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

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

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

tangent of

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

y = x - 1

Solution steps

Find the tangent point:

(0, - 1)

Find the slope of

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

Compute the slope of

f (x) = 2 x^{2} + x - 1

x = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, - 1) : y = x - 1

y = x - 1

\frac{d}{d x} (lo g_{10} (\frac{x}{x ^{2} + 1}))

\frac{\partial}{\partial x} (15 - 2 x^{2} - y^{2})

(3 + x^{4})^{\frac{4}{5}}

x \to 0 + lim (\frac{10}{x ^{2}})

f (x) = 17 e^{x} - e x^{e}

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