What is the tangent of y=x^2 ?

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

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tangent of y=x^2

tangent of

y = x^{2}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = x^{2}

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

Find the line with slope m=

2 a_{0}

and passing through

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

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

y = 5

(53, 5)

f (x) = e^{3 x}

f (x) = x^{2} + 2

y = 1 + \frac{1}{x}