Question

Please help me do this mathematics assignment

Answer 1

dy = -3

dx

equation of normal=

y=1/3x+1

Explanation:

dy/dx=10x-3

=10(0)-3

= -3

equation of normal:

grad=1/3

y-1 =1/3

x-0

3y-3=x

y= 1/3x+1

Answer 2

`\underline{\textbf{First part:}} \\\\\text{Given that,}\\\\~~~~~~y= 5x^2-3x+6\\\\\\\implies (dy)/(dx) = (dy)/(dx) (5x^2 -3x +6)\\ \\\\~~~~~~~~~~~=5(d)/(dx) x^2-3(d)/(dx) x + (d)/(dx) (6)\\\\\\~~~~~~~~~~~=5\cdot 2x-3+0\\\\\\~~~~~~~~~~~=10x-3\\\\\text{At (0,1)}\\\\(dy)/(dx) = 10(0) -3 \\\\~~~~~=0-3\\\\~~~~~=-3`

`\underline{\textbf{Second part:}}\\\\`

`\text{Equation of normal,}\\\\~~~~~~y-y_1 = -\frac{1}{\tfrac{dy}{dx}} (x-x_1)\\\\\\\implies y - 1 = -\frac 1{-3 }(x-0)\\\\\\\implies y-1 = \frac 13 x\\\\\\\implies y= \frac 13 x +1`