What is the tangent of f(x)=x^2 ?

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

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tangent of f(x)=x^2

tangent of

f (x) = 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

f (x) = x^{2} : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = 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}

(7, - 5), (9, - 1)

f (x) = 5

2 ln (x)

y = x

x ln (x)