What is the tangent of y=x^2+9 ?

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

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tangent of y=x^2+9

tangent of

y = x^{2} + 9

y = 2 a_{0} x - a_{0}^{2} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 9)

Find the slope of

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

Compute the slope of

y = x^{2} + 9

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

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} + 9) : y = 2 a_{0} x - a_{0}^{2} + 9

y = 2 a_{0} x - a_{0}^{2} + 9

\frac{x - x}{x ^{\frac{1}{4}}}

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\int cos (x + \frac{π}{3}) d x

\int_{5}^{6} (\frac{7}{x ^{2}} - 1) d x

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