I have tried substituting the coordinate points (2,4) and the formula for the turning point (-b/2a, etc.) into the equation and solving it simultaneously, however this has not seemed to work. The resultant equation gave me a y-value of 4 when x=2, but it was a positive quadratic rather than a negative as is clearly shown in the diagram. Could someone kindly point me in the right direction?
As you said, we substitute the coordinates of Q into the equation and we get $4=4a+4b$ hence $a+b=1$. The slope of the tangent is just $\frac{y_{Q}-y_{U}}{x_{Q}-x_{U}}=-1$. Hence this means that the derivative of $y=ax^{2}+bx$ at x=2 is -1. y'(x)=2ax+b. $y'(2)=4a+b=-1$ you have a system of two equations with two uknowns, which you can now determine!