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{\partial}{\partial x} (x^{8} y^{9} - x^{7} y^{8})

\frac{d}{d x} (2 (4 e^{x^{2}} x^{3} + 6 e^{x^{2}} x))

f (x) = sec (x)

y (θ) = \frac{π}{7} sin (θ)

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

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