What is the normal of y=5+sqrt(4-x),(0,8) ?

The normal of y=5+sqrt(4-x),(0,8) is y=4.11309…x+8

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normal of y=5+sqrt(4-x),(0,8)

normal of

y = 5 + 4 - x, (0, 8)

y = 4.11309 \dots x + 8

Solution steps

Slope of the normal line at

(0, 8) : m = \frac{- x + 4 - 3}{x}

Find the slope of

y = 5 + 4 - x : m_{y} = - \frac{1}{2 - x + 4}

Compute the slope of the normal line:

m = 2 - x + 4

Solve

\frac{- x + 4 - 3}{x} = 2 - x + 4 : x \approx - 0.22937 \dots

Find the normal line at

x = - 0.22937 \dots : y = 4.11309 \dots x + 8

y = 4.11309 \dots x + 8

\int \frac{x + 1}{x} d x

y^{4}

(2 sin (t))^{^{'}}

x \to 2 lim (x^{2} + 7 - x)

\frac{\partial}{\partial x} (y^{4} + 5 x y^{3})

What is the normal of y=5+sqrt(4-x),(0,8) ?
The normal of y=5+sqrt(4-x),(0,8) is y=4.11309…x+8