Answer
$y=-3x+\dfrac{13}{4}$ and $\dfrac{d^2y}{dx^2}=6$
Work Step by Step
Here, $\dfrac{dy}{dx}=\dfrac{-3}{2}t$
At $t=2$
Then $ \dfrac{dy}{dx}=\dfrac{-3}{2}(2)=-3$
Now, $y+\dfrac{1}{2}=(-3)(x-\dfrac{5}{4})\implies y=-3x+\dfrac{13}{4}$
Now, $\dfrac{d^2y}{dx^2}=\dfrac{dy'/dt}{dx/dt}=\dfrac{3}{4}t^3$
At $t=2$
$\dfrac{d^2y}{dx^2}=\dfrac{3}{4} (2)^3=6$
