What is the tangent of x^2-5x+5(0.5) ?

The tangent of x^2-5x+5(0.5) is y=(2a_{0}-5)x-a_{0}^2+2.5

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tangent of x^2-5x+5(0.5)

tangent of

x^{2} - 5 x + 5 (0.5)

y = (2 a_{0} - 5) x - a_{0}^{2} + 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} + 2.5)

Find the slope of

y = x^{2} - 5 x + 5 \cdot 0.5 : \frac{d y}{d x} = 2 x - 5

Compute the slope of

y = x^{2} - 5 x + 5 \cdot 0.5

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} + 2.5) : y = (2 a_{0} - 5) x - a_{0}^{2} + 2.5

y = (2 a_{0} - 5) x - a_{0}^{2} + 2.5

h \to 0 lim (\frac{( 3 ln ( e + h ) - 3 ln ( e ) )}{h})

\int \frac{4}{x ^{3} - 5 x ^{2} + 7 x - 3} d x

\int ln (x) - 1 d x

\frac{3}{8} x^{2}, 1, 2

(- 9, - 6), (3, - 9)

What is the tangent of x^2-5x+5(0.5) ?
The tangent of x^2-5x+5(0.5) is y=(2a_{0}-5)x-a_{0}^2+2.5