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

The normal of y=sqrt(3x^2+2x),(2,4) is y=-4/7 x+36/7

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

normal of

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

y = - \frac{4}{7} x + \frac{36}{7}

Solution steps

Find the slope of

y = 3 x^{2} + 2 x : \frac{d y}{d x} = \frac{3 x + 1}{3 x ^{2} + 2 x}

Compute the slope of

y = 3 x^{2} + 2 x

(2, 4) : m = \frac{7}{4}

Compute the slope of the perpendicular line:

m_{p} = - \frac{4}{7}

Find the line with slope m=

- \frac{4}{7}

and passing through

(2, 4) : y = - \frac{4}{7} x + \frac{36}{7}

y = - \frac{4}{7} x + \frac{36}{7}

\int sin^{2} (x co) s^{2} x d x

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 6, y^{^{'}} (0) = 0

f (x) = \frac{sin ( x )}{2 + cos ( x )}

x \to 0 lim (x^{2} + 1)

\frac{d y}{d x} = x + 4

What is the normal of y=sqrt(3x^2+2x),(2,4) ?
The normal of y=sqrt(3x^2+2x),(2,4) is y=-4/7 x+36/7