What is the tangent of f(x)=2+3x-5x^2,\at x=1 ?

The tangent of f(x)=2+3x-5x^2,\at x=1 is y=-7x+7

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tangent of f(x)=2+3x-5x^2,\at x=1

tangent of

f (x) = 2 + 3 x - 5 x^{2}, at x = 1

y = - 7 x + 7

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = 2 + 3 x - 5 x^{2} : \frac{df ( x )}{d x} = - 10 x + 3

Compute the slope of

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

x = 1 : m = - 7

Find the line with slope m=

- 7

and passing through

(1, 0) : y = - 7 x + 7

y = - 7 x + 7

h \to 1 lim (4 h^{2} + x + 3)

\frac{d}{d x} (e^{x} sin (9 x))

e^{- 2 x} + \frac{1}{3} e^{x}

x \to \infty lim (0 x)

\frac{1}{5 s + 1}

What is the tangent of f(x)=2+3x-5x^2,\at x=1 ?
The tangent of f(x)=2+3x-5x^2,\at x=1 is y=-7x+7