What is the tangent of f(x)=3-x^2+2x ?

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

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tangent of f(x)=3-x^2+2x

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = 3 - x^{2} + 2 x : \frac{df ( x )}{d x} = - 2 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 = - 2 a_{0} + 2

Find the line with slope m=

- 2 a_{0} + 2

and passing through

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

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

f (x) = 8 x^{2} + 4 x - 6, at x = - 3

\int \frac{u - 2 u ^{2}}{u} d u

\int a e^{x} d x

\int \frac{8}{e ^{x} - 2} d x

\int_{- \infty}^{15} r e^{\frac{r}{5}} d r

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