What is the tangent of 5x^2-6x+1 ?

The tangent of 5x^2-6x+1 is y=(10a_{0}-6)x-5a_{0}^2+1

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tangent of 5x^2-6x+1

tangent of

5 x^{2} - 6 x + 1

y = (10 a_{0} - 6) x - 5 a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 a_{0}^{2} - 6 a_{0} + 1)

Find the slope of

y = 5 x^{2} - 6 x + 1 : \frac{d y}{d x} = 10 x - 6

Compute the slope of

y = 5 x^{2} - 6 x + 1

x = a_{0} : m = 10 a_{0} - 6

Find the line with slope m=

10 a_{0} - 6

and passing through

(a_{0}, 5 a_{0}^{2} - 6 a_{0} + 1) : y = (10 a_{0} - 6) x - 5 a_{0}^{2} + 1

y = (10 a_{0} - 6) x - 5 a_{0}^{2} + 1

f (x) = 9 x sin (x)

f (x) = e^{- 5 x}, at x = 0

\frac{d}{d s} (\frac{s ^{2} - 8 s + 15}{s ^{2} + 3 s + 2})

2 y y^{^{'}} + 4 = y^{2} + 4 x

x (7 x + 3)^{\frac{1}{3}}

What is the tangent of 5x^2-6x+1 ?
The tangent of 5x^2-6x+1 is y=(10a_{0}-6)x-5a_{0}^2+1