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

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

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

tangent of

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

y = 3 x - 2

Solution steps

Find the tangent point:

(0, - 2)

Find the slope of

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

Compute the slope of

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

x = 0 : m = 3

Find the line with slope m=

3

and passing through

(0, - 2) : y = 3 x - 2

y = 3 x - 2

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

x \to 1 + lim (x^{3})

\int \frac{1 + x ^{2}}{- 9 x} d x

\int tan^{2} (y) d y

y = \frac{1}{x ^{3}}

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