What is the tangent of y=x^2-5 ?

The tangent of y=x^2-5 is y=2a_{0}x-a_{0}^2-5

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tangent of y=x^2-5

tangent of

y = x^{2} - 5

y = 2 a_{0} x - a_{0}^{2} - 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 5)

Find the slope of

y = x^{2} - 5 : \frac{d y}{d x} = 2 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2 a_{0}

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} - 5) : y = 2 a_{0} x - a_{0}^{2} - 5

y = 2 a_{0} x - a_{0}^{2} - 5

x \to - 1 lim (4 x^{2} - 2 x - 6)

d er i v a t i v e sin (2 π t)

\frac{d}{d y} (\frac{350}{x ^{2} + y ^{2}})

\frac{\partial}{\partial x} (6 - 2 x^{2} + 3 y^{2})

t an g e n t f (x) = x^{\frac{1}{2}}, (81, 9)