What is the tangent of f(x)=x^2-5x ?

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

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tangent of f(x)=x^2-5x

tangent of

f (x) = 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

f (x) = x^{2} - 5 x : \frac{df ( x )}{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}

x \to 0 lim (\frac{1}{x} + 3)

\frac{d}{d x} (e^{(x + y)})

\frac{d}{d x} (arcsin (a x))

\frac{( x + 9 )}{( x + 4 )}

3 sin (x), 0, 0.6 π, 4 cos (x)

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