What is the tangent of f(x)=x^2,\at x=-1/4 ?

The tangent of f(x)=x^2,\at x=-1/4 is y=-1/2 x-1/16

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tangent of f(x)=x^2,\at x=-1/4

tangent of

f (x) = x^{2}, at x = - \frac{1}{4}

y = - \frac{1}{2} x - \frac{1}{16}

Solution steps

Find the tangent point:

(- \frac{1}{4}, \frac{1}{16})

Find the slope of

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

Compute the slope of

f (x) = x^{2}

x = - \frac{1}{4} : m = - \frac{1}{2}

Find the line with slope m=

- \frac{1}{2}

and passing through

(- \frac{1}{4}, \frac{1}{16}) : y = - \frac{1}{2} x - \frac{1}{16}

y = - \frac{1}{2} x - \frac{1}{16}

y^{^{'}} - 2 x y - 8 = 0

e^{3 (t + 1)}

y = x (3 x - 2)^{4}

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

What is the tangent of f(x)=x^2,\at x=-1/4 ?
The tangent of f(x)=x^2,\at x=-1/4 is y=-1/2 x-1/16