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

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

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tangent of y=x^2-5x

tangent of

y = x^{2} - 5 x

y = (2 a_{0} - 5) 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} - 5 a_{0})

Find the slope of

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

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} - 5

Find the line with slope m=

2 a_{0} - 5

and passing through

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

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

\frac{4}{s ( s + 2 ) ^{2}}

\frac{d}{d x} (lo g_{5} (tan (x)))

x \to 4 - lim (c x^{2} + 3 x)

\int 2 x^{3} - 20 x^{- 5} d x

n = 0 \sum \infty \frac{2 ^{n}}{8 ^{n}}

What is the tangent of y=x^2-5x ?
The tangent of y=x^2-5x is y=(2a_{0}-5)x-a_{0}^2