What is the normal of y=x^2-3x,(3,0) ?

The normal of y=x^2-3x,(3,0) is y=-1/3 x+1

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normal of y=x^2-3x,(3,0)

normal of

y = x^{2} - 3 x, (3, 0)

y = - \frac{1}{3} x + 1

Solution steps

Find the slope of

y = x^{2} - 3 x : \frac{d y}{d x} = 2 x - 3

Compute the slope of

y = x^{2} - 3 x

(3, 0) : m = 3

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{3}

Find the line with slope m=

- \frac{1}{3}

and passing through

(3, 0) : y = - \frac{1}{3} x + 1

y = - \frac{1}{3} x + 1

t (t - 4) y^{^{'}} + y = 0

x \to 2 lim (9 x^{2})

\frac{s}{( s + 2 ) ^{2}}

\frac{\partial}{\partial x} (y e^{x} - x e^{y})

x \to 0 + lim (\frac{1}{e ^{x} + 1})

What is the normal of y=x^2-3x,(3,0) ?
The normal of y=x^2-3x,(3,0) is y=-1/3 x+1