What is the normal of y=x^4+6e^x,(0,6) ?

The normal of y=x^4+6e^x,(0,6) is y=-1/6 x+6

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normal of y=x^4+6e^x,(0,6)

normal of

y = x^{4} + 6 e^{x}, (0, 6)

y = - \frac{1}{6} x + 6

Solution steps

Find the slope of

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

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 = 6

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{6}

Find the line with slope m=

- \frac{1}{6}

and passing through

(0, 6) : y = - \frac{1}{6} x + 6

y = - \frac{1}{6} x + 6

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

(x - 2)^{2}, x

f (x) = 3 + 4 x

(x + 2 y) y^{^{'}} = - y - 2 x

\frac{( 5 + 5 s )}{( s ^{2} + 25 )}

What is the normal of y=x^4+6e^x,(0,6) ?
The normal of y=x^4+6e^x,(0,6) is y=-1/6 x+6