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

The tangent of f(x)=-3x^2+3x+1,\at x=1 is y=-3x+4

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

tangent of

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

y = - 3 x + 4

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

f (x) = - 3 x^{2} + 3 x + 1 : \frac{df ( x )}{d x} = - 6 x + 3

Compute the slope of

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

x = 1 : m = - 3

Find the line with slope m=

- 3

and passing through

(1, 1) : y = - 3 x + 4

y = - 3 x + 4

y = \frac{2}{1 - 9 x ^{2}}

\int_{0}^{\frac{π}{2}} x cos (\frac{x}{3}) d x

\frac{2 ln ( x )}{x}

\frac{d y}{d x} = 2 y (9 - y)

\frac{d x}{d t} = x (1 - \frac{1}{2500} x)

What is the tangent of f(x)=-3x^2+3x+1,\at x=1 ?
The tangent of f(x)=-3x^2+3x+1,\at x=1 is y=-3x+4