What is the normal of y=x^2-4x,(3,-3) ?

The normal of y=x^2-4x,(3,-3) is y=-1/2 x-3/2

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normal of y=x^2-4x,(3,-3)

normal of

y = x^{2} - 4 x, (3, - 3)

y = - \frac{1}{2} x - \frac{3}{2}

Solution steps

Find the slope of

y = x^{2} - 4 x : \frac{d y}{d x} = 2 x - 4

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(3, - 3) : y = - \frac{1}{2} x - \frac{3}{2}

y = - \frac{1}{2} x - \frac{3}{2}

\frac{d}{d x} (9.75 + 1.5 x^{3})

\int_{0}^{8} \frac{2 x - 1}{x + 1} d x

\int \frac{1}{2 x ^{2} - 2 x + 3} d x

\frac{d}{d t} (t e^{- t})

\int 1 - cos (4 x) d x

What is the normal of y=x^2-4x,(3,-3) ?
The normal of y=x^2-4x,(3,-3) is y=-1/2 x-3/2