What is the tangent of 2x^2-10x+9 ?

The tangent of 2x^2-10x+9 is y=(4a_{0}-10)x-2a_{0}^2+9

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tangent of 2x^2-10x+9

tangent of

2 x^{2} - 10 x + 9

y = (4 a_{0} - 10) x - 2 a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{2} - 10 a_{0} + 9)

Find the slope of

y = 2 x^{2} - 10 x + 9 : \frac{d y}{d x} = 4 x - 10

Compute the slope of

y = 2 x^{2} - 10 x + 9

x = a_{0} : m = 4 a_{0} - 10

Find the line with slope m=

4 a_{0} - 10

and passing through

(a_{0}, 2 a_{0}^{2} - 10 a_{0} + 9) : y = (4 a_{0} - 10) x - 2 a_{0}^{2} + 9

y = (4 a_{0} - 10) x - 2 a_{0}^{2} + 9

\frac{4 x ^{2} - 3}{x - 1}

\frac{2}{3 x + 5}

\int \frac{cot ( x )}{ln ( sin ( x ) )} d x

x \to 3 - lim (\frac{x ^{3}}{x ^{2} - 9})

\int_{0}^{1} 2 x e^{- x^{2}} d x

What is the tangent of 2x^2-10x+9 ?
The tangent of 2x^2-10x+9 is y=(4a_{0}-10)x-2a_{0}^2+9