What is the tangent of f(x)=3x^2+4x+1,\at x=0 ?

The tangent of f(x)=3x^2+4x+1,\at x=0 is y=4x+1

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tangent of f(x)=3x^2+4x+1,\at x=0

tangent of

f (x) = 3 x^{2} + 4 x + 1, at x = 0

y = 4 x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

f (x) = 3 x^{2} + 4 x + 1 : \frac{df ( x )}{d x} = 6 x + 4

Compute the slope of

f (x) = 3 x^{2} + 4 x + 1

x = 0 : m = 4

Find the line with slope m=

4

and passing through

(0, 1) : y = 4 x + 1

y = 4 x + 1

8 cos^{2} (y) d x + csc^{2} (x) d y = 0, y (\frac{π}{12}) = \frac{π}{4}

f (x) = \frac{x + 3}{x}

\frac{d}{d x} (\frac{a + b x + c x ^{2}}{x})

f (x) = x - 1, (17, 4)

n = 0 \sum \infty 4 - sin (n)

What is the tangent of f(x)=3x^2+4x+1,\at x=0 ?
The tangent of f(x)=3x^2+4x+1,\at x=0 is y=4x+1