Question

Suppose $a,b,c$ are such that the curve $y=a{x}^2+bx+c$ has tangent $y=3x-3$ at $(1,0)$ and has tangent $y=x+1$ at $(3,4)$, then the value of $(2a-b-4c)$ equals to

Answer 1

$y=a{x}^2+bx+c$
$\frac{dy}{dx}=2ax+b$
When
$x=1,y=0$
$\Rightarrow a+b+c=0\dots \left(1\right)$
${\left(\frac{dy}{dx}\right)}_{x=1}=3$ and ${\left(\frac{dy}{dx}\right)}_{x=3}=1$
$\Rightarrow 2a+b=3\cdots \left(2\right)$
$\Rightarrow 6a+b=1\cdots \left(3\right)$
Solving $\left(1\right),\left(2\right)$ and $\left(3\right),$ we get
$a=-\frac{1}{2}$, $b=4$, $c=-\frac{7}{2}$
$\therefore 2a-b-4c=-1-4+14=9$