What is the tangent of 4x^2-3x+2 ?

The tangent of 4x^2-3x+2 is y=(8a_{0}-3)x-4a_{0}^2+2

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tangent of 4x^2-3x+2

tangent of

4 x^{2} - 3 x + 2

y = (8 a_{0} - 3) x - 4 a_{0}^{2} + 2

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} - 3 a_{0} + 2)

Find the slope of

y = 4 x^{2} - 3 x + 2 : \frac{d y}{d x} = 8 x - 3

Compute the slope of

y = 4 x^{2} - 3 x + 2

x = a_{0} : m = 8 a_{0} - 3

Find the line with slope m=

8 a_{0} - 3

and passing through

(a_{0}, 4 a_{0}^{2} - 3 a_{0} + 2) : y = (8 a_{0} - 3) x - 4 a_{0}^{2} + 2

y = (8 a_{0} - 3) x - 4 a_{0}^{2} + 2

\int_{1}^{3} x^{4} d x

f^{^{'}} (x) = 2 x + 3 x^{1.7}

\frac{d}{d x} (\frac{1}{x ( ln ( x ) ) ^{\frac{2}{3}}})

\frac{d}{d θ} (\frac{sin ( θ )}{1 + cos ( θ )})

\frac{\partial}{\partial z} (2 x z^{2} cos (y))

What is the tangent of 4x^2-3x+2 ?
The tangent of 4x^2-3x+2 is y=(8a_{0}-3)x-4a_{0}^2+2