What is the tangent of f(x)=-3x^2+6x+4,\at x=1 ?

The tangent of f(x)=-3x^2+6x+4,\at x=1 is y=7

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tangent of f(x)=-3x^2+6x+4,\at x=1

tangent of

f (x) = - 3 x^{2} + 6 x + 4, at x = 1

y = 7

Solution steps

Find the tangent point:

(1, 7)

Find the slope of

f (x) = - 3 x^{2} + 6 x + 4 : \frac{df ( x )}{d x} = - 6 x + 6

Compute the slope of

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

x = 1 : m = 0

Find the line with slope m=

0

and passing through

(1, 7) : y = 7

y = 7

x \to 0 lim (\frac{ln ( x + 2 )}{x})

\frac{d y}{d x} = \frac{y}{y - x}

y = x^{2}, y = 9

x^{\frac{7}{2}}

\frac{d}{d t} (t e^{- 2 t})

What is the tangent of f(x)=-3x^2+6x+4,\at x=1 ?
The tangent of f(x)=-3x^2+6x+4,\at x=1 is y=7