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

The tangent of y=x^2+3 is y=2a_{0}x-a_{0}^2+3

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tangent of y=x^2+3

tangent of

y = x^{2} + 3

y = 2 a_{0} x - a_{0}^{2} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = x^{2} + 3

x = a_{0} : m = 2 a_{0}

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} + 3) : y = 2 a_{0} x - a_{0}^{2} + 3

y = 2 a_{0} x - a_{0}^{2} + 3

\frac{\partial}{\partial y} (41 - x^{2} - 4 y^{2})

y^{^{''}} + \frac{y}{a} = 0

\int \frac{1}{x ^{3} + x ^{2} + x + 1} d x

\int_{0}^{1} π (x^{2} - x^{4}) d x

\frac{\partial}{\partial x} (y \cdot arcsin (\frac{x}{y}))