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

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

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

tangent of

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

y = 7 x - 7

Solution steps

Find the tangent point:

(2, 7)

Find the slope of

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

Compute the slope of

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

x = 2 : m = 7

Find the line with slope m=

7

and passing through

(2, 7) : y = 7 x - 7

y = 7 x - 7

\int_{0}^{1} e^{- 0.1 t} d t

\frac{d}{d x} y = \frac{2 cos ( x )}{1 + sin ( x )}

\frac{d}{d x} (2 cos^{2} (\frac{x}{2}))

\frac{d y}{d x} + 0.7 y - 18.2 = 0

(7 x y^{2} - 3) d x + (7 x^{2} y + 4) d y = 0

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