What is the tangent of f(x)=-5x^2+2x ?

The tangent of f(x)=-5x^2+2x is y=(-10a_{0}+2)x+5a_{0}^2

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tangent of f(x)=-5x^2+2x

tangent of

f (x) = - 5 x^{2} + 2 x

y = (- 10 a_{0} + 2) x + 5 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 5 a_{0}^{2} + 2 a_{0})

Find the slope of

f (x) = - 5 x^{2} + 2 x : \frac{df ( x )}{d x} = - 10 x + 2

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 = - 10 a_{0} + 2

Find the line with slope m=

- 10 a_{0} + 2

and passing through

(a_{0}, - 5 a_{0}^{2} + 2 a_{0}) : y = (- 10 a_{0} + 2) x + 5 a_{0}^{2}

y = (- 10 a_{0} + 2) x + 5 a_{0}^{2}

f (x) = 5 x^{2} + 3

\frac{d}{d x} (\frac{3 sin ( x ) - 11 cos ( x )}{x - 2})

n = 1 \sum \infty \frac{n ^{n}}{5 ^{n}}

\int \frac{1}{3 - 6 x} d x

\frac{d}{d x} (x^{2} sin (x) tan (x))

What is the tangent of f(x)=-5x^2+2x ?
The tangent of f(x)=-5x^2+2x is y=(-10a_{0}+2)x+5a_{0}^2