What is the normal of y=(-2y+x+2)/(2y-x+4),(-3,-1) ?

The normal of y=(-2y+x+2)/(2y-x+4),(-3,-1) is y=-2.68416…x-9.05248…

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normal of y=(-2y+x+2)/(2y-x+4),(-3,-1)

normal of

y = \frac{- 2 y + x + 2}{2 y - x + 4}, (- 3, - 1)

y = - 2.68416 \dots x - 9.05248 \dots

Solution steps

y = \frac{- 2 y + x + 2}{2 y - x + 4} : y = \frac{x - 6 + x ^{2} - 4 x + 52}{4}, y = \frac{x - 6 - x ^{2} - 4 x + 52}{4}

Slope of the normal line at

(- 3, - 1) : m = \frac{x - x ^{2} - 4 x + 52 - 2}{4 ( x + 3 )}

Find the slope of

y = \frac{x - 6 + x ^{2} - 4 x + 52}{4}, y = \frac{x - 6 - x ^{2} - 4 x + 52}{4} : m_{y} = \frac{1}{4} (- \frac{x - 2}{x ^{2} - 4 x + 52} + 1)

Compute the slope of the normal line:

m = - \frac{4 x ^{2} - 4 x + 52}{- x + 2 + x ^{2} - 4 x + 52}

Solve

\frac{x - x ^{2} - 4 x + 52 - 2}{4 ( x + 3 )} = - \frac{4 x ^{2} - 4 x + 52}{- x + 2 + x ^{2} - 4 x + 52} : x \approx - 1.89671 \dots

Find the normal line at

x = - 1.89671 \dots : y = - 2.68416 \dots x - 9.05248 \dots

y = - 2.68416 \dots x - 9.05248 \dots

f (x) = - sin (3 x) \cdot 3

\frac{\partial}{\partial z} (- yz)

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

\int 8 x + 7 d x

x \to 0 lim (\frac{2 ^{x} - 1}{x})

What is the normal of y=(-2y+x+2)/(2y-x+4),(-3,-1) ?
The normal of y=(-2y+x+2)/(2y-x+4),(-3,-1) is y=-2.68416…x-9.05248…