To find the corresponding y-coordinate, we substitute $x = 5$ back into the original equation of the curve:
$y = \frac{1}{3}(5)^3 - 4(5)^2 + 8(5) + 3$
Calculating gives:
$y = \frac{1}{3}(125) - 4(25) + 40 + 3$
$y = \frac{125}{3} - 100 + 40 + 3$
$y = \frac{125}{3} - 300/3 + 120/3 + 9/3$
$y = \frac{125 - 300 + 120 + 9}{3} = \frac{-46}{3}$
Thus, the coordinates of point Q are:
$(5, \frac{-46}{3})$.