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

The tangent of f(x)=-3x^2+2,\at x=1 is y=-6x+5

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

tangent of

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

y = - 6 x + 5

Solution steps

Find the tangent point:

(1, - 1)

Find the slope of

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

Compute the slope of

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

x = 1 : m = - 6

Find the line with slope m=

- 6

and passing through

(1, - 1) : y = - 6 x + 5

y = - 6 x + 5

f (x) = \frac{4 x ^{3}}{5 - 2 x}

x \to 6 + lim (\frac{4}{x - 6})

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

\int \frac{1}{u} d u

\frac{\partial}{\partial x} (e^{2 x yz - 7})

What is the tangent of f(x)=-3x^2+2,\at x=1 ?
The tangent of f(x)=-3x^2+2,\at x=1 is y=-6x+5