What is the tangent of y=4.5x^2-3x+2 ?

The tangent of y=4.5x^2-3x+2 is y=(9a_{0}-3)x+(-9a_{0}^2+4)/2

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tangent of y=4.5x^2-3x+2

tangent of

y = 4.5 x^{2} - 3 x + 2

y = (9 a_{0} - 3) x + \frac{- 9 a _{0}^{2} + 4}{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4.5 a_{0}^{2} - 3 a_{0} + 2)

Find the slope of

y = 4.5 x^{2} - 3 x + 2 : \frac{d y}{d x} = 9 x - 3

Compute the slope of

y = 4.5 x^{2} - 3 x + 2

x = a_{0} : m = 9 a_{0} - 3

Find the line with slope m=

9 a_{0} - 3

and passing through

(a_{0}, 4.5 a_{0}^{2} - 3 a_{0} + 2) : y = (9 a_{0} - 3) x + \frac{- 9 a _{0}^{2} + 4}{2}

y = (9 a_{0} - 3) x + \frac{- 9 a _{0}^{2} + 4}{2}

\frac{\partial}{\partial x} (e^{\frac{x}{2}})

y^{^{'}} = sec^{2} (x) (1 - 3 y), y (\frac{π}{4}) = 0

f (x) = x^{2} + 16, at x = 3

\int sin^{3} (5 x) cot (5 x) d x

6 y^{^{''}} + 9 y^{^{'}} = 0

What is the tangent of y=4.5x^2-3x+2 ?
The tangent of y=4.5x^2-3x+2 is y=(9a_{0}-3)x+(-9a_{0}^2+4)/2