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

The normal of y=(1+2x)/(3+x),(2,1) is y=-5x+11

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

normal of

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

y = - 5 x + 11

Solution steps

Find the slope of

y = \frac{1 + 2 x}{3 + x} : \frac{d y}{d x} = \frac{5}{( 3 + x ) ^{2}}

Compute the slope of

y = \frac{1 + 2 x}{3 + x}

(2, 1) : m = \frac{1}{5}

Compute the slope of the perpendicular line:

m_{p} = - 5

Find the line with slope m=

- 5

and passing through

(2, 1) : y = - 5 x + 11

y = - 5 x + 11

\frac{\partial}{\partial x} (5 x^{2} + 4 y^{2})

\int (x^{4} + 1) e^{x^{5} + 5 x} d x

n = 1 \sum \infty \frac{1}{( 2 n ) !}

x \to 0 lim ((1 - 6 x)^{\frac{1}{x}})

f (x) = 4 sec (x), at x = \frac{π}{3}

What is the normal of y=(1+2x)/(3+x),(2,1) ?
The normal of y=(1+2x)/(3+x),(2,1) is y=-5x+11