What is the tangent of y=x^2+12x-3 ?

The tangent of y=x^2+12x-3 is y=(2a_{0}+12)x-a_{0}^2-3

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tangent of y=x^2+12x-3

tangent of

y = x^{2} + 12 x - 3

y = (2 a_{0} + 12) x - a_{0}^{2} - 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 12 a_{0} - 3)

Find the slope of

y = x^{2} + 12 x - 3 : \frac{d y}{d x} = 2 x + 12

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2 a_{0} + 12

Find the line with slope m=

2 a_{0} + 12

and passing through

(a_{0}, a_{0}^{2} + 12 a_{0} - 3) : y = (2 a_{0} + 12) x - a_{0}^{2} - 3

y = (2 a_{0} + 12) x - a_{0}^{2} - 3

\int x^{2 n - 2} d x

y = \frac{8 x ^{2} + 4 x + 8}{x}

\int_{- 1}^{1} t^{3} (1 + t^{4})^{3} d t

\frac{d}{d x} (\frac{e ^{x}}{ln ( x )})

\int 1 - \frac{x}{2} d x

What is the tangent of y=x^2+12x-3 ?
The tangent of y=x^2+12x-3 is y=(2a_{0}+12)x-a_{0}^2-3