What is the tangent of f(x)=9x^2-2x ?

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

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tangent of f(x)=9x^2-2x

tangent of

f (x) = 9 x^{2} - 2 x

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = 9 x^{2} - 2 x : \frac{df ( x )}{d x} = 18 x - 2

Compute the slope of

f (x) = 9 x^{2} - 2 x

x = a_{0} : m = 18 a_{0} - 2

Find the line with slope m=

18 a_{0} - 2

and passing through

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

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

f (x) = 7 x + 6

\frac{d}{d x} (5 cos (4 x))

\int \frac{3 x + 5}{x ^{2} + x + 1} d x

\int \frac{t}{1 - t ^{4}} d t

h = (x x^{2} + 2), (x + h, (x + h)^{2} + 2)

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