[tex]\underline{\textbf{First part:}} \\\\\text{Given that,}\\\\~~~~~~y= 5x^2-3x+6\\\\\\\implies \dfrac{dy}{dx} = \dfrac{dy}{dx} (5x^2 -3x +6)\\ \\\\~~~~~~~~~~~=5\dfrac{d}{dx} x^2-3\dfrac{d}{dx} x + \dfrac{d}{dx} (6)\\\\\\~~~~~~~~~~~=5\cdot 2x-3+0\\\\\\~~~~~~~~~~~=10x-3\\\\\text{At (0,1)}\\\\\dfrac{dy}{dx} = 10(0) -3 \\\\~~~~~=0-3\\\\~~~~~=-3[/tex]
[tex]\underline{\textbf{Second part:}}\\\\[/tex]
[tex]\text{Equation of normal,}\\\\~~~~~~y-y_1 = -\dfrac{1}{\tfrac{dy}{dx}} (x-x_1)\\\\\\\implies y - 1 = -\dfrac 1{-3 }(x-0)\\\\\\\implies y-1 = \dfrac 13 x\\\\\\\implies y= \dfrac 13 x +1[/tex]
