What is the tangent of x^2-14x+48 ?

The tangent of x^2-14x+48 is y=(2a_{0}-14)x-a_{0}^2+48

Popular >

tangent of x^2-14x+48

tangent of

x^{2} - 14 x + 48

y = (2 a_{0} - 14) x - a_{0}^{2} + 48

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 14 a_{0} + 48)

Find the slope of

y = x^{2} - 14 x + 48 : \frac{d y}{d x} = 2 x - 14

Compute the slope of

y = x^{2} - 14 x + 48

x = a_{0} : m = 2 a_{0} - 14

Find the line with slope m=

2 a_{0} - 14

and passing through

(a_{0}, a_{0}^{2} - 14 a_{0} + 48) : y = (2 a_{0} - 14) x - a_{0}^{2} + 48

y = (2 a_{0} - 14) x - a_{0}^{2} + 48

\frac{d}{d x} (\frac{e ^{5 x^{2} - 2 x}}{cot ( 7 ^{x} )})

\frac{d}{d x} (8 cos^{3} (x))

arctan (8 x^{2} - 1)

\frac{\partial}{\partial x} (\frac{1}{sin ( x ^{2} - y ^{2} )})

6 x

What is the tangent of x^2-14x+48 ?
The tangent of x^2-14x+48 is y=(2a_{0}-14)x-a_{0}^2+48