What is the normal of y=5+sqrt(1-x),(-8,8) ?

The normal of y=5+sqrt(1-x),(-8,8) is y=6x+56

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normal of y=5+sqrt(1-x),(-8,8)

normal of

y = 5 + 1 - x, (- 8, 8)

y = 6 x + 56

Solution steps

Find the slope of

y = 5 + 1 - x : \frac{d y}{d x} = - \frac{1}{2 - x + 1}

Compute the slope of

y = 5 + 1 - x

(- 8, 8) : m = - \frac{1}{6}

Compute the slope of the perpendicular line:

m_{p} = 6

Find the line with slope m=

6

and passing through

(- 8, 8) : y = 6 x + 56

y = 6 x + 56

\int_{0}^{12} 2 π y (144 - y^{2}) d y

\int cos (x) sec (x) sin (x) d x

\frac{d}{d x} (1 + x e^{x})

x \to 4 + lim (ln (x - 4))

\frac{3 s + 2}{( s + 1 ) ^{3}}

What is the normal of y=5+sqrt(1-x),(-8,8) ?
The normal of y=5+sqrt(1-x),(-8,8) is y=6x+56