What is the tangent of y=0.03x^2+1 ?

The tangent of y=0.03x^2+1 is y=0.06a_{0}x+(-3a_{0}^2+100)/(100)

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tangent of y=0.03x^2+1

tangent of

y = 0.03 x^{2} + 1

y = 0.06 a_{0} x + \frac{- 3 a _{0}^{2} + 100}{100}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 0.03 a_{0}^{2} + 1)

Find the slope of

y = 0.03 x^{2} + 1 : \frac{d y}{d x} = 0.06 x

Compute the slope of

y = 0.03 x^{2} + 1

x = a_{0} : m = 0.06 a_{0}

Find the line with slope m=

0.06 a_{0}

and passing through

(a_{0}, 0.03 a_{0}^{2} + 1) : y = 0.06 a_{0} x + \frac{- 3 a _{0}^{2} + 100}{100}

y = 0.06 a_{0} x + \frac{- 3 a _{0}^{2} + 100}{100}

\int_{2}^{3} \frac{1}{4 - 3 x} d x

\int - 3 cos (x) d x

\frac{d}{d x} (x^{7} (x - 12)^{5})

\frac{d y}{d x} = \frac{1}{x - x ^{3}}

\frac{\partial}{\partial x} (2 x - 3 y + 5 z)

What is the tangent of y=0.03x^2+1 ?
The tangent of y=0.03x^2+1 is y=0.06a_{0}x+(-3a_{0}^2+100)/(100)