Put the linear part into y=mx+c form i.e. y=-2x+b , the gradient part 'm' should equal y' when x=2,
i.e. solve for a when 2a(2) = -2
now you know the initial quadratic equation. Can you finish it?
I didn't really know what to put as a title so I hope it describes the problem relatively well... I have been working on this problem for about 20 minutes and haven't come very far, so I was hoping for someones help in putting me in the right direction!
For what values of a and b is the line tangent to the parabola when x = 2?
I feel I have all the pieces of the puzzle, but there is some trick to putting it together.
I know how to find the equation of a tangent line when given an equation and a point on that equation (just find the derivative and then use y = mx + b and sub in everything to find b). However in this question I don't even know where to start. I know that .
If I sub 2 into either y' or y I get y = 4a(or y' = 4a). How do I use this information to solve for a and b?
This feels like an easy problem that I just can't wrap my head around!
Thanks for your help!
A point which lies on the parabola must have as its coordinates. You want to find the parameters of a tangent line in , so, necessarily, the point must lie on both parabola and the line. This is your first information.
The second information you have is that the slope of the line must be equal to the slope of the parabola at the point . So the slope of the tangent line is , then . The slope of your line is , therefore you just put to find .
Now you just plug this result with the first information on the line equation: , then .