What is the tangent of f(x)=x^2,\at m(x)=-1/2 ?

The tangent of f(x)=x^2,\at m(x)=-1/2 is y=2a_{0}x-a_{0}^2

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tangent of f(x)=x^2,\at m(x)=-1/2

tangent of

f (x) = x^{2}, at m (x) = - \frac{1}{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}

\frac{d}{d x} (r^{2} - 8 x^{2})

\int x (x + 4)^{- \frac{3}{2}} d x

n = 1 \sum \infty (- 1)^{n} n^{- 3}

\int \frac{1}{x ^{2} x ^{2} - 81} d x

\int_{- 1}^{3} (- x^{3} + 3 x^{2} + 1) d x

What is the tangent of f(x)=x^2,\at m(x)=-1/2 ?
The tangent of f(x)=x^2,\at m(x)=-1/2 is y=2a_{0}x-a_{0}^2