What is the tangent of f(x)=2x^2-9x+10 ?

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

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tangent of f(x)=2x^2-9x+10

tangent of

f (x) = 2 x^{2} - 9 x + 10

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

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} + 10)

Find the slope of

f (x) = 2 x^{2} - 9 x + 10 : \frac{df ( x )}{d x} = 4 x - 9

Compute the slope of

f (x) = 2 x^{2} - 9 x + 10

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

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

f^{^{'}} (x) = 6 x + 1 2^{\frac{3}{4}}

\int (e^{x} - e^{- x})^{2} d x

x \to 3 lim (x + 5)

\frac{\partial}{\partial y} (sin (8 x) cos (3 y))

(6 x^{2} - 9 x) e^{x}

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