What is the normal of y=x^2-5x+3,(2,-3) ?

The normal of y=x^2-5x+3,(2,-3) is y=x-5

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normal of y=x^2-5x+3,(2,-3)

normal of

y = x^{2} - 5 x + 3, (2, - 3)

y = x - 5

Solution steps

Find the slope of

y = x^{2} - 5 x + 3 : \frac{d y}{d x} = 2 x - 5

Compute the slope of

y = x^{2} - 5 x + 3

(2, - 3) : m = - 1

Compute the slope of the perpendicular line:

m_{p} = 1

Find the line with slope m=

1

and passing through

(2, - 3) : y = x - 5

y = x - 5

n = 5 \sum \infty 3 (0.9)^{n + 1}

x \to 0 lim (x - ln (x))

y^{^{'}} + y^{5} x + 9 y = 0

x \to - \frac{π}{2} lim (tan (x))

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