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

The normal of y=2x^3-x^2+2,(1,3) is y=-1/4 x+13/4

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normal of y=2x^3-x^2+2,(1,3)

normal of

y = 2 x^{3} - x^{2} + 2, (1, 3)

y = - \frac{1}{4} x + \frac{13}{4}

Solution steps

Find the slope of

y = 2 x^{3} - x^{2} + 2 : \frac{d y}{d x} = 6 x^{2} - 2 x

Compute the slope of

y = 2 x^{3} - x^{2} + 2

(1, 3) : m = 4

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{4}

Find the line with slope m=

- \frac{1}{4}

and passing through

(1, 3) : y = - \frac{1}{4} x + \frac{13}{4}

y = - \frac{1}{4} x + \frac{13}{4}

\int x^{2} cos (nπ x) d x

cos (x^{2})

\int \frac{1}{sec ( x ^{2} )} d x

\int (e^{\frac{x}{a}} + 1)^{\frac{1}{3}} e^{\frac{x}{a}} d x

\int_{2}^{4} \frac{x ( 2 x - 3 )}{6} d x

What is the normal of y=2x^3-x^2+2,(1,3) ?
The normal of y=2x^3-x^2+2,(1,3) is y=-1/4 x+13/4