What is the tangent of x^2-5x+5(6.11) ?

The tangent of x^2-5x+5(6.11) is y=(2a_{0}-5)x-a_{0}^2+30.55

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tangent of x^2-5x+5(6.11)

tangent of

x^{2} - 5 x + 5 (6.11)

y = (2 a_{0} - 5) x - a_{0}^{2} + 30.55

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} + 30.55)

Find the slope of

y = x^{2} - 5 x + 5 \cdot 6.11 : \frac{d y}{d x} = 2 x - 5

Compute the slope of

y = x^{2} - 5 x + 5 \cdot 6.11

x = a_{0} : 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} + 30.55) : y = (2 a_{0} - 5) x - a_{0}^{2} + 30.55

y = (2 a_{0} - 5) x - a_{0}^{2} + 30.55

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What is the tangent of x^2-5x+5(6.11) ?
The tangent of x^2-5x+5(6.11) is y=(2a_{0}-5)x-a_{0}^2+30.55