What is the tangent of y=2x^2-9x+7 ?

The tangent of y=2x^2-9x+7 is y=(4a_{0}-9)x-2a_{0}^2+7

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tangent of y=2x^2-9x+7

tangent of

y = 2 x^{2} - 9 x + 7

y = (4 a_{0} - 9) x - 2 a_{0}^{2} + 7

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{2} - 9 a_{0} + 7)

Find the slope of

y = 2 x^{2} - 9 x + 7 : \frac{d y}{d x} = 4 x - 9

Compute the slope of

y = 2 x^{2} - 9 x + 7

x = a_{0} : m = 4 a_{0} - 9

Find the line with slope m=

4 a_{0} - 9

and passing through

(a_{0}, 2 a_{0}^{2} - 9 a_{0} + 7) : y = (4 a_{0} - 9) x - 2 a_{0}^{2} + 7

y = (4 a_{0} - 9) x - 2 a_{0}^{2} + 7

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

\int (x x + 2) d x

\frac{d}{d x} (2 x - 0.1 x^{2})

y = 9 + 4 x^{2} - 2 x^{3}, (1, 11)

\frac{d}{d x} (\frac{9}{ln ( x )})

What is the tangent of y=2x^2-9x+7 ?
The tangent of y=2x^2-9x+7 is y=(4a_{0}-9)x-2a_{0}^2+7