What is the tangent of y=2x^2+5x ?

The tangent of y=2x^2+5x is y=(4a_{0}+5)x-2a_{0}^2

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tangent of y=2x^2+5x

tangent of

y = 2 x^{2} + 5 x

y = (4 a_{0} + 5) x - 2 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{2} + 5 a_{0})

Find the slope of

y = 2 x^{2} + 5 x : \frac{d y}{d x} = 4 x + 5

Compute the slope of

y = 2 x^{2} + 5 x

x = a_{0} : m = 4 a_{0} + 5

Find the line with slope m=

4 a_{0} + 5

and passing through

(a_{0}, 2 a_{0}^{2} + 5 a_{0}) : y = (4 a_{0} + 5) x - 2 a_{0}^{2}

y = (4 a_{0} + 5) x - 2 a_{0}^{2}

\int (x + \frac{1}{9 x}) d x

\int ln (1 + x^{3}) d x

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 4, y^{^{'}} (0) = 7

\frac{\partial}{\partial z} (x^{2} + y^{2} + z^{2})

y^{^{'}} + 9 x e^{y} = 0

What is the tangent of y=2x^2+5x ?
The tangent of y=2x^2+5x is y=(4a_{0}+5)x-2a_{0}^2