What is the normal of y=4x^2-6x,\at x=2 ?

The normal of y=4x^2-6x,\at x=2 is y=-1/10 x+21/5

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normal of y=4x^2-6x,\at x=2

normal of

y = 4 x^{2} - 6 x, at x = 2

y = - \frac{1}{10} x + \frac{21}{5}

Solution steps

Find the tangent point:

(2, 4)

Find the slope of

y = 4 x^{2} - 6 x : \frac{d y}{d x} = 8 x - 6

Compute the slope of

y = 4 x^{2} - 6 x

x = 2 : m = 10

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{10}

Find the line with slope m=

- \frac{1}{10}

and passing through

(2, 4) : y = - \frac{1}{10} x + \frac{21}{5}

y = - \frac{1}{10} x + \frac{21}{5}

\frac{d y}{d x} = \frac{y ^{2} + x ^{2}}{y x}

\int cos^{9} (x) d x

\int (8 x^{3} - 3 x^{2} + 8) d x

\frac{d}{d x} (x^{5} e^{6 x})

x \to \infty lim (\frac{2}{x})

What is the normal of y=4x^2-6x,\at x=2 ?
The normal of y=4x^2-6x,\at x=2 is y=-1/10 x+21/5