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

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

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

tangent of

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

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

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 = 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 9 sin (3 x) d x

x \to - 2 lim (x^{2} - 4)

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\int_{2 π}^{\infty} sin (x) d x

\int_{1}^{e} ln (18 x) d x

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