What is the tangent of 5x^2-3x-5 ?

The tangent of 5x^2-3x-5 is y=(10a_{0}-3)x-5a_{0}^2-5

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tangent of 5x^2-3x-5

tangent of

5 x^{2} - 3 x - 5

y = (10 a_{0} - 3) x - 5 a_{0}^{2} - 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 a_{0}^{2} - 3 a_{0} - 5)

Find the slope of

y = 5 x^{2} - 3 x - 5 : \frac{d y}{d x} = 10 x - 3

Compute the slope of

y = 5 x^{2} - 3 x - 5

x = a_{0} : m = 10 a_{0} - 3

Find the line with slope m=

10 a_{0} - 3

and passing through

(a_{0}, 5 a_{0}^{2} - 3 a_{0} - 5) : y = (10 a_{0} - 3) x - 5 a_{0}^{2} - 5

y = (10 a_{0} - 3) x - 5 a_{0}^{2} - 5

f (x) = e^{x^{7}}, (- 1, \frac{1}{e})

\frac{d}{d x} (\frac{x ^{2} cos ( e ^{x} )}{( 2 x + 1 ) ^{3}})

\int_{0}^{\frac{π}{2}} 7 cos^{5} (x) d x

\int x^{3} x^{2} - 49 d x

\frac{d}{d x} (ln (1 0^{x}))

What is the tangent of 5x^2-3x-5 ?
The tangent of 5x^2-3x-5 is y=(10a_{0}-3)x-5a_{0}^2-5