What is the tangent of y=9x^2+3x-9 ?

The tangent of y=9x^2+3x-9 is y=(18a_{0}+3)x-9a_{0}^2-9

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tangent of y=9x^2+3x-9

tangent of

y = 9 x^{2} + 3 x - 9

y = (18 a_{0} + 3) x - 9 a_{0}^{2} - 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 9 a_{0}^{2} + 3 a_{0} - 9)

Find the slope of

y = 9 x^{2} + 3 x - 9 : \frac{d y}{d x} = 18 x + 3

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 = 18 a_{0} + 3

Find the line with slope m=

18 a_{0} + 3

and passing through

(a_{0}, 9 a_{0}^{2} + 3 a_{0} - 9) : y = (18 a_{0} + 3) x - 9 a_{0}^{2} - 9

y = (18 a_{0} + 3) x - 9 a_{0}^{2} - 9

- \frac{y ^{2}}{t ^{2}} d t + \frac{2 y}{t} d y = 0

x \to - 3 lim (- h (x))

(5 x^{6} + 4 x^{3})^{4}

\frac{d}{d x} (\frac{1 - x}{1 + x})

x \to 3 lim (6 f (x) + 2 x)

What is the tangent of y=9x^2+3x-9 ?
The tangent of y=9x^2+3x-9 is y=(18a_{0}+3)x-9a_{0}^2-9