methods stationary point prob

[Echelon5]

The Tangent to the curve of y=ax^2+bx+c at the point where x = 2 is parallel to the line y-4x. There us a stationary point at (1,-3). Find the Value if a,b and c.

How please, and whats the best approach to doing these kinds of questions.

[tram]

ok

first, sub the point (1,-3) into the problem. This gives you one enq.

next. differentiate the original funtion. sub in x=1 and let the differentiated function equal 0 as you know when x=0, there is a stationary point. You now have two equations

next, bcs  when x=2, the tangent is parallel to y=4  => when x=2, the gradient is 4. sub this into the differentiated function. there's ur third equation.

you now have three function and three unkowns. Do the hack work and solve (or use ur caculator )

[Echelon5]

Thank you sir

[tram]

np, now get to 50 posts and karma me!

btw, with these type of questions, the important thing is to be reli sure to use ALL of the information you are given, e.g. being told "There is a stationary point at (1,-3)" is actually two pieces of information, not just one.