What is the tangent of y=x^2-7 ?

The tangent of y=x^2-7 is y=2a_{0}x-a_{0}^2-7

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tangent of y=x^2-7

tangent of

y = x^{2} - 7

y = 2 a_{0} x - a_{0}^{2} - 7

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 7)

Find the slope of

y = x^{2} - 7 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} - 7

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

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} - 7) : y = 2 a_{0} x - a_{0}^{2} - 7

y = 2 a_{0} x - a_{0}^{2} - 7

\frac{\partial}{\partial y} (x cos (y) - y cos (x))

sin (t + c)

\int cos^{4} (\frac{x}{5}) d x

\int \frac{x + 3}{x ^{2} 2 x + 3} d x

y^{^{'}} = \frac{x ^{4} y ^{4} - 2 x ^{4}}{y ^{3}}