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

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

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tangent of y=x^2-5x+5

tangent of

y = x^{2} - 5 x + 5

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

Find the slope of

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

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

\int \frac{x ^{2}}{361 - x ^{2}} d x

\int \frac{4 x + 2}{x ^{2} + x + 5} d x

\frac{d ^{2} y}{d x ^{2}} - b y = 0

100 x - \frac{d y}{d x} = x y

d er i v a t i v e cot (x)

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