What is the tangent of ax^2+bx+c ?

The tangent of ax^2+bx+c is y=(2aa_{3}+b)x-aa_{3}^2+c

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tangent of ax^2+bx+c

tangent of

a x^{2} + b x + c

y = (2 a a_{3} + b) x - a a_{3}^{2} + c

Solution steps

Compute the tangent line to the general point

x = a_{3}

Find the tangent point:

(a_{3}, a a_{3}^{2} + b a_{3} + c)

Find the slope of

y = a x^{2} + b x + c : \frac{d y}{d x} = 2 a x + b

Compute the slope of

y = a x^{2} + b x + c

x = a_{3} : m = 2 a a_{3} + b

Find the line with slope m=

2 a a_{3} + b

and passing through

(a_{3}, a a_{3}^{2} + b a_{3} + c) : y = (2 a a_{3} + b) x - a a_{3}^{2} + c

y = (2 a a_{3} + b) x - a a_{3}^{2} + c

\frac{d}{d x} (- 4 cos (4 x))

\frac{\partial}{\partial y} (sin (x) + π cos (y))

2 v^{'} t (x) = v

\int \frac{( 1 - y ) e ^{y}}{y ^{2}} d y

y (8 x - 9 y) d x + 2 x (x - 3 y) d y = 0

What is the tangent of ax^2+bx+c ?
The tangent of ax^2+bx+c is y=(2aa_{3}+b)x-aa_{3}^2+c