What is the tangent of x^2-5(-4.11) ?

The tangent of x^2-5(-4.11) is y=2a_{0}x-a_{0}^2+20.55

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tangent of x^2-5(-4.11)

tangent of

x^{2} - 5 (- 4.11)

y = 2 a_{0} x - a_{0}^{2} + 20.55

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 20.55)

Find the slope of

y = x^{2} - 5 (- 4.11) : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} - 5 (- 4.11)

x = a_{0} : m = 2 a_{0}

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} + 20.55) : y = 2 a_{0} x - a_{0}^{2} + 20.55

y = 2 a_{0} x - a_{0}^{2} + 20.55

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

\frac{d}{d x} (\frac{x + 4}{9 x ^{2} e ^{x}})

\frac{d}{d x} (- \frac{250 x}{( 25 x ^{2} + 1 ) ^{2}})

\frac{d}{d x} ((x - 1) e^{x^{2} - 4})

\frac{\partial}{\partial x} (- y sec^{2} (x))

What is the tangent of x^2-5(-4.11) ?
The tangent of x^2-5(-4.11) is y=2a_{0}x-a_{0}^2+20.55