What is the tangent of-x^2+80x-600 ?

The tangent of-x^2+80x-600 is y=(-2a_{0}+80)x+a_{0}^2-600

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tangent of-x^2+80x-600

tangent of

- x^{2} + 80 x - 600

y = (- 2 a_{0} + 80) x + a_{0}^{2} - 600

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - a_{0}^{2} + 80 a_{0} - 600)

Find the slope of

y = - x^{2} + 80 x - 600 : \frac{d y}{d x} = - 2 x + 80

Compute the slope of

y = - x^{2} + 80 x - 600

x = a_{0} : m = - 2 a_{0} + 80

Find the line with slope m=

- 2 a_{0} + 80

and passing through

(a_{0}, - a_{0}^{2} + 80 a_{0} - 600) : y = (- 2 a_{0} + 80) x + a_{0}^{2} - 600

y = (- 2 a_{0} + 80) x + a_{0}^{2} - 600

sin^{2} (π x)

\int \frac{24 x - 2 x ^{3}}{x ^{4} - 16} d x

\frac{d}{d x} (\frac{8}{x} - \frac{4}{x})

\frac{\partial}{\partial y} (yz - ln (x + z))

x \to 2 lim (3 x^{2} + 1)

What is the tangent of-x^2+80x-600 ?
The tangent of-x^2+80x-600 is y=(-2a_{0}+80)x+a_{0}^2-600