What is the normal of y=3x^2+13x+4,\at x=2 ?

The normal of y=3x^2+13x+4,\at x=2 is y=-1/25 x+1052/25

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normal of y=3x^2+13x+4,\at x=2

normal of

y = 3 x^{2} + 13 x + 4, at x = 2

y = - \frac{1}{25} x + \frac{1052}{25}

Solution steps

Find the tangent point:

(2, 42)

Find the slope of

y = 3 x^{2} + 13 x + 4 : \frac{d y}{d x} = 6 x + 13

Compute the slope of

y = 3 x^{2} + 13 x + 4

x = 2 : m = 25

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{25}

Find the line with slope m=

- \frac{1}{25}

and passing through

(2, 42) : y = - \frac{1}{25} x + \frac{1052}{25}

y = - \frac{1}{25} x + \frac{1052}{25}

(- 5, 0)

(23, - 2)

x ∣ x ∣

f (x) = \frac{1}{x}

(17.1) (- 2.8)

What is the normal of y=3x^2+13x+4,\at x=2 ?
The normal of y=3x^2+13x+4,\at x=2 is y=-1/25 x+1052/25