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

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

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

tangent of

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

y = 5 x + 1

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 5

Find the line with slope m=

5

and passing through

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

y = 5 x + 1

e^{- 2 t} cos (2 t)

f (x) = 7 x^{3}

\frac{\partial}{\partial x} (x e^{(x^{2} + y^{2})})

\int sin^{2} (6 x) cos^{2} (6 x) d x

x^{2}, 12 - x

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