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

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

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

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = 18 a_{0} + 9

Find the line with slope m=

18 a_{0} + 9

and passing through

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

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

f (x) = sin (1 + cos (x))

\int_{0}^{1} \frac{14}{4 y - 1} d y

\frac{d}{d p} (ln (p))

\int x x - 3 d x

\int x^{2} 5 x + 7 d x

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