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

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

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tangent of 2x^2-9x+10

tangent of

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

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

Compute the slope of

y = 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

\frac{2}{x - 4}

\int_{0}^{3} 1

\frac{\partial}{\partial z} (x)

\int_{0}^{\frac{4}{3}} \frac{1}{16 + 9 x ^{2}} d x

\int sin (9 x) cos (4 x) d x

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