What is the tangent of y=x^2-5x+5,(4,1) ?

The tangent of y=x^2-5x+5,(4,1) is y=3x-11

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tangent of y=x^2-5x+5,(4,1)

tangent of

y = x^{2} - 5 x + 5, (4, 1)

y = 3 x - 11

Solution steps

Find the slope of

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

Compute the slope of

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

(4, 1) : m = 3

Find the line with slope m=

3

and passing through

(4, 1) : y = 3 x - 11

y = 3 x - 11

\frac{d}{d x} (\frac{3 x ^{2} e ^{x} - 4}{x ^{2}})

5 y^{^{'}} = y^{2}, y (0) = 1

\int_{0}^{1} \frac{x ^{2}}{x ^{2} + 1} d x

\frac{d}{d x} (1 - \frac{2}{x})

y^{^{''}} + x y^{^{'}} - y = 0

What is the tangent of y=x^2-5x+5,(4,1) ?
The tangent of y=x^2-5x+5,(4,1) is y=3x-11