What is the normal of y=3x^2,(-4,48) ?

The normal of y=3x^2,(-4,48) is y= 1/24 x+289/6

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normal of y=3x^2,(-4,48)

normal of

y = 3 x^{2}, (- 4, 48)

y = \frac{1}{24} x + \frac{289}{6}

Solution steps

Find the slope of

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

Compute the slope of

y = 3 x^{2}

(- 4, 48) : m = - 24

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{24}

Find the line with slope m=

\frac{1}{24}

and passing through

(- 4, 48) : y = \frac{1}{24} x + \frac{289}{6}

y = \frac{1}{24} x + \frac{289}{6}

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

x \to - 3 lim (3 x + 9)

\frac{3}{4} x^{\frac{4}{3}} - \frac{3}{8} x^{\frac{2}{3}} + 7

6 y^{^{''}} - y^{^{'}} - y = 0, y (0) = 10, y^{^{'}} (0) = 3

\frac{d}{d x} (\frac{x - x}{x + x})

What is the normal of y=3x^2,(-4,48) ?
The normal of y=3x^2,(-4,48) is y= 1/24 x+289/6