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

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

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

tangent of

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

y = 3 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

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

x = 0 : m = 3

Find the line with slope m=

3

and passing through

(0, 0) : y = 3 x

y = 3 x

\frac{\partial}{\partial x} (x ln (t))

5 y^{^{'}} - 15 x^{2} = 0

n \to 10 lim ([1 + (\frac{1}{n})]^{n})

n = 0 \sum \infty (\frac{i}{3})^{n}

\frac{d}{d x} (2 (1 - x)^{- 3})

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