What is the tangent of 4x^2+2x-1 ?

The tangent of 4x^2+2x-1 is y=(8a_{0}+2)x-4a_{0}^2-1

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tangent of 4x^2+2x-1

tangent of

4 x^{2} + 2 x - 1

y = (8 a_{0} + 2) x - 4 a_{0}^{2} - 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} + 2 a_{0} - 1)

Find the slope of

y = 4 x^{2} + 2 x - 1 : \frac{d y}{d x} = 8 x + 2

Compute the slope of

y = 4 x^{2} + 2 x - 1

x = a_{0} : m = 8 a_{0} + 2

Find the line with slope m=

8 a_{0} + 2

and passing through

(a_{0}, 4 a_{0}^{2} + 2 a_{0} - 1) : y = (8 a_{0} + 2) x - 4 a_{0}^{2} - 1

y = (8 a_{0} + 2) x - 4 a_{0}^{2} - 1

f (x) = (5 x^{6} + 4 x^{3})^{4}

\frac{\partial}{\partial x} (ln (8 x + 7 y))

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y = x e^{x}

\int_{1}^{5} \frac{x}{x ^{2} + 6 x + 13} d x

\int \frac{1}{1 + e ^{12 x}} d x

What is the tangent of 4x^2+2x-1 ?
The tangent of 4x^2+2x-1 is y=(8a_{0}+2)x-4a_{0}^2-1