What is the tangent of y=4x^2+3x(-3.27) ?

The tangent of y=4x^2+3x(-3.27) is y=(8a_{0}-9.81)x-4a_{0}^2

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tangent of y=4x^2+3x(-3.27)

tangent of

y = 4 x^{2} + 3 x (- 3.27)

y = (8 a_{0} - 9.81) x - 4 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} - 9.81 a_{0})

Find the slope of

y = 4 x^{2} + 3 x (- 3.27) : \frac{d y}{d x} = 8 x - 9.81

Compute the slope of

y = 4 x^{2} + 3 x (- 3.27)

x = a_{0} : m = 8 a_{0} - 9.81

Find the line with slope m=

8 a_{0} - 9.81

and passing through

(a_{0}, 4 a_{0}^{2} - 9.81 a_{0}) : y = (8 a_{0} - 9.81) x - 4 a_{0}^{2}

y = (8 a_{0} - 9.81) x - 4 a_{0}^{2}

x \to 2 lim (\frac{( x + 2 )}{x - 2})

\frac{d}{d y} (\frac{1}{3} (y^{2} + 2)^{\frac{3}{2}})

\int cos (2) d x

\int cos^{4} (x) sin^{2} (x) d x

\int_{0}^{3} sin (x^{2}) d x

What is the tangent of y=4x^2+3x(-3.27) ?
The tangent of y=4x^2+3x(-3.27) is y=(8a_{0}-9.81)x-4a_{0}^2