What is the tangent of f(x)=1+2x-3x^2,\at x=0 ?

The tangent of f(x)=1+2x-3x^2,\at x=0 is y=2x+1

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tangent of f(x)=1+2x-3x^2,\at x=0

tangent of

f (x) = 1 + 2 x - 3 x^{2}, at x = 0

y = 2 x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

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

Compute the slope of

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

x = 0 : m = 2

Find the line with slope m=

2

and passing through

(0, 1) : y = 2 x + 1

y = 2 x + 1

\frac{d}{d x} (\frac{ln ( 1 - x ^{2} )}{x ^{2}})

\frac{d}{d x} (x^{5} + 1)

\int \frac{cot ( ln ( x ^{4} + 1 ) ) x ^{3}}{x ^{4} + 1} d x

\frac{( s ^{2} + 1 )}{s ( s - 1 ) ^{3}}

\frac{\partial}{\partial x} (e^{x y + 1})

What is the tangent of f(x)=1+2x-3x^2,\at x=0 ?
The tangent of f(x)=1+2x-3x^2,\at x=0 is y=2x+1