What is the normal of 2x^2-3x+6,\at x=2 ?

The normal of 2x^2-3x+6,\at x=2 is y=-1/5 x+42/5

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normal of 2x^2-3x+6,\at x=2

normal of

2 x^{2} - 3 x + 6, at x = 2

y = - \frac{1}{5} x + \frac{42}{5}

Solution steps

Find the tangent point:

(2, 8)

Find the slope of

y = 2 x^{2} - 3 x + 6 : \frac{d y}{d x} = 4 x - 3

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 = 5

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{5}

Find the line with slope m=

- \frac{1}{5}

and passing through

(2, 8) : y = - \frac{1}{5} x + \frac{42}{5}

y = - \frac{1}{5} x + \frac{42}{5}

n = 5 \sum \infty n! (x - 4)^{n}

x^{2}

y^{^{'}} y - x^{2} y = - x y

\int tan^{5} (x) sec^{3} (x) d x

\int_{0}^{1} x x^{2} + 4 d x

What is the normal of 2x^2-3x+6,\at x=2 ?
The normal of 2x^2-3x+6,\at x=2 is y=-1/5 x+42/5