What is the tangent of y=x^2+7 ?

The tangent of y=x^2+7 is y=2a_{0}x-a_{0}^2+7

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tangent of y=x^2+7

tangent of

y = x^{2} + 7

y = 2 a_{0} x - a_{0}^{2} + 7

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 7)

Find the slope of

y = x^{2} + 7 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} + 7

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

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} + 7) : y = 2 a_{0} x - a_{0}^{2} + 7

y = 2 a_{0} x - a_{0}^{2} + 7

\frac{\partial}{\partial x} (\frac{1}{cos ( x )})

(ln (2 x + 1))^{^{'}}

(2 x y) d x - (1 + y) d y = 0

y^{^{''}} - 8 y^{^{'}} - 9 y = 21 e^{2 t}

x \to - 1 lim (4 - x)