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

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

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

tangent of

f (x) = 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

f (x) = 4 x^{2} - 3 x + 2 : \frac{df ( x )}{d x} = 8 x - 3

Compute the slope of

f (x) = 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_{0}^{5} x^{2} d x

\frac{d}{d x} (y^{1 - n})

\int - \frac{x ^{3}}{3 e ^{\frac{x}{2}}} d x

\frac{d}{d q} (500 q - \frac{q ^{2}}{ln ( q + 300 )})

\frac{\partial}{\partial x} (tan (9 + 2 x^{2} y^{4} z^{2}))

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