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

The tangent of f(x)=2x^2-x-3 is y=(4a_{0}-1)x-2a_{0}^2-3

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

tangent of

f (x) = 2 x^{2} - x - 3

y = (4 a_{0} - 1) x - 2 a_{0}^{2} - 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = 2 x^{2} - x - 3 : \frac{df ( x )}{d x} = 4 x - 1

Compute the slope of

f (x) = 2 x^{2} - x - 3

x = a_{0} : m = 4 a_{0} - 1

Find the line with slope m=

4 a_{0} - 1

and passing through

(a_{0}, 2 a_{0}^{2} - a_{0} - 3) : y = (4 a_{0} - 1) x - 2 a_{0}^{2} - 3

y = (4 a_{0} - 1) x - 2 a_{0}^{2} - 3

\int_{0}^{1} 8 e^{- 4 x} d x

(0.2) (3.5)

x \to 5 lim (x^{2} - 16)

\int - \frac{y}{x} d x

f (x) = (8 x^{2} - 14 x) e^{x}

What is the tangent of f(x)=2x^2-x-3 ?
The tangent of f(x)=2x^2-x-3 is y=(4a_{0}-1)x-2a_{0}^2-3