What is the normal of y=(3x^2+1)^2,(0,1) ?

The normal of y=(3x^2+1)^2,(0,1) is x=0

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normal of y=(3x^2+1)^2,(0,1)

normal of

y = (3 x^{2} + 1)^{2}, (0, 1)

x = 0

Solution steps

Find the slope of

y = (3 x^{2} + 1)^{2} : \frac{d y}{d x} = 12 x (3 x^{2} + 1)

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 0

Compute the slope of the perpendicular line:

m_{p} = \infty

Find the line with slope m=

\infty

and passing through

(0, 1) : x = 0

x = 0

\int \frac{x ^{2}}{4 - 9 x ^{2}} d x

\frac{\partial}{\partial x} (x^{2} - y^{2} - z^{2})

f (x) = 2 x^{2} - 1

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

y^{^{''}} + 16 y = cos (4 x) + sin (4 x)

What is the normal of y=(3x^2+1)^2,(0,1) ?
The normal of y=(3x^2+1)^2,(0,1) is x=0