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

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

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tangent of 2x^2-x-3

tangent of

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

y = 2 x^{2} - x - 3 : \frac{d y}{d x} = 4 x - 1

Compute the slope of

y = 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

\frac{1}{( s ^{2} + 0.2 s + 1 ) s}

x \to - 3 lim (x^{2} + 3 x + 1)

cos (π x) \frac{1}{2}

\int x tan^{2} (x) d x

(x + 2 y) d x + (2 x - 3 y) d y = 0

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