What is the tangent of y=x^2+1 ?

The tangent of y=x^2+1 is y=2a_{0}x-a_{0}^2+1

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tangent of y=x^2+1

tangent of

y = x^{2} + 1

y = 2 a_{0} x - a_{0}^{2} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 1)

Find the slope of

y = x^{2} + 1 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} + 1

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

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} + 1) : y = 2 a_{0} x - a_{0}^{2} + 1

y = 2 a_{0} x - a_{0}^{2} + 1

f (x) = - 2 x + 2

ln (3) x^{2}

x = - 4

f (x) = \frac{12}{x}, at x = 2

(- 5.13) (6.4)