What is the normal of y=x^2-3x,(5,10) ?

The normal of y=x^2-3x,(5,10) is y=-1/7 x+75/7

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normal of y=x^2-3x,(5,10)

normal of

y = x^{2} - 3 x, (5, 10)

y = - \frac{1}{7} x + \frac{75}{7}

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

(5, 10) : m = 7

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{7}

Find the line with slope m=

- \frac{1}{7}

and passing through

(5, 10) : y = - \frac{1}{7} x + \frac{75}{7}

y = - \frac{1}{7} x + \frac{75}{7}

\int sec (38 x) tan (38 x) d x

\frac{d}{d x} (3^{4})

\int (3 x - 2) d x

\frac{d}{d u} (u + v)

\int 12 lo g_{11} (x) d x

What is the normal of y=x^2-3x,(5,10) ?
The normal of y=x^2-3x,(5,10) is y=-1/7 x+75/7