I am struggling with a question and need some help…
The question is the following:
There is a parabola which joins two tangents (L1 and L2) meeting at a point.
L1 has a slope of -1.8 and L2 a slope of 0.7.
P is given as $(0,0)$, and Q is given as $(30,y)$
I am trying to find the ‘a’ value for $ax^2 + bx + c$, as I have already found that $b=-1.8$ and $c=0$.
Where can I go from here to solve for the a value of the parabola?
Thanks! Much appreciated. (Also this is my first time posting here, so forgive me if I am incorrectly formatting this question)
It looks like you have already set the derivative $f'(x) = 2ax + b$ to $-1.8$ at $x = 0$. Now you just need to use the fact that $f'(30) = 0.7$, as the gradient of the tangent at $x = 30$ is $0.7$.
Doing this you should get $60a - 1.8 = 0.7$ or $a = \frac{5}{120}$.