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

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

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tangent of 2x^2+x-1,\at a=0

tangent of

2 x^{2} + x - 1, at a = 0

y = (4 a + 1) x - 2 a^{2} - 1

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, 2 a^{2} + a - 1)

Find the slope of

y = 2 x^{2} + x - 1 : \frac{d y}{d x} = 4 x + 1

Compute the slope of

y = 2 x^{2} + x - 1

x = a : m = 4 a + 1

Find the line with slope m=

4 a + 1

and passing through

(a, 2 a^{2} + a - 1) : y = (4 a + 1) x - 2 a^{2} - 1

y = (4 a + 1) x - 2 a^{2} - 1

\int_{0}^{3} x^{3} - x d x

f (x) = (4 x^{4} - 7 x^{2} + 6)

(x + \frac{1}{x})^{3}

x \to 2 lim (5 x^{3} - 9 x^{2} + 7)

\int x^{3} \cdot cos (x) d x

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