What is the tangent of f(x)=x^2,\at (-3/8) ?

The tangent of f(x)=x^2,\at (-3/8) is y=2a_{0}x-a_{0}^2

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tangent of f(x)=x^2,\at (-3/8)

tangent of

f (x) = x^{2}, at (- \frac{3}{8})

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}

f (x) = \frac{x ^{2}}{x - 9}

\int_{0}^{2} x + 2 d x

x \to 0 lim (\frac{x}{( x + 0 )})

(4 x^{3} + 5 x^{2})^{5}

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

What is the tangent of f(x)=x^2,\at (-3/8) ?
The tangent of f(x)=x^2,\at (-3/8) is y=2a_{0}x-a_{0}^2